Prasanna – Page 240 – Eureka Math Answers

What is a Quadrilateral? | Definition, Types and Properties

February 1, 2021

The quadrilateral is a figure formed by adding the four-line segments. It consists of 4 sides, 4 vertices, and two diagonals. By going through this entire article you will get a complete idea of Quadrilateral like Types, Sides, Angles, Vertices, Diagonals of it, its Properties etc. If P, Q, R, S are four points and where no three points are collinear and also the line segments PQ, QR, RS, and SP do not intersect at their endpoints. Check the below figure that is the quadrilateral PQRS.

Quadrilateral

In a quadrilateral PQRS
(i) The vertices in a quadrilateral are P, Q, R, S.
(ii) The sides of a quadrilateral are PQ, QR, RS, and SP.
(iii) Also, the angles of quadrilateral are ∠SPQ, ∠PQR, ∠QRS and ∠RSP.
(iv) The line segments are PQ and QS.

Convex Quadrilaterals and Concave Quadrilaterals

If each angle of a quadrilateral is less than 180°, then it is called a convex quadrilateral. Also, if one angle of the quadrilateral is more than 180°, then it is called a concave quadrilateral. A quadrilateral is not a simple closed figure.

Sides, Angles, Vertices, Diagonals of the Quadrilateral

Have a look at the complete details of a Quadrilateral below.

Adjacent Sides of a Quadrilateral

Adjacent Sides of a Quadrilateral are nothing but the sides that have a common endpoint. If PQRS is a Quadrilateral, then (PQ, QR), (QR, RS), (RS, SP), and (SP, PQ) are four pairs of adjacent sides of quadrilateral PQRS.

Opposite Sides of a Quadrilateral

In a given quadrilateral, the two sides are said to be opposite sides when they do not have a common endpoint. If PQRS is a quadrilateral, then (PQ, SR) and (PS, QR) are two pairs of opposite sides of quadrilateral PQRS.

Adjacent Angles of a Quadrilateral

An angle is formed when two rays meeting at a common endpoint. Two angles of a quadrilateral are said to be adjacent when they have a common arm. From the given figure, (∠P, ∠Q), (∠Q, ∠R), (∠R, ∠S), and (∠S, ∠P) are four pairs of adjacent angles of quadrilateral PQRS.

Opposite Angles of a Quadrilateral

Opposite Angles of a Quadrilateral are not adjacent angles. If you consider a quadrilateral PQRS, then (∠P, ∠R) and (∠Q, ∠S) are two pairs of opposite angles of quadrilateral PQRS.

Adjacent Vertices of a Quadrilateral

In a Quadrilateral, if two vertices have a common side are known as adjacent vertices. From the figure, the pairs of adjacent vertices are (P, Q); (Q, R); (R, S), and (S, P).

Opposite Vertices of a Quadrilateral

Opposite Vertices of a Quadrilateral are vertices that do not have a common side. From the figure, the pairs of opposite vertices are (P, R) and (Q, S).

Diagonal of a Quadrilateral

When a line segment of the opposite vertices of a quadrilateral is joined, then the diagonal of the quadrilateral is formed. From the given figure, the two diagonals are PR and QS.

Practice Test on Parallelogram | Multiple Choice Questions on Parallelogram

January 30, 2021

Practice Test on Parallelogram will help you to test your knowledge. Answer on your own for every question given below. Before you take the practice test, make sure you have read the concept completely and solved all problems. It becomes easy if you have a perfect grip on the entire concept. Also, it will avoid you to confuse while choosing the answers. Complete concept and Objective Questions on Parallelogram are given on our website for free of cost.

Tick (✔) the correct answer in each of the following

1. From the given options, which parallelogram two diagonals are not necessarily equal ………………..

(a) square
(b) rectangle
(c) isosceles trapezium
(d) rhombus

Answer:

(d) rhombus

Explanation: rhombus two diagonals are not necessarily equal.

2. A rhombus has diagonals of 16 cm and 12 cm. Find the length of each side?

(a) 8cm
(b) 12cm
(c) 10cm
(d) 9cm

Answer:

Explanation:
Given that One diagonal is 16 and another 12 then half of both is 8 and 6. Diagonal of a rhombus bisect at 90º.
By pythogaurus theorem
h² = 8² + 6²
h² = 64 + 36=100
h = √100 = 10

Side = 10

3. Two adjacent angles of a parallelogram are (4b + 15)° and (2b – 10)°. The value of b is ………………. .

(a) 29.16
(b) 32
(c) 42
(d) 36

Answer:

(a) 29.16

Explanation: sum of the adjacent angles of parallelogram=180
4b + 15 + 2b – 10=180
6b + 5 = 180
6b =180 – 5
6b = 175
b = 175/6

b = 29.16

4. From the given options, which parallelogram two diagonals do not necessarily intersect at right angles ………………..

(a) rhombus
(b) rectangle
(c) kite
(d) parallelogram

Answer:

(d) parallelogram

Explanation: The diagonals do not necessarily intersect at right angles in a parallelogram.

5. The length and breadth of a rectangle are in the ratio 8 : 6. If the diagonal measures 50 cm. Find the perimeter of a rectangle?

(a) 800 cm
(b) 700 cm
(c) 600 cm
(d) 560 cm

Answer:

(b) 700 cm

Explanation: Let m be the common multiple.
Length = 8m
Breadth = 6m
According to Pythagoras theorem,
(8m)² + (6m)²=(50)²
64m²+ 36m² = 2500
100m² = 2500
m² = 2500/100
m = 25
So, Length = 8m = 200 cm
Breadth = 6m = 150 cm
Perimeter = 2 (l × b)
= 2 (200 + 150)
= 700 cm

So, perimeter of rectangle is 700 cm.

6. The bisectors of any two adjacent angles of a parallelogram intersect at ………………..

(a) 90°
(b) 30°
(c) 60°
(d) 45°

Answer:

(a) 90°

Explanation: The bisectors of any two adjacent angles of a parallelogram intersect at 90°.

7. If an angle of a parallelogram is 2/3 of its adjacent angle find the angle of a parallelogram.

(a) 72°
(b) 54°
(c) 108°
(d) 81°

Answer:

(a) 72°

Explanation: Let m and n be the adjacent angles of a parallelogram.
Now, as we know that adjacent angles of a parallelogram are supplementary
Therefore, the sum of angles a and b will be 180º.
m + n = 180º
One angle is 2/3rd of the other.
m = 2/3 . n
2/3 . n + n = 180º
5/3 . n = 180º
n = 108º
m = 2/3 . 108º = 72º

8. The diagonals do not necessarily bisect the interior angles at the vertices in a ………………..

(a) square
(b) rectangle
(c) rhombus
(d) all of these

Answer:

(b) rectangle

Explanation: The diagonals do not necessarily bisect the interior angles at the vertices in a rectangle.

9. In a square PQRS, PQ = (3a + 5) cm and RS = (2a – 2) cm. Find the value of a?

(a) 4
(b) 5
(c) 7
(d) 8

Answer:

Explanation: We know all the sides of a square are equal.
3a + 5 = (2a – 2)
3a – 2a = 5 + 2
a = 7

Hence, solved

10. If one angle if a parallelogram is 24º less than twice the smallest angle then the largest angle if parallelogram is?

(a) 68°
(b) 112°
(c) 102°
(d) 176°

Answer:

(b) 112°

Explanation: Let the smallest angle be a then, the largest angle will be =180−a
but, the same equals to [2a − 24]
so we have [2a − 24] = 180 − a
3a = 204
a = 68

Thus, the largest angle 180 − 68 = 112 degree

Problems on Parallelogram | Questions on Parallelogram with Solutions

January 30, 2021

Problems on Parallelogram are given in this article along with an explanation. It is easy to learn and understand the entire concept of a Parallelogram by solving every problem over here. There are various types of problems included according to the new updated syllabus. Get a good score in the exam and improve your preparation level immediately by working on your difficult topics.

1. Prove that any two adjacent angles of a parallelogram are supplementary?

Solution:
Let us take a parallelogram PQRS.

Then, PS ∥ QR and PQ is a transversal.
The sum of the interior angles on the same side of the transversal is 180°
Therefore, P + Q = 180°
Similarly, ∠R + ∠S = 180°, ∠Q + ∠R = 180°, and ∠S + ∠P = 180°.
Thus, the sum of any two adjacent angles of a parallelogram is 180°.

Hence, any two adjacent angles of a parallelogram are supplementary.

2. Two adjacent angles of a parallelogram PQRS are as 2 : 3. Find the measure of each of its angles?

Solution:
Let us take a parallelogram PQRS.

Then, ∠P and ∠Q are its adjacent angles.
Let ∠P = (2a)° and ∠Q = (3a)°.
The sum of adjacent angles of a parallelogram is 180°
Then, ∠P + ∠Q = 180°
⇒ 2a + 3a = 180
⇒ 5a = 180
⇒ a = 36.
Therefore, ∠P = (2 × 36)° = 72° and ∠Q = (3 × 36°) = 108°.
∠Q and ∠R are adjacent angles. By adding them, we get 180°
Also, ∠Q + ∠R = 180°
= 108° + ∠R = 180° [Since, ∠Q = 108°]
∠R = (180° – 108°) = 72°.
∠R and ∠S are adjacent angles and add up to 180°.
Also, ∠R + ∠S = 180°
⇒ 72° + ∠S = 180°
⇒ ∠S = (180° – 72°) 108°.

Therefore, ∠P = 72°, ∠Q = 108°, ∠R = 72°and ∠S = 108°.

3. In the adjoining figure, PQRS is a parallelogram in which ∠P = 75°. Find the measure of each of the angles ∠Q, ∠R, and ∠S.

Solution:
It is given that PQRS is a parallelogram in which ∠P = 75°.
parallelogram 1
Since the sum of any two adjacent angles of a parallelogram is 180°,
∠P + ∠Q = 180°
⇒ 75° + ∠Q = 180°
⇒∠Q = (180° – 75°) = 105°
∠Q and ∠R are adjacent angles and add up to 180º.
Also, ∠Q + ∠R = 180°
⇒ 105° + ∠R = 180°
⇒ ∠R = (180° – 105°) = 75°.
∠R and ∠S are adjacent angles
Further, ∠R + ∠S = 180°
⇒ 75° + ∠S = 180°
⇒ ∠S = (180° – 75°) = 105°.

Therefore, ∠Q = 105°, ∠R = 75° and ∠S = 105°.

4. In the adjoining figure, PQRS is a parallelogram in which ∠QPS = 75° and ∠SQR = 60°. Calculate:
(i) ∠RSQ and (ii) ∠PSQ.

Solution:
Let us draw a parallelogram PQRS.
parallelogram 2
We know that the opposite angles of a parallelogram are equal.
Therefore, ∠QRS = ∠QPS = 75°.
(i) Now, in ∆ QRS, we have
The sum of the angles of a triangle is 180°
∠RSQ + ∠SQR + ∠QRS = 180°
⇒ ∠RSQ + 60° + 75° = 180°
⇒ ∠RSQ + 135° = 180°
⇒ ∠RSQ = (180° – 135°) = 45°.
(ii) PS ∥ QR and QS are the transversals.
Therefore, ∠PSQ = ∠SQR = 60° [alternate interior angles]

Hence, ∠PSQ = 60°.

5. In the adjoining figure, PQRS is a parallelogram in which ∠RPS = 40°, ∠QPR = 35°, and ∠ROS = 65°.
Calculate: (i) ∠PQS (ii) ∠QSR (iii) ∠PRQ (iv) ∠RQS.
parallelogram 3

Solution:
(i) ∠POQ = ∠ROS = 65° (vertically opposite angles)
Now, from ∆OPQ, we can write as:
The sum of the angles of a triangle is 180°
∠OPQ + ∠PQO + ∠POQ =180°
⇒ 35°+ ∠PQO + 65° = 180°
⇒ ∠PQO + 100° = 180°
⇒ ∠PQO = (180° – 100°) = 80°
⇒ ∠PQS = ∠PQO = 80°.
(ii) PQ ∥ SR and QS is a transversal.
Therefore, ∠QSR = ∠PQS = 80° [alternate interior angles]
Hence, ∠QSR = 80°.
(iii) PS ∥ QR and PR is a transversal.
Therefore, ∠PRQ = ∠RPS = 40° [alternate interior angles]
Hence, ∠PRQ = 40°.
(iv) ∠QRS = ∠QPS = (35° + 40°) = 75° [opposite angles of a parallelogram]
Now, in ∆RQS, we have
The sum of the angles of a triangle is 180°.
∠QSR + ∠QRS + ∠RQS = 180°
⇒ 80° + 75° + ∠RQS = 180°
⇒ 155° + ∠RQS = 180°
⇒ ∠RQS = (180° – 155°) = 25°.
Hence, ∠RQS = 25°.

6. In the adjoining figure, PQRS is a parallelogram, PO and QO are the bisectors of ∠P and ∠Q respectively. Prove that ∠POQ = 90°.
parallelogram 4

Solution:
We know that the sum of two adjacent angles of a parallelogram is 180°
Therefore, ∠P + ∠Q = 180° ……………. (i)
Since PO and QO are the bisectors of ∠P and ∠Q, respectively, we have
∠OPQ = 1/2∠P and ∠PQO = 1/2∠Q.
From ∆OPQ, we have
The sum of the angles of a triangle is 180°
∠OPQ + ∠POQ + ∠PQO = 180°
⇒ ¹/₂∠P + ∠PQO + ¹/₂∠Q = 180°
⇒ ¹/₂(∠P + ∠Q) + ∠POQ = 180°
⇒ (¹/₂ × 180°) + ∠POQ = 180° [from equation (i)]
⇒ 90° + ∠POQ = 180°
⇒ ∠POQ = (180° – 90°) = 90°.

Hence, ∠POQ = 90°.

7. The ratio of two sides of a parallelogram is 5: 4. If its perimeter is 54 cm, find the lengths of its sides?

Solution:
Let the lengths of two sides of the parallelogram be 5a cm and 4a cm respectively.
Find the perimeter using given values.
Then, its perimeter = 2(5a + 4a) cm = 2 (9a) cm = 18a cm.
Therefore, 18a = 54 ⇔ a = 54/18 = 3.

Therefore, one side = (5 × 3) cm = 15 cm and other side = (4 × 3) cm = 12 cm.

8. The length of a rectangle is 16 cm and each of its diagonals measures 20 cm. Find its breadth?

Solution:
Let PQRS be the given rectangle in which length PQ = 16 cm and diagonal PR = 20 cm.

parallelogram 5

Since each angle of a rectangle is a right angle, we have
∠PQR = 90°.
From the right ∆PQR, we have
PQ² + QR² = PR² [From Pythagoras’ Theorem]
⇒ QR² = (PR² – PQ²) = {(20)² – (16)²} = (400 – 256) = 144

⇒ QR = √144 = 12 cm.

Hence, breadth = 12 cm.

9. In the below figure, PQRS is a rhombus whose diagonals PR and QS intersect at a point O. If side PQ = 20 cm and diagonal QS = 32 cm, find the length of diagonal PR.
parallelogram 6

Solution:
We know that the diagonals of a rhombus bisect each other at right angles.
Therefore, QO = ¹/₂QS = (¹/₂ × 32) cm = 16 cm, PQ = 20 cm and ∠POQ = 90°.
From right ∆OPQ, we have PQ² = PO² + QO²
⇒ PO² = (PQ² – QO²) = {(20) ² – (16)²} cm²
= (400 – 256) cm²
= 144 cm²
⇒ PO = √144 cm = 12 cm.

Therefore, PR = 2 × PO = (2 × 12) cm = 24 cm.

Properties of a Rectangle Rhombus and Square | Special Parallelograms Properties

January 30, 2021

Properties of a Rectangle Rhombus and Square is always a confusing concept for students. Learning every individual topic is important to score good marks in the exam. So, we have explained every individual topic clearly in a detailed manner in this article. Therefore, those who wish to learn the concepts of Parallelogram and its properties, problems, can completely learn the Parallelogram concepts on our website.

Rectangle

A rectangle is said to be a parallelogram when it has all 4 angles having equal measure.

Properties of Rectangle

The Opposite sides of a rectangle are parallel.
Also, the Opposite sides of a rectangle are equal in length.
Diagonals are equal in length.
The interior angles are 90 degrees each.
Diagonals bisect each other.
It has horizontal and vertical lines of symmetry.
Each of the diagonal bisects the rectangle into 2 congruent triangles.
If you combine the 4 sides of a rectangle, then the mid-points of it form a rhombus.

Rectangle Formulas

If l is the length of the rectangle and b is the breadth of the rectangle, then
Area = lb square units
Perimeter = 2 (l+b) units.

Diagonal Properties of a Rectangle

Prove that the diagonals of a rectangle are equal and bisect each other.

Proof:
Let PQRS be a rectangle that has diagonals PQ and QS intersect at the point O.

rectangle

From ∆ PQR and ∆ QPS,
PQ = QP (common)
∠PQR = ∠QPS (each equal to 90º)
QR = PS (opposite sides of a rectangle).
Therefore, ∆ PQR ≅ ∆ QPS (by SAS congruence)
⇒ PR = QS.
Hence, the diagonals of a rectangle are equal.
From ∆ OPQ and ∆ ORS,
∠OPQ = ∠ORS (alternate angles)
∠OQP = ∠OSR (alternate angles)
PQ = RS (opposite sides of a rectangle)
Therefore, ∆OPQ ≅ ∆ ORS. (by ASA congruence)
⇒ OP = OR and OQ = OS.
This shows that the diagonals of a rectangle bisect each other.

Hence, the diagonals of a rectangle are equal and bisect each other.

Rhombus

The rhombus is a quadrilateral that consists of four sides with equal lengths.

Properties of Rhombus

The Rhombus consists of parallel and equal opposite sides. As it consists of parallel and equal opposite sides, it is said to be a parallelogram.
All available sides (4 sides) are equal.
Also, opposite angles in a rhombus are equal.
Diagonals bisect each other.
Diagonals of a rhombus intersect each other at right angles.
Furthermore, Diagonals bisect opposite vertex angles.
Every diagonal divides the rhombus into 2 congruent triangles.

Rhombus Formula

If b is the side, a and b are the two diagonals of the rhombus, then
Area = ab/2 Square units.
Perimeter = 4b units

Diagonal Properties of a Rhombus

Prove that the diagonals of a rhombus bisect each other at right angles.

Proof:
Let PQRS be a rhombus whose diagonals AC and BD intersect at point O.
Rhombus

The diagonals of a parallelogram bisect each other. Also, we know that every rhombus is a parallelogram.
So, the diagonals of a rhombus bisect each other.
Therefore, OP = OR and OQ = OS
From ∆ ROQ and ∆ ROS,
RQ = RS (sides of a rhombus)
RO = RO (common).
OQ = OS (proved)
Therefore, ∆ ROQ ≅ ∆ ROS (by SSS congruence)
⇒ ∠ROQ = ∠ROS
But, ∠ROQ + ∠ROS = 2 right angles (linear pair)
Therefore, ∠ROQ = ∠ROS = 1 right angle.

Hence, the diagonals of a rhombus bisect each other at right angles.

Square

A square is a rectangle that has all equal sides.

Properties of Square

The opposite sides of a square are parallel.
All 4 sides are equal in length.
Diagonals are equal in length.
Diagonals bisect opposite vertex angles.
The interior angles of a square measure 90 degrees each.
Diagonals bisect each other at right angles.
It has 4 lines of symmetry – a horizontal, a vertical, and 2 diagonals.
Each diagonal bisects the square into 2 congruent triangles.

Square Formula

If b is the side of the square, then
Area = b² square units
Perimeter = 4b units.

Diagonal Properties of a Square

Prove that the diagonals of a square are equal and bisect each other at right angles.

Proof:
We know that the diagonals of a rectangle are equal.
Also, every square is a rectangle.
Therefore, the diagonals of a square are equal.
Again, the diagonals of a rhombus bisect each other at right angles. But, every square is a rhombus.
So, the diagonals of a square bisect each other at right angles.

Hence, the diagonals of a square are equal and also bisect each other at right angles.

Note 1: If the diagonals of a quadrilateral are equal but it is not necessary to be a rectangle.
Note 2: If the diagonals of a quadrilateral interest at a point with right angles then also it is not necessary to become a rhombus.

Examples on Multiplication of Integers | Questions on Multiplication of Integers

January 29, 2021

Do you find it difficult to understand the Integers Multiplication? Here is the best solution for you all. We are providing Examples on Multiplication of Integers here. Integers Multiplication is an important concept which helps you to score more marks in the exam. Questions on Multiplication of Integers with Answers are available so that you can practice them regularly. Learn How to Multiply Integers by referring to the further modules.

Worked out Multiplication of Integers Problems

Before going to solve the Integers Multiplication problems, know all the definitions, rules, formulae etc. In the upcoming sections, you will find all the details and also problem-solving techniques and tips. To be more precise, you can only solve the problems if you know all the details regarding integers. Know various properties of integers beforehand and how they work while multiplying the integers.

Key Points to Remember on Properties of Integers Multiplication

The Closure Integer Property of multiplication defines that the product value of two or three integer numbers will be an integer number.
The commutative Integer Multiplication property defines that swapping two or three integers will not differ the value of the final result.
The associative Integer Multiplication property defines that the grouping of integer values together will not affect the final result.
The distributive Integer property of multiplication defines that the distribution concept of 1 operation value on other mathematical integer values within the given braces.
Multiplication by zero defines the product value of any negative or positive integer number by zero
Multiplicative Integer defines the final result as 1 when any integer number is multiplied with 1.

Integer Multiplication Rules on Problems

Question 1:

The temperature in an area drops by 4 degrees for 4 hours. How much is the total drop in the temperature?

Solution:

As given in the question, the temperature drops by 4 degrees. Therefore, the temperature is a negative factor.

Also. given that it decreases for 4 hours.

The total drop in temperature is (-4) * (4) = -16

Therefore, the total drop in temperature is 16 degrees C

Thus, the final result is -16 degree C

Question 2:

Jason borrowed $2 a day to buy a launch. She now owes $60. How many days did Jason borrow $2?

Solution:

As given in the question, Jason borrowed to buy a launch = $2

After buying she owes $60

No of days Jason borrowed money = 60/2 = 30

Therefore, the total days = 30 days

Thus, the final result is 30 days.

Question 3: A football team 12 yards on each of the four consecutive plays. What was the team’s total change in position for four plays?

Solution:

As given in the question, A football team lost yards = -12 yards

No of plays = 4

Team total change in position for 4 plays = (-12) * (4) = 48

Therefore, the total change = -48 yards

Thus, the final answer is -48 yards.

Question 4: On a certain day, the temperature changed at a rate of -2 degrees F per hour. If this happened for continuous 5 days. For how many days there was a change in temperature?

Solution:

As per the given question, The temperature changed at the rate = -2 degree F

The change happened for days = 5 days

No of days there was a change = (-2)*5 = -10

Therefore, there was a change for days = 10 days

Thus, the final solution is 10 days.

Question 5: Flora made 6 deposits $ 7 each from her bank account. What was the overall change in her account?

Solution:

As per the given question,

Flora made no of deposits = 6

Amount of deposited money = $7

The overall change in the account = 6 * ($7) = 42

Therefore, the change in money = 42

Thus, the final solution is $42

Questions on Multiplication of Integers

Question 6: A winter coat was priced at $200. Each month for three months, the price was reduced by $15. How much was the coat reduced in price?

Solution:

As per the given question,

The price of the winter coat is reduced by $15, Therefore it is negative = -$15

No of times it is reduced = 3

The absolute values of |3| and |-15| are 3 and 15

The coat reduced in price = 3*15 = 45

Therefore, the total change in price = $45

Thus the final answer = -$45

Question 7:

Netflix charges $9 per month for their streaming plan to watch movies. If they automatically bill a customer for 6 months, How much will be deducted from the customer’s bank account?

Solution:

As per the given question,

Netflix charges $9 per month. Therefore, it is negative.

Given, the bill will be deducted automatically for months = 6

The absolute values of |6| and |9| are 6 and 9.

The amount of money deducted from customers bank account = 6*9 = 54

Therefore, the total amount after determining the signs = -$54

Hence, the final solution is -$54

Question 8: Lisa decided her hair was too long. In June and again in July. she cut 3 inches off. Then, in August, September, and October she cut off 2 inches. Write an equation to represent the change in the length of her hair?

Solution:

As given in the question,

In 2 months, she cut 3 inches off her hair. Cutting her hair made the length shorter, therefore it is negative.

In 3 months, she cut 2 inches off her hair. This is also negative.

For the month of June and July, the length of the hair she cut = 2 * (-3) = -6

For the months August, September, and October, the length of the hair she cut = 3 * (-2) = -6

Therefore, the total length = (-6) + (-6) = -12 inches

Thus. the complete length she cut = 12 inches

Hence, the final solution is -12 inches

Question 9: The depth of the water in a pool decreases an average of two inches each week during the summer. What will be the change in the depth of water of four weeks?

Solution:

As given in the question,

The depth of the water in a pool decreases each week = 2 inches

As the water level decreases, it will be negative.

The decrease in water for weeks = 4

The change in depth of water = (-2)*4 = -8

Therefore, the water level decreases by 8 inches.

Thus, the final solution is -8 inches

Question 10:

For every 1000 feet, you gain in elevation, the temperature drops by 3 degrees. If you increase your elevation by 5000 feet, How would the temperature change?

Solution:

As per the given question,

The temperature drops by 3 degrees. Therefore, it will be negative.

Also given for every 1000 feet it is 3 degrees. Thus for every 5000 feet, it is 5 degrees.

The temperature change = (-3)*5 = -15

Thus, for every 5000 feet, the temperature changes by -15 degrees.

Hence, the final solution is -15 degrees.

Examples on Division of Integers | Dividing Integers Problems with Solutions

January 28, 2021

Get the complete practice test questions and worksheet here. Follow the step by step procedure to solve examples on the division of integers all the problems. Know the shortcuts, tricks, and steps involved in solving Division of Integers problems. Also, find the definitions, formulae before going to start the practice sessions. Go through the below sections to know the detailed information regarding formulas, definitions, and problems.

Integers Division Rules

Rule 1: The quotient value of a positive integer number and a negative integer number is negative.

Rule 2: The quotient value of two positive integer numbers is a positive number.

Rule 3: The quotient value of two negative integer numbers is a positive number.

Division of Integers Rules and Examples

Question 1: Find the value of ||-17|+17| / ||-25|-42|

Solution:

||-17|+17| / ||-25|-42|

= |17+17| / |25–42|

= |34| / |-17|

= 34 / 17

= 2

Question 2: Simplify: {36 / (-9)} / {(-24) / 6}

Solution:

{36 / (-9)} / {(-24) / 6}

= {36/-9} / {-24/6}

= – (36/9) / – (24/6)

= -4/-4

= 4/4

Question 3: Find the value of [32 + 2 x 17 + (-6)] ÷ 15

Solution:

[32+2 x 17+(-6)] / 15

= [32+34+(-6)] / 15

= (66-6) / 15

= 60 / 15

= 4

Question 4: Divide the absolute values of the two given integers?

Solution:

The quotient of the absolute value of integer +24 and the absolute value of integer -8

From the rules given above, dividing the integers with different signs gives the final result as negative.

When positive(+) 24 is divided by negative(-) 8 results negative(-) 3, which can be defined as +24/(-8) = -3

Question 5: Prove that [((-8) / (-4)] ≠ =-8 / [(-4) / (-2)]

Solution:

From the given question

[((-8)/(-4)]/(-2)] ≠ -8/[(-4)/(-2)]

First of all, we will consider the LHS part i.e., [((-8) / (-4)] / (-2)]

To solve the equation, first, we divide 8 by 4, we get the result as 2

As both numbers have a negative sign, it will be positive.

Then we divide the result (2) by -2, then the final result will be -1.

Therefore, the result of the LHS part is -1.

Now, we consider the RHS part i.e., -8 / [(-4) / (-2)]

First of all, we divide -4 by -2, the result will be 2.

As both the numbers have a negative sign, the result will be positive.

Now, divide -8 by the above result 2.

Hence, the result will be -4.

Therefore, the result of RHS is -4.

Hence, LHS ≠ RHS

The above given equation [((-8) / (-4)] / (-2)] ≠ -8 / [(-4) / (-2)] is thus satisfied.

Question 6: In a maths test containing 10 questions, 2 marks are given for every correct answer and (-1) marks are given for every wrong answer. Rohith attempts all the questions and 8 questions answers are correct. What is Rohith’s total score?

Solution:

From the given question,

The marks given for every correct answer = 2 marks

Marks given for 8 correct answers = 2 * 8= 16 marks

Marks given for 1 incorrect answer = -1 marks

Marks given for 2 incorrect answers = -1 * 2 = -2 marks

Rohit’s total score = 16-2 = 14 marks

Therefore, the answer is 14 marks.

Question 7: Priya sells 20 pens and some pencils losing 2 Rs in all. If Priya gains 2 Rs on each pen and loses 1 Rs on each pencil. How many pencils does Priya sell?

Solution:

Suppose that Priya sells x pencils.

Total gain on pens = 2*20 = 40 Rs

Total loss on pencils = 1x = x

Total loss on selling pens and pencils = (-2)

40-x = (-2)

40+2 = x

x = 42

Therefore, Priya sells 42 pencils.

Thus, the answer is 42 pencils.

Question 8: To make ice cream, the room temperature must be decreased from 45degree C at the rate of 5 degrees C per hour. What will be the room temperature 12 hours after the freezing point of the icecream?

Solution:

As given in the question,

Temperature after 12 hours = 12*5 = 60degree C

Room temperature = 45degree C

Hence, the room temperature after freezing process of 12 hours = (45-60)degree C

= -15degree C

Question 9:

A car runs at a rate of 50km/hr. If the car starts at 5 km above the starting point, how long will it take to reach 2505km?

Solution:

As per the question,

Total distance covered by car = (2505-5) km = 2500 km

Rate of car = 50km/hr

From the above values, Distance = 2500 km, speed = 50 km/hr

Therefore, the time required by the car = distance/speed

=2500/50 = 50 hours

Therefore, the car will take 50 hours to travel 2505 km.

Hence, The final solution is 50 hours.

Question 10: Jason borrowed $5 a day to buy launch. She now owes $65 dollars. How many days did Jane borrow $5?

Solution:

As per the question,

Jason borrowed a launch at = $5

Now she owes = $65

No of days = (-65) / (-5)

= 13 days

Therefore, Jane borrows $5 for 13 days.

Hence, The final solution is 13 days.

Division of Integers Word Problems

Question 11: Allen’s score in a video game was changed by -120 points because he missed some target points. He got -15 points for each of the missed targets. How many targets did he miss?

Solution:

As per the given question,

Allen scored points in a video game = -120

He got points for missed targets = -15

No of targets he missed = -120/-15 = 8 targets.

Therefore, he missed 8 targets.

Question 12: Karthik made five of his truck payments late and was fined five late fees. The total change in his savings of late fees was -$30. What integer represents the one late fee?

Solution:

As given in the question, Karthik has made five of his truck payments. Therefore, it is positive.

He was fined -$30 as the late fees.

To find one late fee, we have to divide the fine by no of payments he did.

Therefore, One Late fee = -$30/5

=-$6

Thus, He paid $6 for each payment as late fees.

Hence, the final answer is -$6.

Big Ideas Math Answers Grade 6 Chapter 10 Data Displays

January 23, 2021

Big Ideas Math Answers Grade 6 Chapter 10 Data Displays: Free step by step solutions to Big Ideas Math Answers Grade 6 Chapter 10 Data Displays are available here. You can learn the concepts of Stem and leaf plot, histogram, shapes of distribution, Box and Whisker plots in an easy manner. Hence Download Answer Key of Big Ideas Math 6th Grade Chapter 10 Data Displays for free of cost. Start practicing the Big Ideas Math Answers Grade 6 Data Displays problems and score good marks in the exams.

Big Ideas Math Book 6th Grade Answer Key Chapter 10 Data Displays

Solve the problems on Data Displays listed below and become a master in maths. I know it is difficult for parents to explain the homework problems. So, in order to help them, we are providing the solutions for BIM Math 6th Grade Answer Key Chapter 10 Data Displays. Make use of the Big Ideas Math Grade 6 Solution Key and make your child completer their homework in time.

Performance Task

Data Displays STEAM Video/Performance Task
Data Displays Getting Ready for Chapter 10

Lesson 1 – Stem-and-Leaf Plots

Lesson 10.1 Stem-and-Leaf Plots
Stem-and-Leaf Plots Homework & Practice 10.1

Lesson 2 – Histograms

Lesson 10.2 Histograms
Histograms Homework & Practice 10.2

Lesson 3 – Shapes of Distributions

Lesson 10.3 Shapes of Distributions
Shapes of Distributions Homework & Practice 10.1

Lesson 4 – Choosing Appropriate Measures

Lesson 10.4 Choosing Appropriate Measures
Choosing Appropriate Measures Homework & Practice 10.4

Lesson 5 – Box-and-Whisker Plots

Lesson 10.5 Box-and-Whisker Plots
Box-and-Whisker Plots Homework & Practice 10.5

Data Displays

Data Displays Connecting Concepts
Data Displays Chapter Review
Data Displays Practice Test
Data Displays Cumulative Practice

Data Displays STEAM Video/Performance Task

STEAM Video
Choosing a Dog
Different animals grow at different rates. Given a group of puppies, describe an experiment that you can perform to compare their growth rates. Describe a real-life situation where knowing an animal’s growth rate can be useful.

Watch the STEAM Video “Choosing a Dog.” Then answer the following questions.
1. Using Alex and Tony’s stem-and-leaf plots below, describe the weights of most dogs at 3 months of age and 6 months of age.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 1$

Answer:
Weight of dogs at 3 months:
29, 34, 40, 40, 41, 42, 44, 46, 47, 48, 48, 53
Weight of dogs at 6 months:
57, 58, 61, 61, 63, 64, 65, 65, 65, 66, 67, 73
2. Make predictions about how the stem-and-leaf plot will look after 9 months and after 1 year.
Weight of dogs at 9 months
77, 78, 81, 81, 83, 84, 85, 85, 85, 86, 87, 91
Weight of dogs at 1 year
87, 88, 89, 93, 94, 95, 95, 95, 95, 96, 97, 99

Performance Task
Classifying Dog Breeds by Size
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 2$
After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be given names, breeds, and weights of full-grown dogs at a shelter.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 3$
You will use a data display to make conclusions about the sizes of dogs at the shelter. Why might someone be interested in knowing the sizes of dogs at a shelter?

Answer:
Because they need time to adjust.
You can buy the dog shelter based on the height and weight of the dogs.

Data Displays Getting Ready for Chapter 10

Chapter Exploration
Work with a partner. A famous data set was collected in Scotland in the mid-1800s. It contains the chest sizes(in inches) of 5738 men in the Scottish Militia.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 4$
1. Describe the shape of the bar graph shown above.

Answer: The shape of the above graph is Histogram.

2. Which of the following data sets have a bar graph that is similar in shape to the bar graph shown above? Assume the sample is selected randomly from the population. Explain your reasoning.
a. the heights of 500 women
b. the ages of 500 dogs
c. the last digit of 500 phone numbers
d. the weights of 500 newborn babies

Answer: The last digit of 500 phone numbers is similar in shape to the bar graph shown above.

3. Describe two other real-life data sets, one that is similar in shape to the bar graph shown above and one that is not.

Answer: The height of 500 students in the school and age of students in the classroom.

Vocabulary
The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.
stem-and-leaf plot
box-and-whisker plot
frequency table
five-number summary

Answer:
i. stem-and-leaf plot: A stem-and-leaf display or stem-and-leaf plot is a device for presenting quantitative data in a graphical format, similar to a histogram, to assist in visualizing the shape of a distribution.
ii. A box and whisker plot—also called a box plot—displays the five-number summary of a set of data. The five-number summary is the minimum, first quartile, median, third quartile, and maximum. In a box plot, we draw a box from the first quartile to the third quartile. A vertical line goes through the box at the median.
iii. In statistics, a frequency distribution is a list, table, or graph that displays the frequency of various outcomes in a sample. Each entry in the table contains the frequency or count of the occurrences of values within a particular group or interval.
iv. The five-number summary is a set of descriptive statistics that provides information about a dataset.

Lesson 10.1 Stem-and-Leaf Plots

EXPLORATION 1

Making a Data Display
Workwith a partner. The list below gives the ages of women when they became first ladies of the United States.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 1$
a. The incomplete data display shows the ages of the first ladies in the left column of the list above. What do the numbers on the left represent? What do the numbers on the right represent?
b. This data display is called a stem-and-leaf plot. What numbers do you think represent the stems? leaves? Explain your reasoning.
c. Complete the stem-and-leaf plot using the remaining ages.

Answer:
$Big-Ideas-Math-Answer-Key-Grade-6-Chapter-10-Data-Displays-10.1-1$
The tens place represents the stem and the ones place represents the leaf.
d. REASONING
Write a question about the ages of first ladies that is easier to answer using a stem-and-leaf plot than a dot plot.
Answer: Make the stem and leaf plot to find the ages of the first ladies.
By using the above data you can make the stem and leaf plot easily.

$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 2$

Key Idea
Stem-and-Leaf Plots
A stem-and-leaf plot uses the digits of data values to organize a data set. Each data value is broken into a stem(digit or digits on the left) and a leaf(digit or digits on the right).
A stem-and-leaf plot shows how data are distributed.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 4$

EXAMPLE 1

Making a Stem-and-Leaf Plot
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 5$
Make a stem-and-leaf plot of the lengths of the 12 phone calls.
Step 1: Order the data.
2, 3, 5, 6, 10, 14, 18, 23, 23, 30, 36, 55
Step 2: Choose the stems and the leaves. Because the data values range from 2 to 55, use the tens digits for the stems and the ones digits for the leaves. Be sure to include the key.
Step 3: Write the stems to the left of the vertical line.
Step 4: Write the leaves for each stem to the right of the vertical line.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 6$

Try It
Question 1.
Make a stem-and-leaf plot of the hair lengths.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 7$
Answer:
Step 1: Order the data.
1, 1, 1, 2, 2, 4, 5, 5, 7, 12, 20, 23, 27, 30, 32, 33, 38, 40, 44, 47
Step 2: Choose the stems and the leaves. Because the data values range from 1 to 47, use the tens digits for the stems and the ones digits for the leaves. Be sure to include the key.
Step 3: Write the stems to the left of the vertical line.
Step 4: Write the leaves for each stem to the right of the vertical line.
$Big Ideas Math Answers Grade 6 Chapter 12 Data Displays img_5$

EXAMPLE 2
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 8$
The stem-and-leaf plot shows student quiz scores. (a) How many students scored less than 8 points? (b) How many students scored at least 9 points? (c) How are the data distributed?
a. There are ﬁve scores less than 8 points:
6.6, 7.0, 7.5, 7.7, and 7.8.
Five students scored less than 8 points.10
b. There are four scores of at least 9 points:
9.0, 9.2, 9.9, and 10.0.
Four students scored at least 9 points.
c. There are few low quiz scores and few high quiz scores. So, most of the scores are in the middle, from 8.1 to 8.9 points.

Try It
Question 2.
Use the grading scale at the right.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 9$
a. How many students received a B on the quiz?
Answer: There are 9 students who received a B on the quiz.
b. How many students received a C on the quiz?
Answer: There are 4 students who received a C on the quiz.

Self – Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.
Question 3.
MAKING A STEM-AND-LEAF PLOT
Make a stem-and-leaf plot of the data values 14, 22, 9, 13, 30, 8, 25, and 29.
Answer:
The ones represent the leaf and the tens place represent the stem.
$Big Ideas Math Grade 6 Chapter 10 Data Displays img_6$

Question 4.
WRITING
How does a stem-and-leaf plot show the distribution of a data set?
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 10$
Answer:
02, 03, 1, 21, 26, 30, 34, 36, 44, 45, 48, 48, 49

Explanation:
A stem-and-leaf display or stem-and-leaf plot is a device for presenting quantitative data in a graphical format, similar to a histogram, to assist in visualizing the shape of a distribution.

Question 5.
REASONING
Consider the stem-and-leaf plot shown.
a. How many data values are at most 10?
Answer: By seeing the above stem and leaf plot we can find the data values of at most 10.
The data values less than or equal to 10 are 3.
b. How many data values are at least 30?
Answer: By seeing the above stem and leaf plot we can find the data values of at least 30.
The data values of less than 30 are 5.
c. How are the data distributed?
Answer: The data is distributed according to the stem and leaf plot. The tens place is given to the stem and the ones place is given to the leaf.

Question 6.
CRITICAL THINKING
How can you display data whose values range from 82 through 129 in a stem-and-leaf plot?
Answer:
Given data range from 82 to 129
Considering 9 random values between 82 and 129.
From the data 86, 91, 93, 100, 107, 109, 113, 122, 124, stem and leaf are calculated for each number.
86 is split into 8 (stem) and 6 (leaf)
91 is split into 9 (stem) and 1 (leaf)
93 is split into 9 (stem) and 3 (leaf)
100 is split into 10 (stem) and 0 (leaf)
107 is split into 10 (stem) and 7 (leaf)
109 is split into 10 (stem) and 9 (leaf)
113 is split into 11 (stem) and 3 (leaf)
122 is split into 12 (stem) and 2 (leaf)
124 is split into 12 (stem) and 4 (leaf)

$Big Ideas Math Grade 6 Chapter 10 Data Displays img_7$

EXAMPLE 3
Modeling Real Life
The stem-and-leaf plot shows the heights of several houseplants. Use the data to answer the question, “What is a typical height of a houseplant?
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 11$
Find the mean, median, and mode of the data. Use the measure that best represents the data to answer the statistical question.
Mean: $\frac{162}{15}$ = 10.8
Median: 11
Mode: 11
The mean is slightly less than the median and mode, but all three measures can be used to represent the data.
So, the typical height of a houseplant is about 11 inches.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 12$
Question 7.
DIG DEEPER!
Work with a partner. Use two number cubes to conduct the following experiment. Then use a stem-and-leaf plot to organize your results and describe the distribution of the data.
• Toss the cubes and find the product of the resulting numbers. Record your results.
• Repeat this process 30 times.
Answer:
From the data 15,4,6,30,6,8,36,12,12,6,6,4,15,10,4,2,20,3,15,6,4,6,3,10,3,20,4,12,4,20.
Stem and leaf are calculated for each number.
15 is split into 1 (stem) and 5 (leaf)
04 is split into 0 (stem) and 4 (leaf)
06 is split into 0 (stem) and 6 (leaf)
30 is split into 3 (stem) and 0 (leaf)
06 is split into 0 (stem) and 6 (leaf)
08 is split into 0 (stem) and 8 (leaf)
36 is split into 3 (stem) and 6 (leaf)
12 is split into 1 (stem) and 2 (leaf)
12 is split into 1 (stem) and 2 (leaf)
06 is split into 0 (stem) and 6 (leaf)
06 is split into 0 (stem) and 6 (leaf)
04 is split into 0 (stem) and 4 (leaf)
15 is split into 1 (stem) and 5 (leaf)
10 is split into 1 (stem) and 0 (leaf)
04 is split into 0 (stem) and 4 (leaf)
02 is split into 0 (stem) and 2 (leaf)
20 is split into 2 (stem) and 0 (leaf)
03 is split into 0 (stem) and 3 (leaf)
15 is split into 1 (stem) and 5 (leaf)
06 is split into 0 (stem) and 6 (leaf)
04 is split into 0 (stem) and 4 (leaf)
06 is split into 0 (stem) and 6 (leaf)
03 is split into 0 (stem) and 3 (leaf)
20 is split into 2 (stem) and 0 (leaf)
04 is split into 0 (stem) and 4 (leaf)
12 is split into 1 (stem) and 2 (leaf)
04 is split into 0 (stem) and 4 (leaf)
20 is split into 2 (stem) and 0 (leaf)
$Big ideas Math Grade 6 Chapter 10 Data Displays img_8$

Question 8.
The stem-and-leaf plot shows the weights (in pounds) of several puppies at a pet store. Use the data to answer the question, “How much does a puppy at the pet store weigh?
Answer:
We can use the mean of the data. To find the mean, add the data then divide the sum of the number of data
(8+12+15+17+18+24+24+31)/8 = 149/8 = 18.625
To the nearest pound, a puppy weighs about 19 pounds

Stem-and-Leaf Plots Homework & Practice 10.1

Review & Refresh

Find and interpret the mean absolute deviation of the data.
Question 1.
8, 6, 8, 5, 3, 10, 11, 5, 7
Answer:
First, arrange the given values in the ascending order.
3, 5, 5, 6, 7, 8, 8, 10, 11
We find the mean of the data
mean = (3 + 5 + 5 + 6 + 7 + 8 + 8 + 10 + 11)/9
mean = 7

Question 2.
55, 46, 39, 62, 55, 51, 48, 60, 39, 45
Answer:
First, arrange the given values in the ascending order.
39, 39, 45, 46, 48, 51, 55, 55, 60, 62
We find the mean of the data
mean = (39 + 39 + 45 + 46 + 48 + 51 + 55 + 55 + 60 + 62)/10
mean = 50

Question 3.
37, 54, 41, 18, 28, 32
Answer:
First, arrange the given values in the ascending order.
18, 28, 32, 37, 41, 54
We find the mean of the data
mean = (18+28+32+37+41+54)/6
mean = 35

Question 4.
12, 25, 8, 22, 6, 1, 10, 4
Answer:
First, arrange the given values in ascending order.
1, 4, 6, 8, 10,12, 22, 25
mean = (1+ 4 + 6 + 8 + 10 + 12 + 22 + 25)/8
mean = 11

Use the Distributive Property to simplify the expression.
Question 5.
5(n + 8)
Answer: 5n + 40

Explanation:
5(n + 8) = 5 × n + 5 × 8
5n + 40

Question 6.
7(y – 6)
Answer: 7y – 42

Explanation:
7(y – 6) = 7 × y – 7 × 6
7y – 42

Question 8.
14(2b + 3)
Answer: 28b + 42

Explanation:
14(2b + 3) = 14 × 2b + 14 × 3
28b + 42

Question 9.
11(9 + s)
Answer: 99 + 11s

Explanation:
11(9 + s) = 11 × 9 + 11 × s
99 + 11s

Solve the equation.
Question 9.
$\frac{p}{2}$ = 8
Answer: 16

Explanation:
$\frac{p}{2}$ = 8
p = 8 × 2
p = 16

Question 10.
28 = 6g
Answer: 4.66

Explanation:
28 = 6g
g = 28/6 = 4.66
Thus g = 4.66

Question 11.
3d ÷ 4 = 9
Answer: 12

Explanation:
3d ÷ 4 = 9
3d = 9 × 4
3d = 36
d = 36/3
d = 12
Thus d = 12

Question 12.
10 = $\frac{2z}{3}$
Answer:

Explanation:
10 = $\frac{2z}{3}$
10 × 3 = 2z
2z = 30
z = 30/2
z = 15
So, z = 15

Concepts, Skills, & Problem Solving

REASONING
Write a question that is easier to answer using the stem-and-leaf plot than a dot plot. (See Exploration 1, p. 457.)
Question 13.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 13$
Answer:
Make a stem leaf plot to find the number of customers who visit your shop.
12, 13, 16, 17, 20, 21, 21, 23, 23, 28, 28, 32, 33, 34, 34, 35, 35, 36, 39, 39, 40, 41, 41, 42, 44, 46, 47, 48, 49, 49

Question 14.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 14$
Answer:
Make the stem leaf plot to find the number of text messages you received per hour.
40, 40, 42, 46, 46, 49, 51, 51, 53, 53, 57, 57, 57, 59, 59, 59, 61, 62, 62, 65, 65, 66, 67, 68, 68, 70, 72, 72, 73, 74.

MAKING A STEM-AND-LEAF PLOT Make a stem-and-leaf plot of the data.
Question 15.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 15$
Answer:
We have to write the stem and leaf plot for the above table.
$Big Ideas Math Grade 6 Chapter 10 Data Displays img_9$

Question 16.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 16$
Answer:
We have to write the stem and leaf plot for the above table.
$Big Ideas Math Grade 6 Chapter 10 Data Displays img_10$

Question 17.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 17$
Answer:
We have to write the stem and leaf plot for the above table.
$Big Ideas Math Grade 6 Chapter 10 Data Displays img_10$

Question 18.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 18$
Answer:
We have to write the stem and leaf plot for the above table.
$Big Ideas Math Grade 6 Chapter 10 Data Displays img_12$

Question 19.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 19$
Answer:
We have to write the stem and leaf plot for the above table.
$Big Ideas Math Grade 6 Chapter 10 Data Displays img_13$

Question 20.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 20$
Answer:
We have to write the stem and leaf plot for the above table.
$Big Ideas Math Grade 6 Chapter 10 Data Displays img_14$

Question 21.
YOU BE THE TEACHER
Your friend makes a stem-and-leaf plot of the data. Is your friend correct? Explain your reasoning.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 21$
51, 25, 47, 42, 55, 26, 50, 44, 55
Answer: your friend is correct

Explanation:
In stem and leaf plot the tens place represent stem and the ones place represent the leaf.

MODELING REAL LIFE
The stem-and-leaf plot shows the numbers of confirmed cases of a virus in 15 countries.
Question 22.
How many of the countries have more than 60 confirmed cases?
Answer: 6 countries

Explanation:
By seeing the above stem and leaf plot we can find the number of cases more than 60.
The number of leaf represents the number of countries.
62, 63, 63, 67, 75, 97.
Thus there are 6 countries that have more than 60 confirmed cases.

Question 23.
Find the mean, median, mode, range, and interquartile range of the data.
Answer:
41, 41, 43, 43, 45, 50, 52, 53, 54, 62, 63, 63, 67, 75, 97
In its simplest mathematical definition regarding data sets, the mean used is the arithmetic mean, also referred to as mathematical expectation, or average.
Mean:
mean = (41+41+43+43+45+50+52+53+54+62+63+63+67+75+97)/15
mean = 56.6
Median:
In the odd cases where there are only two data samples or there is an even number of samples where all the values are the same, the mean and median will be the same.
41, 41, 43, 43, 45, 50, 52, 53, 54, 62, 63, 63, 67, 75, 97
So, the median of the given data is 53.
Mode:
The mode is the value in a data set that has the highest number of recurrences.
41, 41, 43, 43, 45, 50, 52, 53, 54, 62, 63, 63, 67, 75, 97
mode = 41, 43, 63 (Repeated 2 times)

Question 24.
How are the data distributed?
Answer:
The distribution of a data set is the shape of the graph when all possible values are plotted on a frequency graph. Usually, we are not able to collect all the data for our variable of interest.

Question 25.
Which data value is an outlier? Describe how the outlier affects the mean.
Answer:
Outliers affect the mean value of the data but have little effect on the median or mode of a given set of data.
Example: A student receives a zero on a quiz and subsequently. has the following scores: 0, 70, 70, 80, 85, 90, 90, 90, 95, 100. Outlier: 0.

Question 26.
REASONING
Each stem-and-leaf plot below has a mean of 39. Without calculating, determine which stem-and-leaf plot has the lesser mean absolute deviation. Explain your reasoning.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 23$
Answer:
i. 23, 27, 30, 32, 36, 39, 41, 42, 45, 48, 51, 54
Mean = (23+27+30+32+36+39+41+42+45+48+51+54)/12
Mean = 39
The mean absolute deviation is 7.833
ii. 22, 24, 25, 28, 29, 33, 38, 45, 53, 56, 57, 58
Mean = (22+24+25+28+29+33+38+45+53+56+57+58)/12
Mean = 39
The mean absolute deviation is 12.333
Thus the first stem and leaf plot has the lesser mean absolute deviation.

Question 27.
DIG DEEPER!
The stem-and-leaf plot shows the daily high temperatures (in degrees Fahrenheit) for the first 15 days of June.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 24$
a. When you include the daily high temperatures for the rest of the month, the mean absolute deviation increases. Draw a stem-and-leaf plot that could represent all of the daily high temperatures for the month.

Answer:
$Big Ideas Math Answers Grade 6 Chapter 12 Data Displays img_6$
b. Use your stem-and-leaf plot from part(a) to answer the question, “What is a typical daily high temperature in June?”
Answer: 89°F is the high temperature in the month of June.

Question 28.
CRITICAL THINKING
The back-to-back stem-and-leaf plot shows the 9-hole golf scores for two golfers. Only one of the golfers can compete in a tournament as your teammate. Use measures of center and measures of variation to support choosing either golfer.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 25$
Answer:
The scores of Rich are
35, 37, 41, 42, 43, 44, 45, 48
The scores of Will are
42, 43, 44, 44, 46, 47, 47, 48, 49
Will can compete in the tournament.

Lesson 10.2 Histograms

EXPLORATION 1

Performing an Experiment
Work with a partner.
a. Make the airplane shown from a single sheet of 8$\frac{1}{2}$ by-11-inch paper. Then design and make your own paper airplane.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 1$
b. PRECISION
Fly each airplane 20 times. Keep track of the distance flown each time. Specify Units. What units will you use to measure the distance flown? Will the units you use affect the results in your frequency table? Explain.
c. A frequency table groups data values into intervals. The frequency is the number of values in an interval. Use a frequency table to organize the results for each airplane.
d. MODELING Represent the data in the frequency tables graphically. Which airplane flies farther? Explain your reasoning.
Answer:

$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 3$

Key Idea
Histograms
p. 463 frequency, A histogram is a bar graph that shows the frequencies of data values in intervals of the same size.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 4$
The height of a bar represents the frequency of the values in the interval.

EXAMPLE 1
Making a Histogram
The frequency table shows the numbers of laps that people in a swimming class completed today. Display the data in a histogram.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 5$
Step 1: Draw and label the axes.
Step 2: Draw a bar to represent the frequency of each interval.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 6$

Try It
Question 1.
The frequency table shows the ages of people riding a roller coaster. Display the data in a histogram.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 7$
Answer:
BIM Grade 6 Answers Chapter 10 Data Displays img_21

EXAMPLE 2
Using a Histogram
The histogram shows winning speeds at the Daytona 500.
(a) Which interval contains the most data values?
(b) How many of the winning speeds are less than 140 miles per hour?
(c) How many of the winning speeds are at least 160 miles per hour?
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 8.1$
a. The interval with the tallest bar contains the most data values.
So, the 150−159 miles per hour interval contains the most data values.
b. One winning speed is in the 120−129 miles per hour interval, and eight winning speeds are in the 130−139 miles per hour interval.
So, 1 + 8 = 9 winning speeds are less than 140 miles per hour.
c. Eight winning speeds are in the 160−169 miles per hour interval, and five winning speeds are in the 170−179 miles per hour interval.
So, 8 + 5 =13 winning speeds are at least 160 miles per hour.

Try It
Question 2.
The histogram shows the numbers of hours that students in a class slept last night.
a. How many students slept at least 8 hours?
b. How many students slept less than 12 hours?
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 8$

Answer:

A. Number of students who slept for 8 to 11 hours is 8.
Number of students who slept for 12 to 15 hours is 3.
The total number of students who slept for atleast 8 hours is 8.

B. Number of students who slept for 8 to 11 hours is 8.
Number of students who slept for 4 to 7 hours is 8.
Number of students who slept for 0 to 3 hours is 2.
Thus the number of students who slept for less than 12 hours is 8 + 8 + 2 = 18 students

EXAMPLE 3
Comparing Data Displays
The data displays show how many push-ups students in a class completed for a physical fitness test. Which data display can you use to find how many students are in the class? Explain.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 9$
You can use the histogram because it shows the number of students in each interval. The sum of these values represents the number of students in the class. You cannot use the circle graph because it does not show the number of students in each interval.

Try It
Question 3.
Which data display should you use to describe the portion of the entire class that completed 30−39 push-ups? Explain.
Answer: You should use the percentage of the number of students in the interval of 30-39 to find the completed push-ups.
The portion of the entire class that completed 30−39 push-ups is 24%

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal
Question 4.
MAKING A HISTOGRAM
The table shows the numbers of siblings of students in a class.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 10$
a. Display the data in a histogram.b. Explain how you chose reasonable intervals for your histogram in part
Answer:

Question 5.
NUMBER SENSE
Can you find the range and the interquartile range of the data in the histogram? If so, find them. If you cannot find them, explain why not.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 11$
Answer:

EXAMPLE 4
Modeling Real Life made using the data displays in Example 3?
A. Twelve percent of the class completed 9 push-ups.
B. Five students completed at least 10 and at most 19 push-ups.
C. At least one student completed more than 39 push-ups.
D. Less than $\frac{1}{4}$ of the class completed 30 or more push-ups.
The circle graph shows that12% completed 0−9 push-ups, but you cannot determine how many completed exactly 9. So, Statement A cannot be made.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 12$
In the histogram, the bar height for the 10−19 interval is 5, and the bar height for the 40−49 interval is 1. So, Statements B and C can be made.
The circle graph shows that24% completed 30−39 push-ups, and 4% completed 40−49 push-ups. So, 24% + 4% =28% completed 30 or more push-ups. Because $\frac{1}{4}$ = 25% and 28% > 25%, Statement D cannot be made.
The correct answers are A and D.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 13$
Question 6.
The histogram shows the numbers of rebounds per game for a middle school basketball player in a season.
a. Which interval contains the most data values?
b. 54 How many games did the player play during the season?
c. In what percent of the games did the player have 4 or more rebounds?
Answer:

Question 7.
Determine whether you can make each statement by using the histogram in the previous exercise.Explain.Rebounds
a. The basketball player had 2 rebounds in 6 different games.
b. The basketball player had more than 1 rebound in 9 different games
Answer:

Histograms Homework & Practice 10.2

Review & Refresh

Make a stem-and-leaf plot of the data.
Question 1.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 14$
Answer:
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays img_11$

Question 2.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 15$
Answer:
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays img_12$

Find the percent of the number.
Question 3.
25% of 180
Answer: 45

Explanation:
25% = 25/100
25/100 × 180
We get 45
So, 25% of 180 is 45.

Question 4.
30% of 90
Answer: 27

Explanation:
30% = 30/100
30/100 × 90 = 27
So, 30% of 90 is 27

Question 5.
16% of 140
Answer: 22.4

Explanation:
16% = 16/100
16/100 × 140 = 22.4
So, 16% of 140 is 22.4

Question 6.
64% of 807.
Answer: 516.48

Explanation:
64% = 64/100
64/100 × 807 = 516.48
So, 64% of 807 is 516.48

Question 7.
What is the least common multiple of 7 and 12?
A. 28
B. 42
C. 84
D. 168
Answer: 84

Explanation:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 7:
7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98
Multiples of 12:
12, 24, 36, 48, 60, 72, 84, 96, 108
Therefore,
LCM(7, 12) = 84
Thus the correct answer is option c.

Concepts, Skills, & Problem Solving
MAKING A FREQUENCY TABLE Organize the data using a frequency table. (See Exploration 1, p. 463.)
Question 8.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 16$
Answer:

BIM Grade 6 Answer Key Chapter 10 Data Displays img_13

Question 9.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 17$
Answer:
BIM Grade 6 Answer Key Chapter 10 Data Displays img_14

MAKING A HISTOGRAM Display the data in a histogram.
Question 10.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 18$
Answer:
BIM Grade 6 Answer Key Chapter 10 Data Displays img_15

Question 11.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 19$
Answer:
BIM Grade 6 Answer Key Chapter 10 Data Displays img_16

Question 12.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 20$
Answer:
BIM Grade 6 Answer Key Chapter 10 Data Displays img_17

Question 13.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 21$
Answer:
BIM Grade 6 Answer Key Chapter 10 Data Displays img_18

Question 14.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 22$
Answer:
BIM Grade 6 Answer Key Chapter 10 Data Displays img_19

Question 15.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 23$
Answer:
BIM Grade 6 Answer Key Chapter 10 Data Displays img_20

Question 16.
YOU BE THE TEACHER
Your friend displays the data in a histogram. Is your friend correct? Explain your reasoning.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 24$
Answer: yes your friend is correct.
The frequency table matches the histogram.

Question 17.
MODELING REAL LIFE
The histogram shows the numbers of magazines read last month by the students in a class.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 25$
a. Which interval contains the fewest data values?
Answer: The interval 4-5 has the fewest data values.
b. How many students are in the class?
Answer:
0-1 = 2
2-3 = 15
4-5 = 0
6-7 = 3
2 + 15 + 3 = 20
c. What percent of the students read fewer than six magazines?
Answer: By seeing the above histogram we can say that 25% of the students read fewer than six magazines.

Question 18.
YOU BE THE TEACHER
Your friend interprets the histogram. Is your friend correct? Explain your reasoning.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 26$
Answer:
Compare your friend with the above histogram.
By seeing the above histogram we can say that it took 12 seconds to download songs.
So, your friend is correct.

Question 19.
REASONING
The histogram shows the percent of the voting-age population in each state who voted in a presidential election. Explain whether the graph supports each statement.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 27$
a. Only 40% of one state voted.
b. In most states, between 50% and 64.9% voted.
c. The mode of the data is between 55% and 59.9%
Answer:

Question 20.
PROBLEM SOLVING
The histograms show the areas of counties in Pennsylvania and Indiana. Which state do you think has the greater area? Explain.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 28$
Answer:

Question 21.
MODELING REAL LIFE
The data displays show how many pounds of garbage apartment residents produced in 1 week. Which data display can you use to find how many residents produced more than 25 pounds of garbage? Explain.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 29$
Answer:

Question 22.
REASONING
Determine whether you can make each statement by using the data displays in Exercise 21. Explain your reasoning.
a. One resident produced 10 pounds of garbage.
b. Twelve residents produced between 20 and 29 pounds of garbage.
Answer:

Question 23.
DIG DEEPER!
The table shows the lengths of some whales in a marine sanctuary.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 30$
a. Make a histogram of the data starting with the interval 51−55.
b. Make another histogram of the data using a different-sized interval.
c. Compare and contrast the two histograms.
Answer:

Question 24.
LOGIC
Can you find the mean or the median of the data in Exercise 17? Explain.
Answer:

Lesson 10.3 Shapes of Distributions

$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 1$

EXPLORATION
Describing Shapes of Distributions
Work with a partner. The lists show the first three digits and last four digits of several phone numbers in the contact list of a cell phone.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 2$
a. Compare and contrast the distribution of the last digit of each phone number to the distribution of the first digit of each phone number. Describe the shapes of the distributions.
b. Describe the shape of the distribution of the data in the table below. Compare it to the distributions in part(a).
Answer:

You can use dot plots and histograms to identify shapes of distributions.

Key Ideas
Symmetric and Skewed Distributions
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 3$

EXAMPLE 1
Describing Shapes of Distributions

Describe the shape of each distribution.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 4$

Try It
Question 1.
Describe the shape of the distribution.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 5$
Answer:
BIM 6th Grade Chapter 10 Data Displays Answer Key img_4
A symmetric distribution has a graph in which the left side is a mirror image of the right side.
A skewed distribution has a graph in which a “tail” extends to the left and most data are on the right OR a “tail” extends to the right and most data are on the left.

EXAMPLE 2
Describing the Shape of a Distribution
The frequency table shows the ages of people watching a comedy in a theater. Display the data in a histogram. Then describe the shape of the distribution.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 6$
Draw and label the axes. Then draw a bar to represent the frequency of each interval.
Most of the data are on the right, and the tail extends to the left.
So, the distribution is skewed left.
Answer:
For a distribution that is skewed right, the tail extends to the right and most of the data are on the left side of the graph.

Try It
Question 2.
The frequency table shows the ages of people watching a historical movie in a theater. Display the data in a histogram. Describe the shape of the distribution.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 7$
Answer:

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.
Question 3.
WRITING
Explain in your own words what it means for a distribution to be (a) skewed left, (b) symmetric, and (c) skewed right.
Answer:

Question 4.
DESCRIBING A DISTRIBUTION
Display the data shown in a histogram. Describe the shape of the distribution.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 8$
Answer:

Question 5.
WHICH ONE DOESN’T BELONG?
Which histogram does not belong with the other three? Explain your reasoning.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 9$
Answer:

EXAMPLE 3
Modeling Real Life
The histogram shows the ages of people watching an animated movie in the same theater as in Example 2. Which movie has an older audience?
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 10$
Understand the problem
You are given histograms that display the ages of people watching two movies. You are asked to determine which movie has an older audience.

Make a plan
Use the intervals and distributions of the data to determine which movie has an older audience.

Solve and check
The intervals in the histograms are the same. Most of the data for the animated movie are on the left, while most of the data for the comedy are on the right. This means that the people watching the comedy are generally older than the people watching the animated movie.

So, the comedy has an older audience.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 11$
Check Reasonableness
The movies have similar attendance. However,only4 people watching the comedy are 17 or under. A total of 35 people watching the animated movie are 17 or under. So, it is reasonable to conclude that the comedy has an older audience.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.
Question 6.
The frequency table shows the numbers of visitors each day to parks in Aurora and Grover in one month. Which park generally has more daily visitors? Justify your answer.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 12$
Answer:

Question 7.
DIG DEEPER!
The frequency tables below show the ages of guests on two cruises. Can you make accurate comparisons of the ages of the guests? Explain your reasoning.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 13$
Answer:

Shapes of Distributions Homework & Practice 10.1

Review & Refresh

Display the data in a histogram.
Question 1.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 14$
Answer:
$Big Ideas Math Answers Grade 6 Chapter 12 Data Displays img_1$
On the vertical axis, place frequencies. Label this axis “Frequency”.
On the horizontal axis, place the lower value of each interval.
Draw a bar extending from the lower value of each interval to the lower value of the next interval.

Question 2.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 15$
Answer:
BIM Grade 6 Answer Key Chapter 10 Data Displays img_2

On the vertical axis, place frequencies. Label this axis “Frequency”.
On the horizontal axis, place the lower value of each interval.
Draw a bar extending from the lower value of each interval to the lower value of the next interval.

Question 3.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 16$
Answer:
$Big Ideas Math Answers Grade 6 Chapter 12 Data Displays img_3$

Write a unit rate for the situation.
Question 4.
$200 per 8 days
Answer:
200/8 = 25
Thus $25 per day.

Question 5.
60 kilometers for every 1.5 hours
Answer:
Your average speed is 60 km per 1.5 hours.
60/1.5 = 40 km/hr

Concepts, Skills, &Problem Solving

DESCRIBING SHAPES OF DISTRIBUTIONS Describe the shape of the distribution of the data in the table. (See Exploration 1, p. 471.)
Question 6.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 17$
Answer:
Step 1:
Order the data
0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6

Question 7.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 18$
Answer:
Step 1:
Order the data
12, 12, 12, 13, 13, 13, 14, 14, 14, 15, 15, 15, 16, 16, 16.

DESCRIBING SHAPES OF DISTRIBUTIONS

Describe the shape of the distribution.
Question 8.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 19$
Answer: 25, 26, 27, 27, 28, 28, 28, 28, 29, 29, 29, 29, 29, 30, 30, 30

Question 9.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 20$
Answer:

15, 16, 16, 17, 17, 17, 18, 18, 18, 18, 18, 19, 19, 19, 20, 20, 21

Question 10.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 21$
Answer:

Question 11.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 22$
Answer:

Question 12.
MODELING REAL LIFE
The frequency table shows the years of experience for the medical states in Jones County and Pine County. Display the data for each county in a histogram. Which county’s medical state has less experience? Explain.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 23$
Answer:

Question 13.
REASONING
What is the shape of the distribution of the restaurant waiting times? Explain your reasoning.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 24$
Answer:

Question 14.
LOGIC
Are all distributions either approximately symmetric or skewed? Explain. If not, give an example.
Answer:

Question 15.
REASONING
Can you use a stem-and-leaf plot to describe the shape of a distribution? Explain your reasoning.
Answer:

Question 16.
DIG DEEPER!
The table shows the donation amounts received by a charity in one day.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 25$
a. Make a histogram of the data starting with the interval 0–14. Describe the shape of the distribution.
b. A company adds $5 to each donation. Make another histogram starting with the same interval as in part(a). Compare the shape of this distribution with the distribution in part(a). Explain any differences in the distributions.
Answer:

Question 17.
CRITICAL THINKING
Describe the shape of the distribution of each bar graph. Match the letters A, B, and C with the mean, the median, and the mode of each data set. Explain your reasoning.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 26$
Answer:

Lesson 10.4 Choosing Appropriate Measures

EXPLORATION 1
Using Shapes of Distributions
Work with a partner.
In Section 10.3 Exploration 1(a), you described the distribution of the first digits of the numbers at the right. In Exploration 1(b), you described the distribution of the data set below.
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 1$
What do you notice about the measures of center, measures of variation, and the shapes of the distributions? Explain.
b. Which measure of center best describes each data set? Explain your reasoning.
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 2$
c. which measures best describe the data. Which measure of variation best describes each data set? Explain your reasoning.
Answer:

You can use a measure of center and a measure of variation to describe the distribution of a data set.e shape of the distribution can help you choose which measures are the most appropriate to use.

Key Idea

Choosing Appropriate Measures
The mean absolute deviation (MAD) uses the mean in its calculation. So, when a data distribution is symmetric,
• use the mean to describe the center and
• use the MAD to describe the variation.

The interquartile range (IQR) uses quartiles in its calculation. So, when a data distribution is skewed,
• use the median to describe the center and
• use the IQR to describe the variation.

EXAMPLE 1
Choosing Appropriate Measures
The frequency table shows the number of states that border each state in the United States. What are the most appropriate measures to describe the center and the variation?
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 3$
To see the distribution of the data, display the data in a histogram.
The left side of the graph is approximately a mirror image of the right side of the graph. The distribution is symmetric.
So, the mean and the mean absolute deviation are the most appropriate measures to describe the center and the variation.

Try It
Question 1.
The frequency table shows the gas mileages of several motorcycles made by a company. What are the most appropriate measures to describe the center and the variation?
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 4$
Answer:
To see the distribution of the data, display the data in a histogram.
$Big Ideas Math Grade 6 Chapter 10 Data Displays img_15$

EXAMPLE 2
Describing a Data Set
The dot plot shows the average numbers of hours students in a class sleep each night. Describe the center and the variation of the data set.
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 5$
Most of the data values are on the right, clustered around 9, and the tail extends to the left. The distribution is skewed left, so the median and the interquartile range are the most appropriate measures to describe the center and the variation.
The median is 8.5 hours. The first quartile is 7.5, and the third quartile is 9. So, the interquartile range is 9 − 7.5 = 1.5 hours.
The data are centered around 8.5 hours. The middle half of the data varies by no more than 1.5 hours.

Try It
Question 2.
The dot plot shows the numbers of hours people spent at the gym last week. Describe the center and the variation of the data set.
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 6$
Answer:
Most of the data values are on the right, clustered around 6, and the tail extends to the left. The distribution is skewed left, so the median and the interquartile range are the most appropriate measures to describe the center and the variation.
The median is 5 hours. The first quartile is 2, and the third quartile is 4.
So, the interquartile range is 4 – 2 = 2 hours
The data are centered around 5 hours. The middle half of the data varies by no more than 2 hours.

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.
Question 3.
OPEN-ENDED
Construct a dot plot for which the mean is the most appropriate measure to describe the center of the distribution.
Answer:

CHOOSING APPROPRIATE MEASURES
Choose the most appropriate measures to describe the center and the variation. Explain your reasoning. Then find the measures you chose.
Question 4.
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 7$
Answer:
20, 28, 32, 32, 36, 36, 40, 40, 40, 40, 44, 44, 44, 48
Order your data set from lowest to highest values
Find the median. This is the second quartile Q2.
At Q2 split the ordered data set into two halves.
The lower quartile Q1 is the median of the lower half of the data.
Q1 = 32
The upper quartile Q3 is the median of the upper half of the data.
Q3 = 44
Median is the average of the data values.
So, the median, Q2 is 40.
Interquartile Range = Q3 – Q1
IQR = 44 – 32
IQR = 12

Question 5.
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 8$
Answer:
8, 10, 10, 12, 12, 12, 14, 14, 14, 14, 16, 16, 16, 18, 18, 20
8, 10, 12, 14, 16, 18, 20
Order your data set from lowest to highest values
Find the median. This is the second quartile Q2.
At Q2 split the ordered data set into two halves.
The lower quartile Q1 is the median of the lower half of the data.
Q1 = 12
The upper quartile Q3 is the median of the upper half of the data.
Q3 = 16
Median is the average of the data values.
So, the median, Q2 is 14
Interquartile Range = Q3 – Q1
IQR = 16 – 12
IQR = 4

Question 6.
WRITING
Explain why the most appropriate measures to describe the center and the variation of a data set are determined by the shape of the distribution.
Answer:
You can use a measure of center and a measure of variation to describe the distribution of a data set. The shape of the distribution can help you choose which measures are the most appropriate to use. The dot plot shows the average number of hours students in a class sleep each night.

EXAMPLE 3
Modeling Real Life
Two baskets each have16 envelopes with money inside, as shown in the tables. How much does a typical envelope in each basket contain? Why might a person want to pick from Basket B instead of Basket A?
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 9$
In each graph, the left side is a mirror image of the right side. Because both distributions are symmetric, the mean and the mean absolute deviation are the most appropriate measures to describe the center and the variation.
The mean of each data set is $\frac{800}{16}$ = $50. The MAD of Basket A is $\frac{320}{16}$ = $20, and the MAD of Basket B is $\frac{120}{16}$ = $7.50. So, Basket A has more variability.

A typical envelope in each basket contains about $50. A person may choose from Basket B instead of Basket A because there is less variability. This means it is more likely to get an amount near $50 by choosing an envelope from Basket B than by choosing an envelope from Basket A.
Answer:

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.
Question 7.
Why might a person want to pick from Basket A instead of Basket B in Example 3? Explain your reasoning.
Answer:

Question 8.
In a video game, two rooms each have 12 treasure chests containing gold coins. The tables show the numbers of coins in each chest. You pick one chest and are rewarded with the coins inside. From which room would you choose? Explain your reasoning.
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 10$
Answer:

Question 9.
Create a dot plot of the numbers of pets that students in your class own. Describe the center and the variation of the data set.
Answer:

Choosing Appropriate Measures Homework & Practice 10.4

Review & Refresh

Describe the shape of the distribution.
Question 1.
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 11$
Answer:
Order the data
5, 6, 7, 7, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10
The shape of the distribution for the above dot plot is
$Big-Ideas-Math-Solutions-Grade-6-Chapter-10-Data-Displays-10.4-11$

Question 2.
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 12$
Answer:

Find the median, first quartile, third quartile, and interquartile range of the data.
Question 3.
68, 74, 67, 72, 63, 70, 78, 64, 76
Answer:
Order the data
63, 64, 67, 68, 70, 72, 74, 76, 78
The median is nothing but the average value of the data.
70 is the average of the data values.
Order your data set from lowest to highest values
Find the median. This is the second quartile Q2.
Thus the second quartile Q2 is 70.
At Q2 split the ordered data set into two halves.
The lower quartile Q1 is the median of the lower half of the data.
Thus Q1 is 65.5
The upper quartile Q3 is the median of the upper half of the data.
Q3 is 75
Interquartile Range IQR = 9.5
If the size of the data set is odd, do not include the median when finding the first and third quartiles.
If the size of the data set is even, the median is the average of the middle 2 values in the data set. Add those 2 values, and then divide by 2. The median splits the data set into lower and upper halves and is the value of the second quartile Q2.

Question 4.
39, 48, 33, 24, 30, 44, 36, 41, 28, 53
Answer:
Order the data
24, 28, 30, 33, 36, 39, 41, 44, 48, 53
If the size of the data set is even, the median is the average of the middle 2 values in the data set. Add those 2 values, and then divide by 2. The median splits the data set into lower and upper halves and is the value of the second quartile Q2.
Median is (36+39)/2 = 37.5
Order your data set from lowest to highest values
Find the median. This is the second quartile Q2.
Thus the second quartile Q2 is 37.5
At Q2 split the ordered data set into two halves.
The lower quartile Q1 is the median of the lower half of the data.
Thus Q1 is 30
The upper quartile Q3 is the median of the upper half of the data.
Q3 is 44
Interquartile Range IQR = 14
If the size of the data set is odd, do not include the median when finding the first and third quartiles.

Divide. Write the answer in simplest form.
Question 5.
4$\frac{2}{5}$ ÷ 2
Answer: 2 $\frac{1}{5}$

Explanation:
Convert any mixed numbers to fractions.
4$\frac{2}{5}$ = $\frac{22}{5}$
$\frac{22}{5}$ × $\frac{1}{2}$ = $\frac{22}{10}$
Now convert from improper fraction to the mixed fraction.
$\frac{22}{10}$ = 2 $\frac{1}{5}$

Question 6.
5$\frac{1}{8}$ ÷ $\frac{7}{8}$
Answer: 5 $\frac{6}{7}$

Explanation:
Convert any mixed numbers to fractions.
5$\frac{1}{8}$ = $\frac{41}{8}$
$\frac{41}{8}$ ÷ $\frac{7}{8}$
$\frac{41}{8}$ × $\frac{8}{7}$ = $\frac{328}{56}$
Now convert from improper fraction to the mixed fraction.
$\frac{328}{56}$ = 5 $\frac{6}{7}$

Question 7.
2$\frac{3}{7}$ ÷ 1$\frac{1}{7}$
Answer: 2 $\frac{1}{8}$

Explanation:
Convert any mixed numbers to fractions.
2$\frac{3}{7}$ = $\frac{17}{7}$
1$\frac{1}{7}$ = $\frac{8}{7}$
$\frac{17}{7}$ ÷ $\frac{8}{7}$ = $\frac{119}{56}$
Simplify the fraction
$\frac{119}{56}$ = 2 $\frac{1}{8}$

Question 8.
$\frac{4}{5}$ ÷ 7$\frac{1}{2}$
Answer: $\frac{8}{75}$

Explanation:
Convert any mixed numbers to fractions.
7$\frac{1}{2}$ = $\frac{15}{2}$
$\frac{4}{5}$ ÷ $\frac{15}{2}$ = $\frac{8}{75}$

Concepts, Skills, & Problem Solving

USING SHAPES OF DISTRIBUTIONS
Find the mean and the median of the data set. Which measure of center best describes the data set? Explain your reasoning. (See Exploration 1, p. 477.)
Question 9.
9, 3, 7, 7, 9, 2, 8, 9, 6, 7, 8, 9
Answer:

Question 10.
24, 25, 27, 27, 23, 29, 26, 26, 26, 25, 28
Answer:

CHOOSING APPROPRIATE MEASURES
Choose the most appropriate measures to describe the center and the variation.
Question 11.
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 13$
Answer:

Question 12.
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 14$
Answer:

Question 13.
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 15$
Answer:

Question 14.
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 16$
Answer:

Question 15.
DESCRIBING DATA SETS
Describe the centers and the variations of the data sets in Exercises 11 and 12.
Answer:

Question 16.
MODELING REAL LIFE
The frequency table shows the numbers of eggs laid by several octopi. What are the most appropriate measures to describe the center and the variation? Explain your reasoning.
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 17$
Answer:

Question 17.
MODELING REAL LIFE
The dot plot shows the vertical jump heights (in inches) of several professional athletes. Describe the center and the variation of the data set.
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 18$
Answer:

Question 18.
OPEN-ENDED
Describe a real-life situation where the median and the interquartile range are likely the best measures of center and variation to describe the data. Explain your reasoning.
Answer:

Question 19.
PROBLEM SOLVING
You play a board game in which you draw from one of two piles of cards. Each card has a number that says how many spaces you will move your piece forward on the game board. The tables show the numbers on the cards in each pile. From which pile would you choose? Explain your reasoning.
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 19$
Answer:

Question 20.
DIG DEEPER!
The frequency table shows the numbers of words that several students can form in 1 minute using the letters P, S, E, D, A. What are the most appropriate measures to describe the center and variation? Can you find the exact values of the measures of center and variation for the data? Explain.
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 20$
Answer:

Question 21.
REASONING
A bag contains 20 vouchers that can be redeemed for different numbers of tokens at an arcade, as shown in the table.
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 21$
a. Find the most appropriate measure to describe the center of the data set.
b. You randomly select a voucher from the bag. How many tokens are you most likely to receive? Explain.
c. Are your answers in parts (a) and (b) the same? Explain why or why not.
Answer:

Lesson 10.5 Box-and-Whisker Plots

EXPLORATION 1
Drawing a Box-and-Whisker Plot
Work with a partner. Each student in a sixth-grade class is asked to choose a number from 1 to 20. The results are shown below.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 1$
a. The box-and-whisker plot below represents the data set. Which part represents the box? the whiskers? Explain.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 2$
b. What does each of the five plotted points represent?
c. In your own words, describe what a box-and-whisker plot is and what it tells you about a data set.
d. Conduct a survey in your class. Have each student write a number from 1 to 20 on a piece of paper. Collect all of the data and draw a box-and-whisker plot that represents the data. Compare the data with the box-and-whisker plot in part(a).
Answer:

$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 3.1$

Key Idea
Box-and-Whisker Plot
A box-and-whisker plot represents a data set along a number line by using the least value, the greatest value, and the quartiles of the data. A box-and-whisker plot shows the variability of a data set.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 3$
The five numbers that make up the box-and-whisker plot are called the five-number summary of the data set.

EXAMPLE 1
Making a Box-and-Whisker Plot
Make a box-and-whisker plot for the ages(in years) of the spider monkeys at a zoo.
15, 20, 14, 38, 30, 36, 30, 30, 27, 26, 33, 35
Step 1: Order the data. Find the quartiles.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 4$
Step 2: Draw a number line that includes the least and greatest values. Graph points above the number line that represent the five-number summary.
Step 3: Draw a box using the quartiles. Draw a line through the median. Draw whiskers from the box to the least and the greatest values.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 5$
Answer:

Try It
Question 1.
A group of friends spent 1, 0, 2, 3, 4, 3, 6, 1, 0, 1, 2, and 2 hours online last night.Make a box-and-whisker plot for the data.
Answer:

The figure shows how data are distributed in a box-and-whisker plot.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 6$

EXAMPLE 2
Analyzing a Box-and-Whisker Plot
The box-and-whisker plot shows the body mass index (BMI) of a sixth-grade class.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 7$
a. What fraction of the students have a BMI of at least 22?
The right whisker represents students who have a BMI of at least 22.
So, about $\frac{1}{4}$ of the students have a BMI of at least 22.
b. Are the data more spread out below the first quartile or above the third quartile? Explain.
The right whisker is longer than the left whisker.
So, the data are more spread out above the third quartile than below the first quartile.
c. Find and interpret the interquartile range of the data.
interquartile range = third quartile − first quartile
= 22 – 19 = 3
So, the middle half of the students’ BMIsvaries by no more than 3.

Try It
Question 2.
The box-and-whisker plot shows the heights of the roller coasters at an amusement park.
(a) What fraction of the roller coasters are between 120 feet tall and 220 feet tall?
(b) Are the data more spread out below or above the median? Explain.
(c) Find and interpret the interquartile range of the data.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 8$
Answer:

A box-and-whisker plot also shows the shape of a distribution.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 9$

EXAMPLE 3
Identifying Shapes of Distributions
The double box-and-whisker plot represents the life spans of crocodiles and alligators at a zoo. Identify the shape of the distribution of the lifespans of alligators.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 10$
For alligator life spans, the whisker lengths are equal. The median is in the middle of the box. The left side of the box-and-whisker plot is a mirror image of the right side of the box-and-whisker plot.
So, the distribution is symmetric.
Answer:

Try It
Question 3.
Identify the shape of the distribution of the life spans of crocodiles.
Answer:

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.
Question 4.
VOCABULARY
Explain how to find the five-number summary of a data set.
Answer:

MAKING A BOX-AND-WHISKER PLOT
Make a box-and-whisker plot for the data. Identify the shape of the distribution.
Question 5.
Ticket prices (dollars): 39, 42, 40, 47, 38, 39, 44, 55, 44, 58, 45
Answer:

Question 6.
Number of sit-ups: 20, 20, 23, 25, 25, 26, 27, 29, 30, 30, 32, 34, 37, 38
Answer:

Question 7.
NUMBER SENSE
In a box-and-whisker plot, what fraction of the data is greater than the first quartile?
Answer:

EXAMPLE 4
Modeling Real Life
The double box-and-whisker plot represents the prices of snowboards at two stores.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 11$
a. Which store’s prices are more spread out? Explain. Both boxes appear to be the same length. So, the interquartile range of each data set is equal. The range of the prices in Store B, however, is greater than the range of the prices in Store A.
So, the prices in Store B are more spread out.
b. Which store’s prices are generally higher? Explain.
For Store A,the distribution is symmetric with about one-half of the prices above $300.
For Store B, the distribution is skewed right with about three-fourths of the prices above $300.
So, the prices in Store B are generally higher.
Answer:

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.
Question 8.
The tables at the left show the test scores of two sixth-grade achievement tests. The same group of students took both tests. The students took one test in the fall and the other in the spring.
a. Analyze each distribution. Then compare and contrast the test results.
b. Which table likely represents the results of which test? Explain your reasoning.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 12$
Answer:

Question 9.
Make a box-and-whisker plot that represents the heights of the boys in your class. Then make a box-and-whisker plot that represents the heights of the girls in your class. Compare and contrast the distributions.
Answer:

Box-and-Whisker Plots Homework & Practice 10.5

Review & Refresh

Choose the most appropriate measures to describe the center and the variation.
Question 1.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 13$
Answer:

Question 2.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 14$
Answer:

Copy and complete the statement using < or >.
Question 3.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 15$
Answer: –$\frac{2}{3}$ > –$\frac{3}{4}$

Explanation:
Compare fractions to find which fraction is larger and which is smaller.
The least common denominator (LCD) is 12
Rewriting as equivalent fractions with the LCD:
$\frac{2}{3}$ = $\frac{8}{12}$
$\frac{3}{4}$ = $\frac{9}{12}$
Now compare the fractions
–$\frac{8}{12}$ >-$\frac{9}{12}$
Thus we can say that –$\frac{2}{3}$ > –$\frac{3}{4}$

Question 4.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 16$
Answer: -2 $\frac{1}{5}$ < -2 $\frac{1}{6}$

Explanation:
Compare fractions to find which fraction is larger and which is smaller.
Rewriting these inputs as fractions:
2 $\frac{1}{5}$ = $\frac{11}{5}$
2 $\frac{1}{6}$ = $\frac{13}{6}$
The LCM is 30
Rewriting as equivalent fractions with the LCD
$\frac{11}{5}$ = $\frac{66}{30}$
$\frac{13}{6}$ = $\frac{65}{30}$
– $\frac{66}{30}$ < – $\frac{65}{30}$
-2 $\frac{1}{5}$ < -2 $\frac{1}{6}$

Question 5.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 17$
Answer: -5.3 > -5.5

Explanation:
Compare fractions to find which fraction is larger and which is smaller.
The smallest number with the negative sign will be the greater number
Thus -5.3 > -5.5

Factor the expression using the GCF.
Question 6.
42 + 14
Answer

Question 7.
12x – 18
Answer:

Question 8.
28n + 20
Answer:

Question 9.
60g – 25h
Answer:

Concepts, Skills, & Problem Solving

COMPARING DATA Compare the data in the box-and-whisker plots. (See Exploration 1, p. 483.)
Question 10.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 18$
Answer:

Question 11.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 19$
Answer:

MAKING A BOX-AND-WHISKER PLOT
Make a box-and-whisker plot for the data.
Question 12.
Ages of teachers (in years): 30, 62, 26, 35, 45, 22, 49, 32, 28, 50, 42, 35
Answer:

Question 13.
Quiz scores: 8, 12, 9, 10, 12, 8, 5, 9, 7, 10, 8, 9, 11
Answer:

Question 14.
Donations (in dollars): 10, 30, 5, 15, 50, 25, 5, 20, 15, 35, 10, 30, 20
Answer:

Question 15.
Science test scores: 85, 76, 99, 84, 92, 95, 68, 100, 93, 88, 87, 85
Answer:

Question 16.
Shoe sizes: 12, 8.5, 9, 10, 9, 11, 11.5, 9, 9, 10, 10, 10.5, 8
Answer:

Question 17.
Ski lengths (in centimeters): 180, 175, 205, 160, 210, 175, 190, 205, 190, 160, 165, 195
Answer:

Question 18.
YOU BE THE TEACHER
Your friend makes a box-and-whisker plot for the data shown. Is your friend correct? Explain your reasoning.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 20$
2, 6, 4, 3, 7, 4, 6, 9, 6, 8, 5, 7
Answer:

Question 19.
MODELING REAL LIFE
The numbers of days 12 friends went camping during the summer are 6, 2, 0, 10, 3, 6, 6, 4, 12, 0, 6, and 2. Make a box-and-whisker plot for the data. What is the range of the data?
Answer:

Question 20.
ANALYZING A BOX-AND-WHISKER PLOT
The box-and-whisker plot represents the numbers of gallons of water needed to fill different types of dunk tanks offered by a company.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 21$
a. What fraction of the dunk tanks requires at least 500 gallons of water?
b. Are the data more spread out below the first quartile or above the third quartile? Explain.
c. Find and interpret the interquartile range of the data.
Answer:

Question 21.
MODELING REAL LIFE
The box-and-whisker plot represents the heights (in meters) of the tallest buildings in Chicago.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 22$
a. What percent of the buildings are no taller than 345 meters?
b. Is there more variability in the heights above 345 meters or below 260.5 meters? Explain.
c. Find and interpret the interquartile range of the data.
Answer:

Question 22.
CRITICAL THINKING
The numbers of spots on several frogs in a jungle are shown in the dot plot.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 23$
a. Make a box-and-whisker plot for the data.
b. Compare the dot plot and the box-and-whisker plot. Describe the advantages and disadvantages of each data display.
Answer:

SHAPES OF BOX-AND-WHISKER PLOTS
Identify the shape of the distribution. Explain.
Question 23.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 24$
Answer:

Question 24.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 25$
Answer:

Question 25.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 26$
Answer:

Question 26.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 27$
Answer:

Question 27.
MODELING REAL LIFE
The double box-and-whisker plot represents the start times of recess for classes at two schools.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 28$
a. Identify the shape of each distribution.
b. Which school’s start times for recess are more spread out? Explain.
c. You randomly pick one class from each school. Which class is more likely to have recess before lunch? Explain.
Answer:

MAKING A BOX-AND-WHISKER PLOT
Make a box-and-whisker plot for the data.
Question 28.
Temperatures (in °C): 15, 11, 14, 10, 19, 10, 2, 15, 12, 14, 9, 20, 17, 5
Answer:

Question 29.
Checking account balances (in dollars): 30, 0, 50, 20, 90, −15, 40, 100, 45, −20, 70, 0
Answer:

Question 30.
REASONING
The data set in Exercise 28 has an outlier. Describe how removing the outlier affects the box-and-whisker plot.
Answer:

Question 31.
OPEN-ENDED
Write a data set with 12 values that has a symmetric box-and-whisker plot.
Answer:

Question 32.
CRITICAL THINKING
When does a box-and-whisker plot not have one or both whiskers?
Answer: A simpler formulation is this: no whisker will be visible if the lower quartile is equal to the minimum, or if the upper quartile is equal to the maximum.

Question 33.
STRUCTURE Draw a histogram that could represent the distribution shown in Exercise 25.
Answer:

Question 34
DIG DEEPER!
The double box-and-whisker plot represents the goals scored per game by two lacrosse teams during a 16-game season.
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 29$
a. Which team is more consistent? Explain.
b. Team 1 played Team 2 once during the season. Which team do you think won? Explain.
c. Can you determine the number of games in which Team 2 scored 10 goals or less? Explain your reasoning.
Answer:

Question 35.
CHOOSE TOOLS
A market research company wants to summarize the variability of the SAT scores of graduating seniors in the United States. Should the company use a stem-and-leaf plot, a histogram, or a box-and-whisker plot? Explain.
Answer:

Data Displays Connecting Concepts

Using the Problem-Solving Plan
1. The locations of pitches in an at-bat are shown in the coordinate plane, where the coordinates are measured in inches. Describe the location of a typical pitch in the at-bat.
Understand the problem
You know the locations of the pitches. You are asked to find the location of a typical pitch in the at-bat.

Make a plan
First, use the coordinates of the pitches to create two data sets, one for the x-coordinates of the pitches and one for the y-coordinates of the pitches. Next, make a box-and-whisker plot for each data set. Then use the most appropriate measure of center for each data set to find the location of a typical pitch.

Solve and check
Use the plan to solve the problem. Then check your solution.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays cc 1$

2. A set of 20 data values is described below. Sketch a histogram that could represent the data set. Explain.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays cc 2$
• least value: 10
• first quartile: 25
• mean: 29
• third quartile: 34
• greatest value: 48
• MAD: 7

3. The chart shows the dimensions (in inches) of several flat-rate shipping boxes. Each box is in the shape of a rectangular prism. Describe the distribution of the volumes of the boxes. Then find the most appropriate measures to describe the center and the variation of the volumes.

Performance Task
Classifying Dog Breeds by Size
At the beginning of this chapter, you watched a STEAM Video called “Choosing a Dog.” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays cc 3$

Data Displays Chapter Review

Review Vocabulary
Write the definition and give an example of each vocabulary term.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays cr 1$

Graphic Organizers
You can use an Information Frame to help you organize and remember concepts. Here is an example of an Information Frame for the vocabulary term histogram.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays cr 2$
Choose and complete a graphic organizer to help you study the concept.
1. stem-and-leaf plot
2. frequency table
3. shapes of distributions
4. box-and-whisker plot
Answer:

Chapter Self-Assessment
As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays cr 3$

10.1 Stem-and-Leaf Plots
Learning Target: Display and interpret data in stem-and-leaf plots.

Make a stem-and-leaf plot of the data.
Question 1.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 1$
Answer:
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays img_16$

Question 2.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 2$
Answer:
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays img_17$

Question 3.
The stem-and-leaf plot shows the weights (in pounds) of yellowfin tuna caught during a fishing contest.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 3$
a. How many tuna weigh less than 90 pounds?
b. Find the mean, median, mode, range, and interquartile range of the data.
c. How are the data distributed?
Answer:

Question 4.
The stem-and-leaf plot shows the body mass index (BMI) for adults at a recreation center. Use the data to answer the question, “What is the typical BMI for an adult at the recreation center?” Explain.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 4$
Answer:

Question 5.
Write a statistical question that can be answered using the stem-and-leaf plot.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 5$
Answer:

10.2 Histograms (pp. 463-470)
Learning Target: Display and interpret data in histograms.

Display the data in a histogram.
Question 6.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 6$
Answer:

Question 7.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 7$
Answer:
$Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays img_19$

Question 8.
The histogram shows the number of crafts each member of a craft club made for a fundraiser.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 8$
a. Which interval contains the most data values?
b. Frequency How many members made at least 6 crafts?
c. Can you use the histogram to determine the total number of crafts made? Explain.
Answer:

10.3 Shapes of Distributions (pp. 471–476)
Learning Target: Describe and compare shapes of distributions.

Question 9.
Describe the shape of the distribution.
Answer: The shape of a distribution is described by its number of peaks and by its possession of symmetry, its tendency to skew, or its uniformity.

Question 10.
The frequency table shows the math test scores for the same class of students as Exercise 9. Display the data in a histogram. Which test has higher scores?
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 10$
Answer:

Question 11.
The table shows the numbers of neutrons for several elements in the nonmetal group of the periodic table. Make a histogram of the data starting with the interval 0–9. Describe the shape of the distribution.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 11$
Answer:

10.4 Choosing Appropriate Measures (pp. 477–482)
Learning Target: Use the shape of the distribution of a data set to determine which measures of center and variation best describe the data.

Choose the most appropriate measures to describe the center and the variation. Students’ Heights
Question 12.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 12$
Answer:

Question 13.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 13$
Answer:

Question 14.
Describe the center and the variation of the data set in Exercise 13.
Answer:

10.5 Box-and-Whisker Plots (pp. 483–490)
Learning Target: Display and interpret data in box-and-whisker plots.

Make a box-and-whisker plot for the data.
Question 15.
Ages of volunteers at a hospital:
14, 17, 20, 16, 17, 14, 21, 18
Answer:

Question 16.
Masses (in kilograms) of lions:
120, 200, 180, 150, 200, 200, 230, 160
Answer:

Question 17.
The box-and-whisker plot represents the lengths (in minutes) of movies being shown at a theater.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 17$
a. What percent of the movies are no longer than 120 minutes?
b. Is there more variability in the movie lengths longer than 130 minutes or shorter than 110 minutes? Explain.
c. Find and interpret the interquartile range of the data.
Answer:

Question 18.
The double box-and-whisker plot represents the heights of students in two math classes.
$Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 18$
a. Identify the shape of each distribution.Height(cm)
b.Which class has heights that are more spread out? Explain.
c.You randomly pick one student from each class. Which student is more likely to be taller than 170 centimeters? Explain.
Answer:

Data Displays Practice Test

Make a stem-and-leaf plot of the data.
Question 1.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays pt 1$
Answer:
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays img_8$

Question 2.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays pt 2$
Answer:

$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays img_9$

Question 3.
Find the mean, median, mode, range, and interquartile range of the data.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays pt 3$
Answer:
Given data 35, 38, 40, 41, 48, 50, 54, 54, 54, 55, 59, 60
Mean:
The mean refers to an intermediate value between a discrete set of numbers, namely, the sum of all values in the data set, divided by the total number of values.
x̄ = (35+38+40+41+48+50+54+54+54+55+59+60)/12
x̄ = 49
Thus mean of the given data is 49.
Median:
Given data 35, 38, 40, 41, 48, 50, 54, 54, 54, 55, 59, 60
In the case where the total number of values in a data sample is odd, the median is simply the number in the middle of the list of all values. When the data sample contains an even number of values, the median is the mean of the two middle values.
Median = (50+54)/2 = 104/2 = 52
Thus the median of the given data is 52.
Mode:
The mode is the value in a data set that has the highest number of recurrences.
35, 38, 40, 41, 48, 50, 54, 54, 54, 55, 59, 60
Mode = 54 (repeated 3 times)

Question 4.
Display the data in a histogram. How many people watched less than 20 hours of television per week?
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays pt 4$
Answer:
BIM Grade 6 Chapter 10 Data Displays Answer Key img_10
By seeing the above histogram we can find the number of people who watched less than 20 hours of television per week.
14 + 16 = 30
Therefore 30 people watched less than 20 hours per week.

Question 5.
The dot plot shows the numbers of glasses of water Water Consumed that the students in a class drink in one day.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays pt 5$
a. Describe the shape of the distribution.
b. Choose the most appropriate measures to describe the center and the variation. Find the measures you chose.
Answer:

Question 6.
Make a box-and-whisker plot for the lengths (in inches) of fish in a pond: 12, 13, 7, 8, 14, 6, 13, 10.
Answer:

Question 7.
The double box-and-whisker plot compares the battery lives (in hours) of two brands of cell phones.
$Big Ideas Math Answers Grade 6 Chapter 10 Data Displays pt 7$
a. What is the range of the upper 75% of battery life for each brand of cell phone?
b. Which brand of cell phone typically has a longer battery life? Explain.
c. In the box-and-whisker plot, there are 190 cell phones of Brand A that have at most 10.5 hours of battery life. About how many cell phones are represented in the box-and-whisker plot for Brand A?
Answer:

Data Displays Cumulative Practice

Question 1.
Research scientists are measuring the numbers of days lettuce seeds take to germinate. In a study, 500 seeds were planted. Of these,473 seeds germinated. The box-and-whisker plot summarizes the numbers of days it took the seeds to germinate. What can you conclude from the box-and-whisker plot?
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays cp 1$
A. The median number of days for the seeds to germinate is 12.
B. 50% of the seeds took more than 8 days to germinate.
C. 50% of the seeds took less than 5 days to germinate.
D. The median number of days for the seeds to germinate was 6.
Answer:

Question 2.
Find the interquartile range of the data.
15 7 5 8 9 20 12 7 11 7 15
F. 8
G. 11
H. 12
I. 20
Answer: 8

Question 3.
There are seven different integers in a set. When they are listed from least to greatest, the middle integer is −1. Which statement below must be true?
A. There are three negative integers in the set.
B. There are three positive integers in the set.
C. There are four negative integers in the set.
D. The integer in the set after −1 is positive.
Answer:

Question 4.
What is the mean number of seats?
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays cp 4$
F. 2.4 seats
G. 5 seats
H. 6.5 seats
I. 7 seats5.
Answer:

Question 5.
On Wednesday, a town received 17 millimeters of rain. This was x millimeters more rain than the town received on Tuesday. Which expression represents the amount of rain, in millimeters, the town received on Tuesday?
A. 17x
B. 17 – x-c
C. x + 17
D. x – 17
Answer:

Question 6.
One of the leaves is missing in the stem-and-leaf plot.
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays cp 6$
The median of the data set represented by the stem-and-leaf plot is 38. What is the value of the missing leaf?
Answer:

Question 7.
Which property is demonstrated by the equation?
723 + (y + 277) = 723 + (277 + y)
F. Associative Property of Addition
G. Commutative Property of Addition
H. Distributive Property
I. Addition Property of Zero
Answer: Associative Property of Addition

Explanation:
Associative property of addition: Changing the grouping of addends does not change the sum
Thus the correct answer is option F.

Question 8.
A student took five tests and had a mean score of 92. Her scores on the first 4 tests were 90, 96, 86, and 92. What was her score on the fifth test?
A. 92
B. 93
C. 96
D. 98
Answer: 86

Explanation:
Given that,
A student took five tests and had a mean score of 92.
Her scores on the first 4 tests were 90, 96, 86, and 92.
(90+96+86+92+s)/5=90
(364+s)/5=90
364+s=450
s=86
So she scored an 86 on the fifth test.

Question 9.
At the end of the school year, your teacher counted the number of absences for each student. The results are shown in the histogram. How many students had fewer than 10 absences?
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays cp 9$
Answer:

Question 10.
The ages of the 16 members of a camera club are listed below.
40, 22, 24, 58, 30, 31, 37, 25, 62, 40, 39, 37, 28, 28, 51, 44
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays cp 10.1$
$Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays cp 10$
Part A Order the ages from youngest to oldest.
Part B Find the median of the ages.
Part C Make a box-and-whisker plot for the ages.
Answer:

Conclusion:

I wish the information prevailed in Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays is beneficial for all. Tap the links and kickstart your preparation. Share the Big Ideas Math Answers Grade 6 Chapter 10 Data Displays pdf to your friends and help them to overcome their difficulties. Stay tuned to our site to get the solutions of all Big Ideas Math Answers Grade 6 Chapters.

Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors

January 13, 2021

Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors: Students of Grade 6 should be familiar with the factor pairs, prime numbers, and composite numbers. You must remind the concept of adding and subtracting the fractions, mixed fractions, and vocabulary terms. With the help of the Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors, you can know what is exponent, power, perfect square, prime numbers, and composite numbers. All the questions and answers are provided in the pdf format in Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors.

Big Ideas Math Book 6th Grade Answer Key Chapter 1 Numerical Expressions and Factors

Easy learning is possible with BIM 6th Grade Chapter 1 Numerical Expression and Factors Answer key. So, Download Big Ideas Math Book 6th Grade Answer Key Chapter 1 Numerical Expressions and Factors pdf and kickstart your preparation. We have prepared the solutions for all the questions in an easy manner. The topics covered in this chapter are Powers and Exponents, Order of Operations, Prime Factorization, Greatest Common Factor, and Least Common Multiple. Hit the links and learn the problems as per the topics.

Performance Task

Numerical Expressions and Factors Steam Video/Performance Task
Numerical Expressions and Factors Getting Ready for Chapter 1

Lesson 1: Powers and Exponents

Lesson 1.1 Powers and Exponents
Powers and Exponents Practice 1.1

Lesson 2: Order of Operations

Lesson 1.2 Order of Operations
Order of Operations Practice 1.2

Lesson 3: Prime Factorization

Lesson 1.3 Prime Factorization
Prime Factorization Practice 1.3

Lesson 4: Greatest Common Factor

Lesson 1.4 Greatest Common Factor
Greatest Common Factor Practice 1.4

Lesson 5: Least Common Multiple

Lesson 1.5 Least Common Multiple
Least Common Multiple Practice 1.5

Chapter: 1 – Numerical Expressions and Factors

Numerical Expressions and Factors Connecting Concepts
Numerical Expressions and Factors Chapter Review
Numerical Expressions and Factors Practice Test
Numerical Expressions and Factors Cumulative Practice

Numerical Expressions and Factors Steam Video/Performance Task

Filling Piñatas

Common factors can be used to make identical groups of objects. Can you think of any situations in which you would want to separate objects into equal groups? Are there any common factors that may be more useful than others? Can you think of any other ways to use common factors?
watch the STEAM Video “Filling Piñatas.” Then answer the following questions. The table below shows the numbers of party favors that Alex and Enid use to make piñatas.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 1$
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 1.1$

Question 1.
When ﬁnding the number of identical piñatas that can be made, why is it helpful for Alex and Enid to list the factors of each number given in the table?

Answer: By using the list of the factors of all the numbers Alex and Enid can make identical groups of the objects.

Question 2.
You want to create 6 identical piñatas. How can you change the numbers of party favors in the table to make this happen? Can you do this without changing the total number of party favors?

Answer: You can change the number of party favors to create 6 identical pinatas.
There are 100 Mints. So divide it into two identical groups.
Change the number of mints to 50. And add 50 to new identical pinatas.

Performance Task

Setting the Table

After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be asked to plan a fundraising event with the items below.
72 chairs
48 balloons
24 flowers
36 candles
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 2$
You will ﬁnd the greatest number of identical tables that can be prepared, and what will be in each centerpiece. When making arrangements for a party, should a party planner always use the greatest number of identical tables possible? Explain why or why not.

Answer:
72 chairs = 2 × 36
= 2 × 2 × 18
= 2 × 2 × 2 × 9
= 2 × 2 × 2 × 3 × 3
2, 2, 2, 3, 3
Therefore, 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72 are the factors of 72.
48 balloons = 2 × 24
= 2 × 2 × 12
= 2 × 2 × 2 × 6
= 2 × 2 × 2 × 2 × 3
The positive Integer factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
The factors of number 24 are 1, 2, 3, 4, 6, 8, 12, 24.
The factors of number 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36
The greatest number of identical tables possible are 1, 2, 3, 4, 6, 12.

Numerical Expressions and Factors Getting Ready for Chapter 1

Chapter Exploration

Work with a partner. In Exercises 1–3, use the table.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 3$

Cross out the multiples of 2 that are greater than 2. Do the same for 3, 5, and 7.
The numbers that are not crossed out are called prime numbers. The numbers that are crossed out are called composite numbers. In your own words, describe the characteristics of prime numbers and composite numbers.
MODELING REAL LIFE Work with a partner. Cicadas are insects that live underground and emerge from the ground after x or x + 4 years. Is it possible that both x and x +4 are prime? Give some examples.

Answer:
$Big-Ideas-Math-Answer-Key-Grade-6-Chapter-1-Numerical-Numerical-Expressions-and-Factors-3$
The numbers that are not crossed are 2, 3, 5, 7, 9, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59,61, 67, 71, 73, 79, 83, 89, 97.
These are not multiples of any numbers. So, the above numbers are the prime numbers.

Vocabulary

The following vocabulary terms are deﬁned in this chapter. Think about what each term might mean and record your thoughts.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 4$

Answer:
i. First, we solve any operations inside of parentheses or brackets. Second, we solve any exponents. Third, we solve all multiplication and division from left to right. Fourth, we solve all addition and subtraction from left to right.
ii. “Factors” are numbers we can multiply together to get another number. When we find the factors of two or more numbers, and then find some factors are the same, then they are the “common factors”.
iii. A common multiple is a whole number that is a shared multiple of each set of numbers. The multiples that are common to two or more numbers are called the common multiples of those numbers. The smallest positive number is a multiple of two or more numbers.

Lesson 1.1 Powers and Exponents

Exploration 1
Writing Expressions Using Exponents
Work with a partner. Copy and complete the table.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 5$

Answer:
$Big-Ideas-Math-Answer-Key-Grade-6-Chapter-1-Numerical-Numerical-Expressions-and-Factors-5$
i. In your own words, describe what the two numbers in the expression 3⁵ mean.

Answer: 3⁵ means the number 3 repeats 5 times.
3 × 3 × 3 × 3 × 3 = 243

EXPLORATION 2
Using a Calculator to Find a Pattern

Work with a partner. Copy the diagram. Use a calculator to ﬁnd each value. Write one digit of the value in each box. Describe the pattern in the digits of the values.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 6$

Answer:
$Big-Ideas-Math-Answer-Key-Grade-6-Chapter-1-Numerical-Numerical-Expressions-and-Factors-6$

1.1 Lesson

A power is a product of repeated factors. The base of a power is the repeated factor. The exponent of a power indicates the number of times the base is used as a factor.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 7$
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 8$

Try It
Write the product as a power.
Question 1.
2 × 2 × 2

Answer: 2³ = 8
Two cubed or three to the two. Here 2 is repeated three times.

Question 2.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 13$

Answer: 46656 = 6⁶

Six to the six. Here 6 is repeated six times.

Question 3.
15 × 15 × 15 × 15

Answer: 56025 = 15⁴

15 to the power 4. Here 15 is repeated four times.

Question 4.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 14$

Answer: 1280000000 = 20⁷

20 to the power 7. Here 20 is repeated seven times.

Try It
Find the value of the power.

Question 5.
6³

Answer: 6 × 6 × 6 = 216
The value of the power 6³is 216.

Question 6.
9²

Answer: 9 × 9 = 81
The value of the power 9² is 81

Question 7.
3⁴

Answer: 3 × 3 × 3 × 3
The value of the power 3⁴is 81.

Question 8.
18²

Answer: 18 × 18
The value of the power 18² is 324.

Try It
Determine whether the number is a perfect square.

Question 9.
25

Answer: 5²
Yes, 25 is the perfect square.
A perfect square is a number, from a given number system, that can be expressed as the square of a number from the same number system.

Question 10.
2

Answer: 2 is not a perfect square. 2 cannot be expressed as the square of a number from the same number system.

Question 11.
99

Answer: 99 is not a perfect square. 99 cannot be expressed as the square of a number from the same number system.

Question 12.
36

Answer: 6²
36 is a perfect square
A perfect square is a number, from a given number system, that can be expressed as the square of a number from the same number system.

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.
FINDING VALUES OF POWERS
Find the value of the power.

Question 13.
8²

Answer: 8 × 8 = 64
The value of the power 8²is 64.

Question 14.
3⁵

Answer: 3 × 3 × 3 × 3 × 3 = 243
The value of the power 3⁵is 243.

Question 15.
11³

Answer: 11 × 11 × 11 = 1331
The value of the power 11³is 1331.

Question 16.
VOCABULARY
How are exponents and powers different?

Answer:
An expression that represents repeated multiplication of the same factor is called a power. The number 5 is called the base, and the number 2 is called the exponent. The exponent corresponds to the number of times the base is used as a factor.

Question 17.
VOCABULARY
Is 10 a perfect square? Is 100 a perfect square? Explain.

Answer: 10 is not a perfect square.
A perfect square is a number that is generated by multiplying two equal integers by each other.
100 is a perfect square. Because 10 × 10 = 100.

Question 18.
WHICH ONE DOESN’T BELONG?
Which one does not belong with the other three? Explain your reasoning.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 15$

Answer:
2⁴ = 2 × 2 × 2 × 2 = 16
3² = 3 × 3 = 9
3 + 3 + 3 + 3 = 3 × 4
5.5.5 = 125
The 3rd option does not belong to the other three expressions.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 19.
A square solar panel has an area of 16 square feet. Write the area as a power. Then ﬁnd the side lengths of the panel.

Answer: 4 feet

Explanation:
Given that,
A square solar panel has an area of 16 square feet.
A = s × s
16 = s²
4² = s²
s = 4
Thus the side length of the panel is 4 feet.

Question 20.
The four-square court shown is a square made up of four identical smaller squares. What is the area of the court?

$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 16$

Answer:
Given,
The four-square court shown is a square made up of four identical smaller squares.
The side of each square is 6 feet.
6 + 6 = 12 feet
The area of the court is 12 ft × 12 ft
A = 144 square feet
Thus the area of the court is 144 square feet.

Question 21.
DIG DEEPER!
Each face of a number cube is a square with a side length of 16 millimeters. What is the total area of all of the faces of the number cube?

Answer:
Given that,
Each face of a number cube is a square with a side length of 16 millimeters.
Area of the cube = 6 a²
A = 6 × 16 × 16
A = 1536 sq. mm

Powers and Exponents Practice 1.1

Review & Refresh

Multiply.

Question 1.
150 × 2

Answer: 300

Explanation:
Multiply the two numbers 150 and 2.
First multiply 2 with ones place 2 × 0 = 0
Next multiply with tens place 2 × 50 = 100
Next multiply with hundreds place 2 × 100 = 200
200 + 100 = 300

Question 2.
175 × 8

Answer: 1400

Explanation:
Multiply the two numbers 175 and 8.
First, multiply 2 with ones place 8 × 5 = 40
Next multiply with tens place 8 × 70= 560
Next multiply with hundreds place 8 × 100 = 800
800 + 560 + 40 = 1400

Question 3.
123 × 3

Answer: 369

Explanation:
Multiply the two numbers 123 and 3.
First multiply 2 with ones place 3 × 3 = 9
Next multiply with tens place 3 × 20 = 60
Next multiply with hundreds place 3 ×100 = 300
300 + 60 + 9 = 369

Question 4.
151 × 9

Answer: 1359

Explanation:
Multiply the two numbers 151 and 9.
First multiply 2 with ones place 9 × 1 = 9
Next multiply with tens place 9 × 50 = 450
Next multiply with hundreds place 9 × 100 = 900
900 + 450 + 9 = 1359

Write the sentence as a numerical expression.

Question 5.
Add 5 and 8, then multiply by 4.

Answer: The numerical expression for the above sentence is 5 + 8 × 4

Question 6.
Subtract 7 from 11, then divide by 2.

Answer: The numerical expression for the above sentence is 11 – 7 ÷ 2
Round the number to the indicated place value.

Question 7.
4.03785 to the tenths

Answer: The number 4.03785 nearest to the tenths is 4.0

Question 8.
12.89503 to the hundredths

Answer: The number 12.89503 nearest to the hundredths is 12.90

Complete the sentence.

Question 9.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 17$

Answer: 3

Explanation:
(1/10) × 30 = 30/10 = 3
The product of 1/10 and 30 is 3.

Question 10.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 18$

Answer: 20

Explanation:
(4/5) × 25 = 4 × 5 = 20
The product of 4/5 and 25 is 20.

Concepts, Skills, & Problem Solving

WRITING EXPRESSIONS USING EXPONENTS
Copy and complete the table. (See Exploration 1, p. 3.)
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 19$

Answer:
$Big-Ideas-Math-Answer-Key-Grade-6-Chapter-1-Numerical-Numerical-Expressions-and-Factors-19$
WRITING EXPRESSIONS AS POWERS
Write the product as a power.

Question 15.
9 × 9

Answer: The exponential form of the given expression is 9²

Question 16.
13 × 13

Answer: The exponential form of the given expression is 13²

Question 17.
15 × 15 × 15

Answer: The exponential form of the given expression is 15³

Question 18.
2.2.2.2.2

Answer: The exponential form of the given expression is 2⁵

Question 19.
14 × 14 × 14

Answer: The exponential form of the given expression is 14³

Question 20.
8.8.8.8

Answer: The exponential form of the given expression is 8⁴

Question 21.
11 × 11 × 11 × 11 × 11

Answer: The exponential form of the given expression is 11⁵

Question 22.
7.7.7.7.7.7

Answer: The exponential form of the given expression is 7⁶

Question 23.
16.16.16.16

Answer: The exponential form of the given expression is 16⁴

Question 24.
43 × 43 × 43 × 43 × 43

Answer: The exponential form of the given expression is 43⁵

Question 25.
167 × 167 × 167

Answer: The exponential form of the given expression is 167³

Question 26.
245.245.245.245

Answer: The exponential form of the given expression is 245⁴

FINDING VALUES OF POWERS
Find the value of the power.

Question 27.
5²

Answer: The value of the powers 5²is 5 × 5 = 25

Question 28.
4³

Answer: The value of the powers 4³ is 4 × 4 × 4 = 64

Question 29.
6²

Answer: The value of the powers 6² is 6 × 6 = 36

Question 30.
1⁷

Answer: The value of the powers 1⁷ is 1 × 1 × 1 × 1 × 1 × 1 × 1 = 1

Question 31.
0³

Answer: The value of the powers 0³ is 0 × 0 × 0 = 0

Question 32.
8⁴

Answer: The value of the powers 8⁴ is 8 × 8 × 8 × 8 = 4096

Question 33.
2⁴

Answer: The value of the powers 2⁴ is 2 × 2 × 2 × 2 = 64

Question 34.
12²

Answer: The value of the powers 12² is 12 × 12 = 144

Question 35.
7³

Answer: The value of the powers 7³is 7 × 7 × 7 = 343

Question 36.
5⁴

Answer: The value of the powers 5⁴is 5 × 5 × 5 × 5 = 625

Question 37.
2⁵

Answer: The value of the powers 2⁵ is 2 × 2 × 2 × 2 × 2 = 32

Question 38.
14²

Answer: The value of the powers 14² is 14 × 14 = 196

USING TOOLS
Use a calculator to ﬁnd the value of the power.

Question 39.
7⁶

Answer: 7 × 7 × 7 × 7 × 7 × 7 = 117649

Question 40.
4⁸

Answer: 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 = 256

Question 41.
12⁴

Answer: 12 × 12 × 12 × 12 = 20736

Question 42.
17⁵

Answer: 17 × 17 × 17 × 17 × 17 = 1419857

Question 43.
YOU BE THE TEACHER
Your friend ﬁnds the value of 8³. Is your friend correct? Explain your reasoning.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 20$

Answer: Your friend is incorrect
8³ is nothing but 8 repeats 3 times.
8³ = 8 × 8 × 8 = 512

IDENTIFYING PERFECT SQUARES
Determine whether the number is a perfect square.

Question 44.
8

Answer: 8 is not the perfect square. 8 cannot be expressed as the square of a number from the same number system.

Question 45.
4

Answer: 4 is a perfect square.
A perfect square is a number, from a given number system, that can be expressed as the square of a number from the same number system.

Question 46.
81

Answer: 81 perfect square
A perfect square is a number, from a given number system, that can be expressed as the square of a number from the same number system.

Question 47.
44

Answer: 44 is not the perfect square. 44 cannot be expressed as the square of a number from the same number system

Question 48.
49

Answer: 49 is a perfect square
A perfect square is a number, from a given number system, that can be expressed as the square of a number from the same number system.

Question 49.
125

Answer: 125 is not the perfect square. 125 cannot be expressed as the square of a number from the same number system

Question 50.
150

Answer: 150 is not the perfect square. 150 cannot be expressed as the square of a number from the same number system

Question 51.
144

Answer: 144 is the perfect square
A perfect square is a number, from a given number system, that can be expressed as the square of a number from the same number system.

Question 52.
MODELING REAL LIFE
On each square centimeter of a person’s skin, there are about 39² bacteria. How many bacteria does this expression represent?

Answer:
Given,
On each square centimeter of a person’s skin, there are about 39² bacteria.
39² = 39 × 39 = 1521 centimeters
Thus the bacteria represents 1521 centimeters.

Question 53.
REPEATED REASONING
The smallest ﬁgurine in a gift shop is 2 inches tall. The height of each ﬁgurine is twice the height of the previous ﬁgurine. What is the height of the tallest ﬁgurine?

$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 21$

Answer:
Given that,
The smallest ﬁgurine in a gift shop is 2 inches tall. The height of each ﬁgurine is twice the height of the previous ﬁgurine.
The second ﬁgurine is twice that of the first ﬁgurine = 2 × 2 = 4 inches
The third ﬁgurine is twice that of the second ﬁgurine = 4 × 4 = 16 inches
The fourth ﬁgurine is twice that of the third ﬁgurine = 16 × 16 = 256 inches
Thus the height of the tallest ﬁgurine is 256 inches.

Question 54.
MODELING REAL LIFE
A square painting measures 2 meters on each side. What is the area of the painting in square centimeters?

Answer:
Given that,
A square painting measures 2 meters on each side.
Area of the square = s × s
A = 2 m × 2 m = 4 sq. meters
Thus the area of the painting in square centimeters is 4.

Question 55.
NUMBER SENSE
Write three powers that have values greater than 120 and less than 130.

Answer:
11² = 11(11) = 121; this is between 120 and 130.
5³ = 5(5)(5) = 25(5) = 125; this is between 120 and 130.
2⁷ = 2(2)(2)(2)(2)(2)(2) = 4(2)(2)(2)(2)(2) = 8(2)(2)(2)(2) = 16(2)(2)(2) = 32(2)(2) = 64(2) = 128; this is between 120 and 130.

Question 56.
DIG DEEPER!
A landscaper has 125 tiles to build a square patio. The patio must have an area of at least 80 square feet.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 22$
a. What are the possible arrangements for the patio?

Answer:
Given that a square patio of at least 80 square feet is to be built from 125 tiles of length 12 inches or 1 foot.
Since there are 125 tiles and the patio has a shape of a square of at least 80 square feet, then the possible dimensions of the patio are
9 ft × 9 ft = 81 ft
10 ft × 10 ft = 100 ft, and
11 ft × 11 ft = 121 ft.

b. How many tiles are not used in each arrangement?

Answer:
For a patio of dimensions, 9ft by 9ft, the number of tiles that will not be used is given by 125 – 81 = 44
For a patio of dimensions, 10ft by 10ft, the number of tiles that will not be used is given by 125 – 100 = 25
For a patio of dimensions, 11ft by 11ft, the number of tiles that will not be used is given by 125 – 121 = 4

Question 57.
PATTERNS
Copy and complete the table. Describe what happens to the value of the power as the exponent decreases. Use this pattern to ﬁnd the value of 4⁰.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 23$

Answer:
$Big-Ideas-Math-Answer-Key-Grade-6-Chapter-1-Numerical-Numerical-Expressions-and-Factors-23$
4⁰ = 1
Thus the value of 4⁰ is 1.

Question 58.
REPEATED REASONING
How many blocks do you need to add to Square 6 to get Square 7? to Square 9 to get Square 10? to Square 19 to get Square 20? Explain.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 24$

Answer:
You need to add 14 blocks to get square 7. The square 7 contains 7 × 7 = 49 blocks
You need to add 32 blocks to get square 9. The square 9 contains 9 × 9 = 81 blocks
You need to add 19 blocks to get square 10. The square 10 contains 10 × 10 = 100 blocks
You need to add 261 blocks to get square 19. The square 19 contains 19 × 19 = 361 blocks
You need to add 39 blocks to get square 20. The square 20 contains 20 × 20 = 400 blocks

Lesson 1.2 Order of Operations

Order of Operations

EXPLORATION 1
Comparing Different Orders

Work with a partner. Find the value of each expression by using different orders of operations. Are your answers the same?

$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 25$

Answer:
The answers for all the expressions are not the same. The values of each expression will change if you change the order of operations.
a. 3 + 2 × 2
5 × 2 = 10
Multiply, then add
3 + 2 × 2
3 + 4 = 7
b. Subtract then multiply
18 – 3 × 3
15 × 3 = 45
Multiply, then subtract
18 – 3 × 3
18 – 9 = 9
c. Multiply, then subtract
8 × 8 – 2
64 – 2 = 62
Subtract, then Multiply
8 × 8 – 2
8 × 6 = 48
d. Multiply, then add
6 × 6 + 2
36 + 2 = 38
Add, then multiply
6 × 6 + 2
6 × 8 = 48

EXPLORATION 2
Determining Order of Operations
Work with a partner.
a. Scientiﬁc calculators use a standard order of operations when evaluating expressions. Why is a standard order of operations needed?

Answer: The order of operations is a rule that tells you the right order in which to solve different parts of a math problem. The order of operations is important because it guarantees that people can all read and solve a problem in the same way.

b. Use a scientiﬁc calculator to evaluate each expression in Exploration 1. Enter each expression exactly as written. For each expression, which order of operations is correct?

Answer:
a. 3 + 2 × 2 – Multiply, then add
b. 18 – 3 × 3 – Multiply, then subtract
c. 8 × 8 – 2 – Multiply, then subtract
d. 6 × 6 + 2 – Multiply, then add

c. What order of operations should be used to evaluate 3 + 2², 18 − 3², 8² − 2, and 6² + 2?
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 26$

Answer:
Solve the expressions by using the calculator.
a. 3 + 2 × 2
3 + 4 = 7
b. 18 – 3 × 3
18 – 9 = 9
c. 8 × 8 – 2
64 – 2 = 62
d. 6 × 6 + 2
36 + 2 = 38
d. Do 18 ÷ 3.3 and 18 ÷ 3² have the same value? Justify your answer.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 27$

Answer: No

Explanation:
18 ÷ 3.3
(18 ÷ 3) × 3
6 × 3 = 18
18 ÷ 3² = 2
By using the calculator you can find the difference.
e. How does evaluating powers ﬁt into the order of operations?

Answer:
When an expression has parentheses and powers, evaluate it in the following order: contents of parentheses, powers from left to right, multiplication and division from left to right, and addition and subtraction from left to right.

1.2 Lesson

A numerical expression is an expression that contains numbers and operations. To evaluate, or ﬁnd the value of, a numerical expression, use a set of rules called the order of operations.
Key Idea
order of operations

Perform operations in grouping symbols.
Evaluate numbers with exponents.
Multiply and divide from left to right.
Add and subtract from left to right.

Try It
a. Evaluate the expression.

Question 1.
7.5 + 3

Answer: 56

Explanation:
You have to evaluate the expression from left to right.
7(5 + 3) = 7 × 8
= 56

Question 2.
(28 – 20) ÷ 4

Answer: 2

Explanation:
You have to evaluate the expression from left to right.
28 – 20 = 8
8 ÷ 4 = 2

Question 3.
[6 + (15 – 10)] × 5

Answer: 55

Explanation:
You have to evaluate the expression from left to right.
[6 + (15 – 10)] × 5
[6 + 5] × 5
11 × 5 = 55

Try It
Evaluate the expression.

Question 4.
6 + 2⁴ – 1

Answer: 21

Explanation:
You have to evaluate the expression from left to right.
6 + 2⁴ – 1
6 + (16 – 1)
6 + 15 = 21
6 + 2⁴ – 1 = 21

Question 5.
4.3² + 18 – 9

Answer: 45

Explanation:
You have to evaluate the expression from left to right.
4.3² + (18 – 9)
4.3² + 9
4 × 9 + 9
36 + 9 = 45

Question 6.
16 + (5² – 7) ÷ 3

Answer:

Explanation:
You have to evaluate the expression from left to right.
16 + (5² – 7) ÷ 3
16 + (25 – 7) ÷ 3
16 + (18) ÷ 3
16 + (18 ÷ 3)
16 + 6 = 22

Thee symbols × and . are used to indicate multiplication. You can also use parentheses to indicate multiplication. For example, 3(2 +7) is the same as 3 × (2 + 7).

Try It
Evaluate the expression.

Question 7.
50 + 6(12 ÷ 4) – 8²

Answer: 4

Explanation:
You have to evaluate the expression from left to right.
50 + 6(12 ÷ 4) – 8²
50 + 6(3) – 8²
50 + 18 – 8²
50 + 18 – 64
68 – 64
4

Question 8.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 28$

Answer: 24

Explanation:
You have to evaluate the expression from left to right.
5² – 1/5 (10 – 5)
5² – 1/5 (5)
5² – 1
25 – 1
24

Question 9.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 29$

Answer:

Explanation:
You have to evaluate the expression from left to right.
8(2+5) = 8 × 7
(8 × 7)/7 = 8

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

USING ORDER OF OPERATIONS
Evaluate the expression.

Question 10.
7 + 2.4

Answer: 15

Explanation:
You have to evaluate the expression from left to right.
7 + 2 × 4 = 7 + 8 = 15

Question 11.
8 ÷ 4 × 2

Answer: 4

Explanation:
You have to evaluate the expression from left to right.
8 ÷ 4 = 2
2 × 2 = 4

Question 12.
3(5 + 1) ÷ 3²

Answer: 2

Explanation:
You have to evaluate the expression from left to right.
3(5 + 1) ÷ 3²
3 × 6 ÷ 3²
18 ÷ 9 =2

Question 13.
WRITING
Why does 12 − 8 ÷ 2 = 2?

Answer:
12 − 8 ÷ 2
4 ÷ 2 = 2

Question 14.
REASONING
Describe the steps in evaluating the expression 8 ÷ (6 − 4) + 3².

Answer:
8 ÷ (6 − 4) + 3²
8 ÷ 2 + 3²
4 + 9 = 13

Question 15.
WHICH ONE DOESN’T BELONG?
Which expression does not belong with the other three? Explain your reasoning.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 30$

Answer: (5² – 8) × 2 does not belong to the other three. Because the order of operations and expressions are different for the fourth option.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 16.
A square plot of land has side lengths of 40 meters. An archaeologist divides the land into 64 equal parts. What is the area of each part?

Answer:
Given that,
A square plot of land has side lengths of 40 meters.
An archaeologist divides the land into 64 equal parts.
Side of the square field = 40m
Area of the square field = s × s
A = 40m × 40m
A = 1600 sq.m
Area of each part of the square = 1600/64 = 25 sq.m

Question 17.
A glass block window is made of two different-sized glass squares. The window has side lengths of 40 inches. The large glass squares have side lengths of 10 inches. Find the total area of the small glass squares.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 31$

Answer:
Given,
A glass block window is made of two different-sized glass squares.
The window has side lengths of 40 inches. The large glass squares have side lengths of 10 inches.
40 × 10 = 400

Question 18.
DIG DEEPER!
A square vegetable garden has side lengths of 12 feet. You plant ﬂowers in the center portion as shown. You divide the remaining space into 4 equal sections and plant tomatoes, onions, zucchini, and peppers. What is the area of the onion section?
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 32$

Answer:
A square vegetable garden has side lengths of 12 feet.
You plant flowers in the center portion of the garden, a square that has side lengths of 4 feet.
You divide the remaining space into 4 equal sections and plant tomatoes, onions, zucchini, and peppers.
Given that,
→ side of the square vegetable garden is = 12 feet.
So,
→ Area of square vegetable garden = (side)² = (12)² = 144 feet².
Now, given that, inside this area, there is a square of side 4 feet reserved for flowers.
So,
→ The area of the flower section = (side)² = (4)² = 16 feet².
Therefore,
→ The rest of the garden that is intended for vegetables is = The total garden area – The flower section area = 144 – 16 = 128 feet².
Now, this remaining area is to be divided into four equal sections.
So,
→ The area of the onion section = (1/4) of remaining area = (1/4) × 128 = 32 feet².

Order of Operations Practice 1.2

Review & Refresh

Write the product as a power.

Question 1.
11 × 11 × 11 × 11

Answer: The exponent for the product 11 × 11 × 11 × 11 is 11⁴

Question 2.
13 × 13 × 13 × 13 × 13

Answer: The exponent for the product 13 × 13 × 13 × 13 × 13 is 13⁵
Find the missing dimension of the rectangular prism.

Question 3.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 33$

Answer:
Given that,
l = 6 in.
b = 4 in.
h = ?
v = 192 cu. in
Volume of the rectangular prism = lbh
192 = 6 × 4 × h
h = 192/24
h = 8
Thus the height of the rectangular prism is 8 inches.

Question 4.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 34$

Answer:
Given that,
h = 9m
b = 3m
v = 135 cu. m
l = ?
The volume of the rectangular prism = lbh
135 = l × 3 × 9
135 = l × 27
l = 5m

Tell whether the number is prime or composite.

Question 5.
9

Answer: Composite Number
A natural number greater than 1 that is not prime is called a composite number.

Question 6.
11

Answer: Prime Number
A prime number is a natural number greater than 1 that is not a product of two smaller natural numbers.

Question 7.
23

Answer: Prime Number
A prime number is a natural number greater than 1 that is not a product of two smaller natural numbers.

Concepts, Skills, & Problem Solving

COMPARING DIFFERENT ORDERS
Find the value of the expression by using different orders of operations. Are your answers the same? (See Exploration 1, p. 9.)

Question 8.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 35$

Answer:
4 + 6 × 6
10 × 6 = 60
4 + 6 × 6
4 + 36 = 40

Question 9.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 36$

Answer:
5 × 5 – 3
5 × 2 = 10
5 × 5 – 3
25 – 3 = 22

USING ORDER OF OPERATIONS
Evaluate the expression.

Question 10.
5 + 18 ÷ 6

Answer: 8

Explanation:
First, divide then divide.
5 + 3 = 8

Question 11.
(11 – 3) ÷ 2 + 1

Answer:

Explanation:
First, subtract and divide.
(11 – 3) ÷ 2 + 1
8 ÷ 2 + 1
4 + 1 = 5

Question 12.
45 ÷ 9 × 2

Answer: 10

Explanation:
The first divide then multiply.
45 ÷ 9 × 2
5 × 2 = 10

Question 13.
6² – 3.4

Answer: 24

Explanation:
Multiply then subtract
6² – 3.4
36 – 12
24

Question 14.
42 ÷ (15 – 2³)

Answer: 6

Explanation:
Subtract then divide.
42 ÷ (15 – 8)
42 ÷ 7
6

Question 15.
4².2 + 8.7

Answer: 88

Explanation:
Multiply then add
4².2 + 8.7
16 × 2 + 56
32 + 56 = 88

Question 16.
(5² – 2) × 1⁵ + 4

Answer: 27

Explanation:
(5² – 2) × 1⁵ + 4
(25 – 2) × 1 + 4
Add, subtract then multiply
23 + 4 = 27

Question 17.
4 + 2 × 3² – 9

Answer: 13

Explanation:
4 + 2 × 3² – 9
4 + 18 – 9
4 + 9 = 13

Question 18.
8 ÷ 2 × 3 + 4² ÷ 4

Answer: 16

Explanation:
8 ÷ 2 × 3 + 4² ÷ 4
(4 × 3) + (16 ÷ 4)
12 + 4
16

Question 19.
3² + 12 ÷ (6 – 3) × 8

Answer: 41

Explanation:
3² + 12 ÷ (6 – 3) × 8
9 + (12 ÷ (6 – 3)) × 8
9 + (12 ÷ 3) × 8
9 + 4 × 8
9 + 32
41

Question 20.
(10 + 4) ÷ (26 – 19)

Answer: 2

Explanation:
Add, subtract then divide
(10 + 4) ÷ (26 – 19)
14 ÷ 7
2

Question 21.
(5² – 4).2 – 18

Answer: 24

Explanation:
((5² – 4).2) – 18
((25 – 4) × 2) – 18
(21 × 2) – 18
42 – 18
24

Question 22.
2 × [(16 – 8) × 2]

Answer: 32

Explanation:
2 × [(16 – 8) × 2]
2 × [8 × 2]
2 × 16
32

Question 23.
12 + 8 × 3³ – 24

Answer: 204

Explanation:
12 + 8 × 3³ – 24
12 + (8 × 27) – 24
12 + 216 – 24
12 + 192 = 204

Question 24.
6² ÷ [(2 + 4) × 2³]

Answer: 48

Explanation:
6² ÷ [(2 + 4) × 2³]
36 ÷ [(2 + 4) × 2³]
36 ÷ 6 × 8
6 × 8
48

YOU BE THE TEACHER
Your friend evaluates the expression. Is your friend correct? Explain your reasoning.

Question 25.
$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 37$

Answer: Your friend is incorrect.
9 + 3 × 3²
9 + (27)
36

Question 26.
$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 38$

Answer:
19 – 6 + 12
13 + 12
25

Question 27.
PROBLEM SOLVING
You need to read 20 poems in 5 days for an English project. Each poem is 2 pages long. Evaluate the expression 20 × 2 ÷ 5 to ﬁnd how many pages you need to read each day.

Answer:
Given,
You need to read 20 poems in 5 days for an English project. Each poem is 2 pages long.
20 × 2 ÷ 5
40 ÷ 5 = 8
Thus you need to read 8 pages each day.

USING ORDER OF OPERATIONS
Evaluate the expression.

Question 28.
12 – 2(7 – 4)

Answer:
12 -(2 × (7 – 4))
12 – (2 × 3)
12 – 6 = 6

Question 29.
4(3 + 5) – 3(6 -2)

Answer:
4(3 + 5) – 3(6 -2)
4 × 8 – 3 × 4
32 – 12
20

Question 30.
$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 39$

Answer:
6 + 1/4 (12 -8)
6 + 1/4(4)
6 + 1
7

Question 31.
9² – 8(6 + 2)

Answer:
81 – (8(6 + 2))
81 – (8 × 8)
81 – 64
17

Question 32.
4(3 – 1)³ + 7(6) – 5²

Answer:
4(3 – 1)³ + 7(6) – 5²
4(2)³ + 7(6) – 5²
4 × 8 + 42 – 25
32 + 42 – 25 = 49

Question 33.
$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 40$

Answer:
8[(1 1/6 + 5/6) ÷ 4]
[8[7/6 + 5/6] ÷ 4]
8[12/6] ÷ 4
8[2 ÷ 4]
8(1/2)
4

Question 34.
$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 41$

Answer:
49 – 2((11-3)/8)
49 – 2 (8/8)
49 – 2
47

Question 35.
8(7.3 + 3.7 – 8) ÷ 2

Answer:
8(7.3 + 3.7 – 8) ÷ 2
(8(7.3 + 3.7 – 8)) ÷ 2
8 (11 – 8) ÷ 2
8 × 3 ÷ 2
24 ÷ 2
12

Question 36.
2⁴(5.2 – 3.2) ÷ 4

Answer:
2⁴(5.2 – 3.2) ÷ 4
16 (5.2 – 3.2) ÷ 4
16 (2) ÷ 4
32 ÷ 4
8

Question 37.
$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 42$

Answer:
36(3+5)/4
36 × 8/4
36 × 2
72

Question 38.
$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 43$

Answer:
(144 – 24 + 1)/121
121/121
1

Question 39.
$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 44$

Answer:
26 ÷ 2 + 5 = 18
18/6 = 3

Question 40.
PROBLEM SOLVING
Before a show, there are 8 people in a theater. Five groups of 4 people enter, and then three groups of 2 people leave. Evaluate the expression 8 + 5(4) − 3(2) to ﬁnd how many people are in the theater.
$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 45$

Answer:
Given,
Before a show, there are 8 people in a theater. Five groups of 4 people enter, and then three groups of 2 people leave.
8 + (5 × 4) – (3 × 2)
8 + 20 – 6
28 – 6
22

Question 41.
MODELING REAL LIFE
The front door of a house is painted white and blue. Each window is a square with a side length of 7 inches. What is the area of the door that is painted blue?

Answer:
Given,
The front door of a house is painted white and blue. Each window is a square with a side length of 7 inches.
Area of the square = s × s
A = 7 in × 7 in
A = 49 sq. inches
Therefore the area of the door that is painted blue is 49 sq. inches.

Question 42.
PROBLEM SOLVING
You buy 6 notebooks, 10 folders, 1 pack of pencils, and 1 lunch box for school. After using a $10 gift card, how much do you owe? Explain how you solved the problem.
$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 46$

Answer:
Given,
You buy 6 notebooks, 10 folders, 1 pack of pencils, and 1 lunch box for school.
Cost of 1 notebook = $2
6 notebooks = 6 × $2 = $12
Cost of 1 folder = $1
10 folders = 10 × $1 = $10
Cost of 1 pack of pencils = $3
Cost of 1 lunch box = $8
So the total cost is $11 + $10 + $3 + $8 = $31
You used $10 gift card.
$31 – $10 = $21
Thus you ow $21.

Question 43.
OPEN-ENDED
Use all four operations and at least one exponent to write an expression that has a value of 100.

Answer: You need to use +, -, ×, ÷ operations to write the expressions that have the value of 100.
(34 – 1) × 3 + 3² ÷ 9 = 100

Question 44.

$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 47$

REPEATED REASONING
A Petri dish contains 35 cells. Every day, each cell in the Petri dish divides into 2 cells in a process called mitosis. How many cells are there after 14 days? Justify your answer.

Answer:
Given,
A Petri dish contains 35 cells. Every day, each cell in the Petri dish divides into 2 cells in a process called mitosis
35 ÷ 2 = 17.5
1 day = 0.5 + 0.5 cells
14 days = 14 × 1 = 14 cells
17.5 – 14 = 3.5
Thus there are 3.5 cells after 14 days.

Question 45.
REASONING
Two groups collect litter along the side of a road. It takes each group 5 minutes to clean up a 200-yard section. How long does it take both groups working together to clean up 2 miles? Explain how you solved the problem.

Answer:
Given,
Two groups collect litter along the side of a road. It takes each group 5 minutes to clean up a 200-yard section.
To convert 2 miles to yards, you have to multiply 2 by 1760, because 1 mile equals to 1760 yards:
2 × 1760 = 3520 yards.
If you would like to know how long does it take to clean up 2 miles, you can calculate this using the following steps:
5 × 3520 = 200 × x
17600 = 200 × x /200
x = 17600 / 200
x = 88 minutes

Question 46.
NUMBER SENSE
Copy each statement. Insert +, −, ×, or ÷ symbols to make each statement true.
$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 48$

Answer:
You can find the value by using the calculator by inserting the suitable operations.
$Big-Ideas-Math-Answers-6th-Grade-Chapter-1-Numerical-Numerical-Expressions-and-Factors-48$

Lesson 1.3 Prime Factorization

Prime Factorization

EXPLORATION 1
Rewriting Numbers as Products of Factors

Work with a partner. Two students use factor trees to write 108 as a product of factors, as shown below.

$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 49$
a. Without using 1 as a factor, can you write 108 as a product with more factors than each student used? Justify your answer. Math Practice
$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 50$

Answer: Yes you can find the factors by using the prime factorization.
108 = 2 × 54
= 2 × 2 × 27
= 2 × 2 × 3 × 9
= 2 × 2 × 3 × 3 × 3

b. Use factor trees to write 80, 162, and 300 as products of as many factors as possible. Do not use 1 as a factor.
$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 50.1$

Answer:
80 = 2 × 40
= 2 × 2 × 20
= 2 × 2 × 2 × 10
= 2 × 2 × 2 × 2 × 5
162 = 2 × 81
= 2 × 3 × 27
= 2 × 3 × 3 × 9
= 2 × 3 × 3 × 3 × 3
300 = 2 × 150
= 2 × 2 × 75
= 2 × 2 × 3 × 25
= 2 × 2 × 3 × 5 × 5

c. Compare your results in parts (a) and (b) with other groups. For each number, identify the product with the greatest number of factors. What do these factors have in common?

Answer: 300 contains the greatest number of factors. (2, 2, 3, 5, 5)

1.3 Lesson

Because 2 is a factor of 10 and 2 . 5 =10, 5 is also a factor of 10. The pair 2, 5 is called a factor pair of 10.

Try It
List the factor pairs of the number.

Question 1.
18

Answer: The factor pairs of 18 are 1, 2, 3, 6, 9, 18

Explanation:
1 × 18 = 18
2 × 9 = 18
3 × 6 = 18
6 × 3 = 18
9 × 2 = 18
18 × 1 = 18

Question 2.
24

Answer: The factor pairs of 1, 2, 3, 4, 6, 8, 12, 24

Explanation:
1 × 24 = 24
2 × 12 = 24
3 × 8 = 24
4 × 6 = 24
6 × 4 = 24
8 × 3 = 24
12 × 2 = 24
24 × 1 = 24

Question 3.
51

Answer: The factor pairs of 1, 3, 17, 51

Explanation:
1 × 51 = 51
3 × 17 = 51
17 × 3 = 51
51 × 1 = 15

Question 4.
WHAT IF?
The woodwinds section of the marching band has 38 members. Which has more possible arrangements, the brass section or the woodwinds section? Explain.

Answer: Brass section. 38 has only two-factor pairs.
38 = 1 × 38
= 2 × 19

Key Idea

Prime Factorization
The prime factorization of a composite number is the number written as a product of its prime factors.
You can use factor pairs and a factor tree to help ﬁnd the prime factorization of a number. The factor tree is complete when only prime factors appear in the product. A factor tree for 60 is shown.
$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 51$
$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 52$

Try It
Write the prime factorization of the number.

Question 5.
20

Answer:
The Prime Factorization is:
2 x 2 x 5
In Exponential Form:
2² x 51
CSV Format:
2, 2, 5

Question 6.
88

Answer:
88 = 2 × 44
= 2 × 2 × 22
= 2 × 2 × 2 × 11
The Prime Factorization is: 2 × 2 × 2 × 11

Question 7.
90

Answer:
90 = 2 × 45
= 2 × 3 × 15
= 2 × 3 × 3 × 5
The Prime Factorization is: 2 × 3 × 3 × 5

Question 8.
462

Answer:
= 2 × 231
= 2 × 3 × 77
= 2 × 3 × 7 × 11
The Prime Factorization is: 2 × 3 × 7 × 11

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

WRITING A PRIME FACTORIZATION
Write the prime factorization of the number.

Question 9.
14

Answer:
14 = 2 × 7
The Prime Factorization is: 2 × 7

Question 10.
86

Answer:
86 = 2 × 43
The Prime Factorization is: 2 × 43

Question 11.
40

Answer:
40 = 2 × 20
= 2 × 2 × 10
= 2 × 2 × 2 × 5
The Prime Factorization is: 2 × 2 × 2 × 5

Question 12.
516

Answer:
516 = 2 × 258
= 2 × 2 × 129
= 2 × 2 × 3 × 43
The Prime Factorization is: 2 × 2 × 3 × 43

Question 13.
WRITING
Explain the difference between prime numbers and composite numbers.

Answer:
A prime number is a number that has exactly two factors i.e. ‘1’ and the number itself. A composite number has more than two factors, which means apart from getting divided by number 1 and itself, it can also be divided by at least one integer or number.

Question 14.
STRUCTURE
Your friend lists the following factor pairs and concludes that there are 6 factor pairs of 12. Explain why your friend is incorrect.
$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 53$

Answer: Your friend is incorrect. Because there are 5-factor pairs of 12.
The factor pairs of 12 are
1 × 12 =12
2 × 6 = 12
3 × 4 = 12
4 × 3 = 12
6 × 2 = 12

Question 15.
WHICH ONE DOESN’T BELONG?
Which factor pair does not belong with the other three? Explain your reasoning.
$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 54$

Answer:
2, 28 = 2 × 28 = 56
4, 14 = 4 × 14 = 56
6, 9 = 6 × 9 = 54
7, 8 = 7 × 56
By this we can say that 6, 9 does not belong to the other three expressions.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 16.
A group of 20 friends plays a card game. The game can be played with 2 or more teams of equal size. Each team must have atleast 2 members. List the possible numbers and sizes of teams.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 55$

Answer:
Given,
A group of 20 friends plays a card game. The game can be played with 2 or more teams of equal size. Each team must have at least 2 members.
20 = 1 × 20
2 × 10
4 × 5
5 × 4
10 × 2
Thus there are 5 possible numbers and size of teams.

Question 17.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 56$
You arrange 150 chairs in rows for a school play. You want each row to have the same number of chairs. How many possible arrangements are there? Are all of the possible arrangements appropriate for the play? Explain.

Answer:
You arrange 150 chairs in rows for a school play. You want each row to have the same number of chairs.
150 = 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150
Thus there are 12 possible arrangements appropriate for the play.

Question 18.
What is the least perfect square that is a factor of 4536? What is the greatest perfect square that is a factor of 4536?

Answer: least perfect square that is a factor of 4536

Explanation:
What is the last number of 4,536? It is this number: 4536. The answer is 6. Is 6 in the list of numbers that are never perfect squares (2, 3, 7 or 8)?
Answer: No, 6 is not in the list of numbers that are never perfect squares. Let’s continue to the next step.
Step 2:
We now need to obtain the digital root of the number. Here’s how you do it:
Split the number up and add each digit together:
4 + 5 + 3 + 6 = 18
If the answer is more than one digit, you would add each digit of the answer together again:
1 + 8 = 9
1 x 4,5362 x 2,2683 x 1,5124 x 1,1346 x 7567 x 6488 x 5679 x 50412 x 37814 x 32418 x 25221 x 21624 x 18927 x 16828 x 16236 x 12642 x 10854 x 8456 x 8163 x 72
We’re looking for a factor combination with equal numbers for X and Y (like 3×3) above. Notice there isn’t an equal factor combination, that when multiplied together, produce the number 4,536. That means 4,536 is NOT a perfect square.

Question 19.
DIG DEEPER!
The prime factorization of a number is 2⁴ × 3⁴ × 5⁴ × 7². Is the number a perfect square? Explain your reasoning.

Answer:
The prime factorization of a number is 2⁴ × 3⁴ × 5⁴ × 7².
16 × 81 × 625 × 49 = 39690000
Yes, 39690000 is a perfect square.

Prime Factorization Practice 1.3

Review & Refresh

Evaluate the expression.

Question 1.
2 + 4²(5 – 3)

Answer: 34

Explanation:
2 + 4²(5 – 3)
2 + 4²(2)
2 + 16(2) = 2 + 32 = 34

Question 2.
2³ + 4 × 3²

Answer: 44

Explanation:
2³ + 4 × 3²
8 + 4 × 9
8 + 36
44

Question 3.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 57$

Answer:

Explanation:
9 × 5 – 16(5/2 – 1/2)
9 × 5 – 16(4/2)
45 – 16(2)
45 – 32
13

Plot the points in a coordinate plane. Draw a line segment connecting the points.

Question 4.
(1, 1) and (4, 3)

Answer:
$Big Ideas Math Grade 6 Chapter 1 img_1$

Question 5.
(2, 3) and (5, 9)

Answer:
$Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_2$

Question 6.
(2, 5) and (4, 8)

Answer:
$Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_2$

Use the Distributive Property to ﬁnd the quotient. Justify your answer.

Question 7.
408 ÷ 4

Answer: 120
Write 408 as 204 and 204
204 ÷ 4 = 51
204 ÷ 4 = 51
51 + 51 = 102
408 ÷ 4 = 102

Question 8.
628 ÷ 2

Answer: 314
608 can be written as 314 and 314
314 ÷ 2 = 157
314 ÷ 2 = 157
157 + 157 = 314
628 ÷ 2 = 314

Question 9.
969 ÷ 3

Answer: 323
969 can be written as 900 and 69
900 ÷ 3 = 300
69 ÷ 3 = 23
300 + 23 = 323
969 ÷ 3 = 323

Classify the triangle in as many ways as possible.

Question 10.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 58$

Answer: Acute
Acute angles measure less than 90 degrees. Right angles measure 90 degrees. Obtuse angles measure more than 90 degrees.

Question 11.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 59$

Answer: Obtuse
An obtuse angle has a measurement greater than 90 degrees but less than 180 degrees. However, A reflex angle measures more than 180 degrees but less than 360 degrees.

Question 12.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 60$

Answer: Right angle
In geometry and trigonometry, a right angle is an angle of exactly 90° (degrees), corresponding to a quarter turn. If a ray is placed so that its endpoint is on a line and the adjacent angles are equal, then they are right angles.

Concepts, Skills, & Problem Solving

REWRITING A NUMBER
Write the number as a product of as many factors as possible. (See Exploration 1, p. 15.)

Question 13.
60

Answer: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

The factors of 60 are:
1 × 60 = 60
2 × 30 = 60
3 × 20 = 60
4 × 15 = 60
5 × 12 = 60
6 × 10 = 60
10 × 6 = 60
12 × 5 = 60
15 × 4 = 60
20 × 3 = 60
30 × 2 = 60
60 × 1 = 60

Question 14.
63

Answer: The factors of 63 are 1, 3, 7, 9, 21, 63

The factors of 63 are:
1 × 63 = 63
3 × 21 = 63
7 × 9 = 63
9 × 7 = 63
21 × 3 = 63
63 × 1 = 63

Question 15.
120

Answer: The factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

The factors of 120 are:
1 × 120 = 120
2 × 60 = 120
3 × 40 = 120
4 × 30 = 120
5 × 24 = 120
6 × 20 = 120
8 × 15 = 120
10 × 12 = 120
12 × 10 = 120
15 × 8 = 120
20 × 6 = 120
24 × 5 = 120
30 × 4 = 120
40 × 3 = 120
60 × 2 = 120
120 × 1 = 120

Question 16.
150

Answer: The factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150

The factors of 150 are:
1 × 150 = 150
2 × 75 = 150
3 × 50 = 150
5 × 30 = 150
6 × 25 = 150
10 × 15 = 150

FINDING FACTOR PAIRS
List the factor pairs of the number.

Question 17.
15

Answer: The factor pairs of 15 are (1, 15), (3, 5)
1 × 15 = 15
3 × 5 = 15
5 × 3 = 15
15 × 1 = 15

Question 18.
22

Answer: The factor pairs of 22 are (1, 22) (2, 11)
1 × 22 = 22
2 × 11 = 22
11 × 2 = 22
22 × 1 = 22

Question 19.
34

Answer: (1, 34) (2,17)

The factor pairs of 34 are
1 × 34 = 34
2 × 17 = 34
17 × 2 = 34
34 × 1 = 34

Question 20.
39

Answer: (1, 39) (3, 13)

The factor pairs of 39 are
1 × 39 = 39
3 × 13 = 39
13 × 3 = 39
39 × 1 = 39

Question 21.
45

Answer: (1, 45) (3, 15) (5, 9)

The factor pairs of 45 are
1 × 45 = 45
3 × 15 = 45
5 × 9 = 45

Question 22.
54

Answer: (1, 54) (2, 27) (3, 18) (6, 9)

The factor pairs of 54 are
1 × 54 = 54
2 × 27 = 54
3 × 18 = 54
6 × 9 = 54

Question 23.
59

Answer: (1, 59)
The factor pairs of 59 are
1 × 59 = 59
59 × 1 = 59

Question 24.
61

Answer: (1, 61)
The factor pairs of 61 are
1 × 61 = 61
61 × 1 = 61

Question 25.
100

Answer: (1, 100) (2, 50) (4, 25) (5, 20) (10, 10)
The factor pairs of 100 are
1 × 100 = 100
2 × 50 = 100
4 × 25 = 100
5 × 20 = 100
10 × 10 = 100

Question 26.
58

Answer: (1, 58) (2, 29)
The factor pairs of 58 are
1 × 58 = 58
2 × 29 = 58

Question 27.
25

Answer: (1, 25) (5, 5)
The factor pairs of 25 are
1 × 25 = 25
5 × 5 = 25

Question 28.
76

Answer: (1, 76) (2, 38) (4, 19)
The factor pairs of 76 are
1 × 76 = 76
2 × 38 = 76
4 × 19 = 76

Question 29.
52

Answer: (1, 52) (2, 26) (4, 13)
The factor pairs of 52 are
1 × 52 = 52
2 × 26 = 52
4 × 13 = 52

Question 30.
88

Answer: (1,88) (2,44) (4, 22) (8, 11)
The factor pairs of 88 are
1 × 88 = 88
2 × 44 = 88
4 × 22 = 88
8 × 11 = 88

Question 31.
71

Answer: (1,71)
The factor pairs of 71 are
1 × 71 = 71

Question 32.
91

Answer: (1, 91) (7, 13)
The factor pairs of 91 are
1 × 91 = 91
7 × 13 = 91

WRITING A PRIME FACTORIZATION
Write the prime factorization of the number.

Question 33.
16

Answer:
16 = 2 × 8
2 × 2 × 4
2 × 2 × 2 × 2

Question 34.
25

Answer:
25 = 5 × 5

Question 35.
30

Answer:
30 = 2 × 15
= 2 × 3 × 5

Question 36.
26

Answer:
26 = 2 × 13

Question 37.
84

Answer:
84 = 2 × 42
2 × 2 × 21
2 × 2 × 3 × 7

Question 38.
54

Answer:
54 = 2 × 27
2 × 3 × 9
2 × 3 × 3 × 3

Question 39.
65

Answer:
65 = 5 × 13

Question 40.
77

Answer:
77 = 7 × 11

Question 41.
46

Answer:
46 = 2 × 23

Question 42.
39

Answer:
39 = 3 × 13

Question 43.
99

Answer:
99 = 3 × 33
3 × 3 × 11

Question 44.
24

Answer:
24 = 2 × 12
2 × 2 × 6
2 × 2 × 2 × 3

Question 45.
315

Answer:
315 = 3 × 105
3 × 3 × 35
3 × 3 × 5 × 7

Question 46.
490

Answer:
490 = 2 × 245
2 × 5 × 49
2 × 5 × 7 × 7

Question 47.
140

Answer:
2 × 70
2 × 2 × 35
2 × 2 × 5 × 7

Question 48.
640

Answer:
640 = 2 × 320
2 × 2 × 160
2 × 2 × 2 × 80
2 × 2 × 2 × 2 × 40
2 × 2 × 2 × 2 × 2 × 20
2 × 2 × 2 × 2 × 2 × 2 × 10
2 × 2 × 2 × 2 × 2 × 2 × 2 × 5

USING A PRIME FACTORIZATION
Find the number represented by the prime factorization.

Question 49.
2².3².5

Answer:
4 × 9 × 5 = 180
We have to find the prime factorization for 180.
180 = 2 × 90
2 × 2 × 45
2 × 2× 3 × 15
2 × 2 × 3 × 3 × 5

Question 50.
3².5².7

Answer:
9 × 25 × 7 = 1575
We have to find the prime factorization for 1575.
1575 = 3 × 525
3 × 3 × 175
3 × 3 × 5 × 35
3 × 3 × 5 × 5 × 7

Question 51.
2³.11².13

Answer:
8 × 11 × 13 = 1144
We have to find the prime factorization for 1144.
1144 = 2 × 572
2 × 2 × 286
2 × 2 × 2 × 143
2 × 2 × 2 × 11 × 13

Question 52.
YOU BE THE TEACHER
Your friend ﬁnds the prime factorization of 72. Is your friend correct? Explain your reasoning.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 61$

Answer:
72 = 2 × 36
2 × 2 × 18
2 × 2 × 2 × 9
2 × 2 × 2 × 3 × 3
Your friend is incorrect because you have to write the prime factorization for 9 also.
Thus the prime factorization for 72 is 2 × 2 × 2 × 3 × 3

USING A PRIME FACTORIZATION
Find the greatest perfect square that is a factor of the number.

Question 53.
250

Answer:
A = Calculate the square root of the greatest perfect square from the list of all factors of 250. The factors of 250 are 1, 2, 5, 10, 25, 50, 125, and 250. Furthermore, the greatest perfect square on this list is 25 and the square root of 25 is 5. Therefore, A equals 5.
B = Calculate 250 divided by the greatest perfect square from the list of all factors of 250. We determined above that the greatest perfect square from the list of all factors of 250 is 25. Furthermore, 250 divided by 25 is 10, therefore B equals 10.
Now we have A and B and can get our answer to 250 in its simplest radical form as follows:
√250 = A√B
√250 = 5√10

Question 54.
275

Answer:
A = Calculate the square root of the greatest perfect square from the list of all factors of 275. The factors of 275 are 1, 5, 11, 25, 55, and 275. Furthermore, the greatest perfect square on this list is 25 and the square root of 25 is 5. Therefore, A equals 5.
B = Calculate 275 divided by the greatest perfect square from the list of all factors of 275. We determined above that the greatest perfect square from the list of all factors of 275 is 25. Furthermore, 275 divided by 25 is 11, therefore B equals 11.
Now we have A and B and can get our answer to 275 in its simplest radical form as follows:
√275 = A√B
√275 = 5√11

Question 55.
392

Answer:
A = Calculate the square root of the greatest perfect square from the list of all factors of 392. The factors of 392 are 1, 2, 4, 7, 8, 14, 28, 49, 56, 98, 196, and 392. Furthermore, the greatest perfect square on this list is 196 and the square root of 196 is 14. Therefore, A equals 14.
B = Calculate 392 divided by the greatest perfect square from the list of all factors of 392. We determined above that the greatest perfect square from the list of all factors of 392 is 196. Furthermore, 392 divided by 196 is 2, therefore B equals 2.
Now we have A and B and can get our answer to 392 in its simplest radical form as follows:
√392 = A√B
√392 = 14√2

Question 56.
338

Answer:
A = Calculate the square root of the greatest perfect square from the list of all factors of 338. The factors of 338 are 1, 2, 13, 26, 169, 338. Furthermore, the greatest perfect square on this list is 324 and the square root of 324 is 18. Therefore, A equals 18.
B = Calculate 338 divided by the greatest perfect square from the list of all factors of 338. We determined above that the greatest perfect square from the list of all factors of 338 is 324.
Now we have A and B and can get our answer to 338 in its simplest radical form as follows:
√338= A√B
√392 = 18√14

Question 57.
244

Answer:
Our first step would be to find out all the factors of 244
Since this is an even number, we divide 2 we get the factors as 244/2 = 122
Next, we again divide by 2 we get the factor as 122/2 = 61
We cannot go down any further since 61 is a prime number
Since we divided by 2, 2 is itself a factor.
Lastly, we divided by 2, twice; hence 2*2 = 4 is also a factor
The factors of 244 are 2,4,61 and 122
Out of 2,4,61 and 122, the only perfect square is 4
So, the greatest perfect square that is a factor of the number 244 should be 4
Therefore, the answer is: 4

Question 58.
650

Answer:
factor 650 and find the pairs
650=2×5×5×13
the pair is 5×5 which is 5² or 25
the greatest perfect square that is a factor of 650 is 25.

Question 59.
756

Answer:
A = Calculate the square root of the greatest perfect square from the list of all factors of 756. The factors of 756 are 1, 2, 3, 4, 6, 7, 9, 12, 14, 18, 21, 27, 28, 36, 42, 54, 63, 84, 108, 126, 189, 252, 378, and 756. Furthermore, the greatest perfect square on this list is 36 and the square root of 36 is 6. Therefore, A equals 6.
B = Calculate 756 divided by the greatest perfect square from the list of all factors of 756. We determined above that the greatest perfect square from the list of all factors of 756 is 36. Furthermore, 756 divided by 36 is 21, therefore B equals 21.
Now we have A and B and can get our answer to 756 in its simplest radical form as follows:
√756 = A√B
√756 = 6√21

Question 60.
1290

Answer:
There is no greatest perfect square that is a factor of 1290.
The factors of 1290 are 1, 2, 3, 5, 6, 10, 15, 30, 43, 86, 129, 215, 258, 430, 645, 1290. There are no perfect squares as factors.

Question 61.
2205

Answer: No, the number 2,205 is not a perfect square.
The factors of 2205 are 1, 3, 5, 7, 9, 15, 21, 35, 45, 49, 63, 105, 147, 245, 315, 441, 735. There are no perfect squares as factors.

Question 62.
1890

Answer: 1890 is not the perfect square.
The factors of 1890 are 1, 3, 9, 27, 67, 201, 603, 1809. There are no perfect squares as factors.

Question 63.
495

Answer: 495 is not the perfect square.
The factors 495 are 1, 3, 5, 9, 11, 15, 33, 45, 55, 99, 165, 495. There are no perfect squares as factors.

Question 64.
4725

Answer: 4725 is not the perfect square.
The factors of 4725 are 1, 3, 5, 7, 9, 15, 21, 25, 27, 35, 45, 63, 75, 105, 135, 175, 189, 225, 315, 525, 675, 945, 1575, 4725. There are no perfect squares as factors.

Question 65.
VOCABULARY
A botanist separates plants into equal groups of 5 for an experiment. Is the total number of plants in the experiment prime or composite? Explain.

Answer: The total number of plants will be a composite.

Explanation:
She is separating them into equal groups of five. So the total number will be a multiple of 5. Five is a prime number. Multiples of prime numbers are composite numbers.

Question 66.
REASONING
A teacher divides 36 students into equal groups for a scavenger hunt. Each group should have at least 4 students but no more than 8 students. What are the possible group sizes?

$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 62$

Answer:
Given,
A teacher divides 36 students into equal groups for a scavenger hunt.
Each group should have at least 4 students but no more than 8 students.
The factors of 36 are 1, 2, 3, 4, 6, 9
So, there are 6 groups of 6, 9 groups of 4.

Question 67.
CRITICAL THINKING
Is 2 the only even prime number? Explain.

Answer:
The definition of a prime number is a positive integer that has exactly two distinct divisors. Since the divisors of 2 are 1 and 2, there are exactly two distinct divisors, so 2 is prime.

Question 68.
LOGIC
One table at a bake sale has 75 cookies. Another table has 60 cupcakes. Which table allows for more rectangular arrangements? Explain.

Answer:
Given,
One table at a bake sale has 75 cookies. Another table has 60 cupcakes.
75 = 3·5², so has 6 divisors. 6 rectangles are possible if you make the distinction between 1×75 and 75×1.
60 = 2²·3·5, so has 12 divisors. 12 rectangles are possible under the same conditions.

Question 69.
PERFECT NUMBERS
A perfect number is a number that equals the sum of its factors, not including itself. For example, the factors of 28 are 1, 2, 4, 7, 14, and 28. Because 1 + 2 + 4 + 7 + 14 = 28, 28 is a perfect number. What are the perfect numbers between 1 and 27?

Answer: Perfect number, a positive integer that is equal to the sum of its proper divisors. The smallest perfect number is 6, which is the sum of 1, 2, and 3. Other perfect numbers are 28, 496, and 8,128.

Question 70.
REPEATED REASONING
Choose any two perfect squares and ﬁnd their product. Then multiply your answer by another perfect square. Continue this process. Are any of the products perfect squares? What can you conclude?

Answer:
2² × 3² = 4×9=36, which is a square number

Question 71.
PROBLEM SOLVING
The stage manager of a school play creates a rectangular stage that has whole number dimensions and an area of 42 square yards. String lights will outline the stage. What is the least number of yards of string lights needed to enclose the stage?

Answer:
Given that the stage manager of a school play creates a rectangular acting area of 42 square yards.
Let the length of the rectangular acting area be x, then the width is given by 42 / x.
The number of yards of string lights that the manager needs to enclose the area is given by the perimeter of the rectangular area.
Recall that the perimeter of a rectangle is given by
P = 2(length + width) = 2(x + 42/x) = 2x + 84/x
The perimeter is minimum when the differentiation of 2x + 84/x is equal to 0.
Therefore, the minimum number of yards of string lights the manager need to enclose in this area is given by
2x – 84/x = 0
2x² – 84 = 0
2x² = 84
x² = 84/2
x² = 42
x ≈ 6.48

Question 72.
DIG DEEPER!
Consider the rectangular prism shown. Using only whole number dimensions, how many different prisms are possible? Explain.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 63$

Answer:
The volume of the rectangular prism is lbh
v = 40 cubic inches
Let the length be 5 inches
breadth be 2 inches
height be 4 inches
V = 5 × 2 × 4
V = 40 cu. inches

Lesson 1.4 Greatest Common Factor

A Venn diagram uses circles to describe relationships between two or more sets. The Venn diagram shows the factors of 12 and 15. Numbers that are factors of both 12 and 15 are represented by the overlap of the two circles.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 64$

Answer: 1 and 3 are overlapped between the two circles.
So, 1 and 3 are the greatest common factors of 12 and 15.
The factors of 12 are 1, 2, 3, 4, 6, 12
The factors of 15 are 1, 3, 5, 15.

EXPLORATION 1

Identifying Common Factors

Work with a partner. In parts (a) – (d), create a Venn diagram that represents the factors of each number and identify any common factors.
a. 36 and 48
b. 16 and 56
c. 30 and 75
d. 54 and 90
e. Look at the Venn diagrams in parts (a)–(d). Explain how to identify the greatest common factor of each pair of numbers. Then circle it in each diagram.

Answer:
a. $Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_3$

b. $Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_4$

c. $Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_5$

d. $Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_6$

e. 36 and 48 have the greatest common factors.

EXPLORATION 2

Using Prime Factors
Work with a partner
a. Each Venn diagram represents the prime factorizations of two numbers. Identify each pair of numbers. Explain your reasoning.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 65$

Answer:
i. Red 2 × 3 × 3 = 18
Green 3 × 3 × 3 = 27
GCF = 9
ii. Yellow – 2 × 2 × 3 × 3 × 5 = 180
Purple – 5 × 11 = 55
GCF = 5
b. Create a Venn diagram that represents the prime factorizations of 36 and 48.

Answer:
$Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_3$
c. Repeat part(b) for the remaining number pairs in Exploration 1.

Answer:
$Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_4$

d. STRUCTURE
Make a conjecture about the relationship between the greatest common factors you found in Exploration 1 and the numbers in the overlaps of the Venn diagrams you just created.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 66$

Answer:
a. $Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_3$
The GCF between the two numbers 36 and 48 are 1,2,3,4,6,12

b. $Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_4$
The GCF between the two numbers 16 and 56 are 1,2,4,8
c. $Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_5$
The GCF between the two numbers 30 and 75 is 15.
d. $Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors img_6$

1.4 Lesson

Try It

Find the GCF of the numbers using lists of factors.

Question 1.
8, 36

Answer:
The factors of 8 are: 1, 2, 4, 8
The factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, 36
Then the greatest common factor is 4.

Question 2.
18, 72

Answer:
The factors of 18 are: 1, 2, 3, 6, 9, 18
The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Then the greatest common factor is 18.

Question 3.
14, 28, 49

Answer:
The factors of 14 are: 1, 2, 7, 14
The factors of 28 are: 1, 2, 4, 7, 14, 28
The factors of 49 are: 1, 7, 49
Then the greatest common factor is 7.

Another way to ﬁnd the GCF of two or more numbers is by using prime factors. The GCF is the product of the common prime factors of the numbers.

Try It
Find the GCF of the numbers using prime factorizations.

Question 4.
20,45

Answer:
Find the prime factorization of 20
20 = 2 × 2 × 5
Find the prime factorization of 45
45 = 3 × 3 × 5
To find the GCF, multiply all the prime factors common to both numbers:
Therefore, GCF = 5

Question 5.
32,90

Answer:
Find the prime factorization of 32
32 = 2 × 2 × 2 × 2 × 2
Find the prime factorization of 90
90 = 2 × 3 × 3 × 5
To find the GCF, multiply all the prime factors common to both numbers:
Therefore, GCF = 2

Question 6.
45,75,120

Answer:
45= 1,3,5,9,15,45
75= 1,3,5,15,25,75
120=1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120
GCF is 15

Try It

Question 7.
Write a pair of numbers whose greatest common factor is 10.

Answer:
Let’s first find the greatest common factor (GCF) of two whole numbers. The GCF of two numbers is the greatest number that is a factor of both of the numbers. Take the numbers 50 and 30.
50 = 10 × 5
30 = 10 × 3
Their greatest common factor is 10. since 10 is the greatest factor that both numbers have in common.

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

FINDING THE GCF
Find the GCF of the numbers.

Question 8.
16, 40

Answer:
The factors of 16 are: 1, 2, 4, 8, 16
The factors of 40 are: 1, 2, 4, 5, 8, 10, 20, 40
Then the greatest common factor is 8.

Question 9.
35, 63

Answer:
The factors of 35 are: 1, 5, 7, 35
The factors of 63 are: 1, 3, 7, 9, 21, 63
Then the greatest common factor is 7.

Question 10.
18, 72, 144

Answer:
The factors of 18 are: 1, 2, 3, 6, 9, 18
The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
The factors of 144 are: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144
Then the greatest common factor is 18.

Question 11.
MULTIPLE CHOICE
Which number is not a factor of 10? Explain.
A. 1
B. 2
C. 4
D. 5

Answer: 4

Explanation:
Factors of 10 are: 1, 2, 5, 10
Thus the correct answer is option C.

Question 12.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 66.1$

Answer:
The Greatest Common Factor of 24 and 32 is 8
The Greatest Common Divisor of 24 and 32 is 4
The Greatest Common Prime Factor of 24 and 32 is 8
The product of common prime factors of 24 and 32 is 8.
The Greatest Common Divisor of 24 and 32 are different from others.

Self – Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 13.
You use 30 sandwiches and 42 granola bars to make identical picnic baskets. You make the greatest number of picnic baskets with no food left over. How many sandwiches and how many granola bars are in each basket?
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 67$

Answer: 5 sandwiches and 7 granola in each basket

Explanation:
Given
Represent Sandwiches with S and Granola with G
S = 30
G = 42
To do this, we simply need to determine the ratio of S to G
S:G = 30:42
S:G = 5:7

Question 14.
You ﬁll bags with cookies to give to your friends. You bake 45 chocolate chip cookies, 30 peanut butter cookies, and 15 oatmeal cookies. You want identical groups of cookies in each bag with no cookies left over. What is the greatest number of bags you can make?

Answer: 15 bags

Explanation:
Given,
You ﬁll bags with cookies to give to your friends. You bake 45 chocolate chip cookies, 30 peanut butter cookies, and 15 oatmeal cookies.
You want identical groups of cookies in each bag with no cookies leftover.
45 chocolate chip cookies
30 peanut butter cookies
15 oatmeal cookies
So, the GCF is 15.

Greatest Common Factor Practice 1.4

Review & Refresh

List the factor pairs of the number.

Question 1.
20

Answer: The factor pairs of 20 are (1, 20) (4,5) (2,10)

Explanation:
1 × 20 = 20
4 × 5 = 20
2 × 10 = 20

Question 2.
16

Answer: The factor pairs of 16 are (1,16), (2, 8) (4,4)

Explanation:
1 × 16 = 16
2 × 8 = 16
4 × 4 = 16

Question 3.
56

Answer: The factor pairs of 56 are (1,56) (7,8) (28,2) (14,4)

Explanation:
1 × 56 = 56
7 × 8 = 56
28 × 2 = 56
14 × 4 = 56

Question 4.
87

Answer: The factor pairs of 87 are (1,87) (3,29)

Explanation:
1 × 87 = 87
3 × 29 = 87

Tell whether the statement is always, sometimes, or never true.

Question 5.
A rectangle is a rhombus.

Answer: sometimes

No, because all four sides of a rectangle don’t have to be equal. However, the sets of rectangles and rhombuses do intersect, and their intersection is the set of square all squares are both a rectangle and a rhombus.

Question 6.
A rhombus is a square.

Answer: true

A square is a special case of a rhombus because it has four equal-length sides and goes above and beyond that to also have four right angles. Every square you see will be a rhombus, but not every rhombus you meet will be a square.

Question 7.
A square is a rectangle.

Answer: not always

A square also fits the definition of a rectangle.

Question 8.
A trapezoid is a parallelogram.

Answer: never true

A trapezoid has one pair of parallel sides and a parallelogram has two pairs of parallel sides. So a parallelogram is also a trapezoid.

Concepts, Skills, & Problem Solving

USING A VENN DIAGRAM
Use a Venn diagram to ﬁnd the greatest common factor of the numbers. (See Exploration 1, p. 21.)

Question 9.
12,30

Answer: 6

$Big-ideas-math-answers-grade-6-chapter-1-img-3$

Question 10.
32,54

Answer: 2
$Big-ideas-math-grade-6-chapter-1-answer-key-img-4$

Question 11.
24,108

Answer: 12

BIM 6th Grade Chapter 1 Answer Key Numerical Expressions and Factors img_5

FINDING THE GCF
Find the GCF of the numbers using lists of factors.

Question 12.
6, 15

Answer: GCF is 3

Explanation:
The factors of 6 are: 1,2,3,6
The factors of 15 are: 1,3,5,15
The common Factors in 6 and 15 is 3.
Thus the greatest common factor is 3.

Question 13.
14, 84

Answer: GCF is 14

Explanation:
The factors of 14 are: 1,2,7,14
The factors of 84 are: 1,2,3,4,6,7,12,14,21,28,42 84
The greatest common factor is 14.

Question 14.
45, 108

Answer: GCF is 9

Explanation:
The factors of 45 are: 1,3,5,9,15,45
The factors of 108 are: 1,2,3,4,6,9,12,18,27,36,54,108
The greatest common factor is 9.

Question 15.
39, 65

Answer: GCF is 13

Explanation:
The factors of 39 are: 1,3,13,39
The factors of 65 are: 1,5,13,65
Thus the greatest common factor is 13.

Question 16.
51, 85

Answer: GCF is 17

Explanation:
The factors of 51 are: 1,3,17,51
The factors of 1,5,17,85
Thus the greatest common factor is 17

Question 17.
40, 63

Answer: GCF is 1

Explanation:
The factors of 40 are: 1,2,4,5,8,10,20,40
The factors of 63 are: 1,3,7,9,21,63
Thus the greatest common factor is 1.

Question 18.
12, 48

Answer: GCF is 12

Explanation:
The factors of 12 are: 1,2,3,4,6,12
The factors of 48 are: 1,2,3,4,6,8,12,16,24,48
Thus the greatest common factor is 12.

Question 19.
24, 52

Answer: GCF is 4

Explanation:
The factors of 24 are: 1,2,3,4,6,8,12,24
The factors of 1,2,4,13,36,52
Thus the greatest common factor is 4.

Question 20.
30, 58

Answer: GCF is 2

Explanation:
The factors of 30 are: 1,2,3,5,6,10,15,30
The factors of 58 are: 1,2,29,58
Thus the greatest common factor is 2.

FINDING THE GCF
Find the GCF of the numbers using prime factorizations.

Question 21.
45, 60

Answer:
The prime factorization of 45 is 3 x 3 x 5
The prime factorization 60 is 2 x 2 x 3 x 5
GCF of 45, 60 is 3 × 5 = 15

Question 22.
27, 63

Answer: 9
The prime factorization of 27 is 3 x 3 x 3
The prime factorization of 63 is 3 x 3 x 7
Thus GCF of 27, 63 is 9

Question 23.
36, 81

Answer: 9
The prime factorization of 36 is 2 x 2 x 3 x 3
The prime factorization of 81 is 3 x 3 x 3 x 3
Thus the GCF of 36, 81 is 9.

Question 24.
72, 84

Answer: 12
The prime factorization of 72 is 2 x 2 x 2 x 3 x 3
The prime factorization of 84 is 2 x 2 x 3 x 7
Thus the GCF of 72, 84 is 12.

Question 25.
61, 73

Answer: 1
The prime factorization of 61 is 1 × 61
The prime factorization 73 is 1 × 73
Thus the GCF of 61, 73 is 1

Question 26.
38, 95

Answer: 19
The prime factorization of 38 is 2 x 19
The prime factorization of 95 is 5 x 19
Thus the GCF of 38, 95 is 19

Question 27.
60, 75

Answer: 15
The prime factorization of 60 is 2 x 2 x 3 x 5
The prime factorization of 75 is 3 x 5 x 5
Thus the GCF of 60, 75 is 15.

Question 28.
42, 60

Answer: 6
The prime factorization 42 is 2 × 3 × 7
The prime factorization 60 is 2 × 2 × 3 × 5
Thus the GCF of 42, 60 is 6

Question 29.
42, 63

Answer: 21
The prime factorization of 42 is 2 × 3 × 7
The prime factorization of 63 is 3 × 3 ×7
Thus the GCF of 42, 63 is 21

Question 30.
24, 96

Answer: 24
The prime factorization of 24 is 2 × 2 × 2 × 3
The prime factorization of 96 is 2 x 2 x 2 x 2 x 2 x 3
Thus the GCF of 24, 96 is 24.

Question 31.
189, 200

Answer: 24
The prime factorization of 189 is 3 x 3 x 3 x 7
The prime factorization of 200 is 2 x 2 x 2 x 5 x 5
Thus the GCF of 189, 200 is 24.

Question 32.
90, 108

Answer: 18
The prime factorization of 90 is 2 x 3 x 3 x 5
The prime factorization of 108 is 2 x 2 x 3 x 3 x 3
Thus the GCF of 90, 108 is 18.

OPEN-ENDED
Write a pair of numbers with the indicated GCF.

Question 33.
5

Answer: 10, 15
The factors of 10 are: 1,2,5,10
The factors of 15 are: 1,3,5,15
Thus 10, 15 are pairs of numbers with the indicated GCF 5.

Question 34.
12

Answer: 72, 84
The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
The factors of 84 are: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
Then the greatest common factor is 12.

Question 35.
37

Answer: 37,74
The factors of 37 are: 1,37
The factors of 74 are: 1,2,37,74
Thus 37,74 are pairs of numbers with the indicated GCF 37.

Question 36.
MODELING REAL LIFE
A teacher is making identical activity packets using 92 crayons and 23 sheets of paper. What is the greatest number of packets the teacher can make with no items left over?

Answer: 23 packets

Explanation:
Factor both numbers:
1. 92=2·46=2·2·23;
2. 23 is a prime number.
Then the greatest common factor GCF (92, 23)=23.
This means that teacher can make 23 packets, each of them will contain 4 crayons and 1 sheet.

Question 37.
MODELING REAL LIFE
You are making balloon arrangements for a birthday party. There are 16 white balloons and 24 red balloons. Each arrangement must be identical. What is the greatest number of arrangements you can make using every balloon?
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 67.1$

Answer:
This is a GCF problem. To find the GCF, list the factors.
24
1×24, 2×12, 4×6, 8×3 …
16
1×16, 2×8, 4×4, …
The greatest common factor is 8 because it is the greatest factor of both numbers.
so 8 arrangements, 2 white balloons and 3 red balloons each.
Because- 8 is the number of groups, and 8×3 =24 so 3 reds, and 2 whites because 8×2=16

YOU BE THE TEACHER
Your friend ﬁnds the GCF of the two numbers. Is your friend correct? Explain your reasoning.

Question 38.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 68$

Answer:
No, your friend is incorrect.
42 = 2 × 3 × 7
154 = 2 × 7 × 11
Thus the GCF is 14.

Question 39.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 69$

Answer: Yes your friend is correct.
Thus the GCF of 36 and 60 is 12.

FINDING THE GCF
Find the GCF of the numbers.

Question 40.
35, 56, 63

Answer: GCF is 7

Explanation:
The factors of 35 are: 1,5,7,35
The factors of 56 are: 1,2,4,7,8,14,28,56
The factors of 63 are: 1,3,7,9,21,63
Thus the greatest common factor is 7.

Question 41.
30, 60, 78

Answer: GCF is 6

Explanation:
The factors of 30 are: 1,2,3,5,6,10,15,30
The factors of 60 are: 1,2,3,4,5,6,10,12,15,20,30,60
The factors of 78 are: 1,2,3,6,13,26,39,78
Thus the greatest common factor is 6.

Question 42.
42, 70, 84

Answer: GCF is 14

The factors of 42 are: 1, 2, 3, 6, 7, 14, 21, 42
The factors of 70 are: 1, 2, 5, 7, 10, 14, 35, 70
The factors of 84 are: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
Then the greatest common factor is 14.

Question 43.
40, 55, 72

Answer: GCF is 1

The factors of 40 are: 1, 2, 4, 5, 8, 10, 20, 40
The factors of 55 are: 1, 5, 11, 55
The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Then the greatest common factor is 1.

Question 44.
18, 54, 90

Answer: GCF is 18

The factors of 18 are: 1, 2, 3, 6, 9, 18
The factors of 54 are: 1, 2, 3, 6, 9, 18, 27, 54
The factors of 90 are: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
Then the greatest common factor is 18.

Question 45.
16, 48, 88

Answer: GCF is 8

The factors of 16 are: 1, 2, 4, 8, 16
The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
The factors of 88 are: 1, 2, 4, 8, 11, 22, 44, 88
Then the greatest common factor is 8.

Question 46.
52, 78, 104

Answer: GCF is 26

The factors of 52 are: 1, 2, 4, 13, 26, 52
The factors of 78 are: 1, 2, 3, 6, 13, 26, 39, 78
The factors of 104 are: 1, 2, 4, 8, 13, 26, 52, 104
Then the greatest common factor is 26.

Question 47.
96, 120, 156

Answer: GCF is 12

The factors of 96 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96
The factors of 120 are: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120
The factors of 156 are: 1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156
Then the greatest common factor is 12.

Question 48.
280, 300, 380

Answer: GCF is 20

The factors of 280 are: 1, 2, 4, 5, 7, 8, 10, 14, 20, 28, 35, 40, 56, 70, 140, 280
The factors of 300 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 25, 30, 50, 60, 75, 100, 150, 300
The factors of 380 are: 1, 2, 4, 5, 10, 19, 20, 38, 76, 95, 190, 380
Then the greatest common factor is 20.

Question 49.
OPEN-ENDED
Write three numbers that have a GCF of 16. What method did you use to ﬁnd your answer?

Answer: 16, 32, 48

The factors of 16 are: 2 × 2 × 2 × 2
The factors of 32 are : 2 × 2 × 2 × 2 × 2
The factors of 48 are: 3 × 2 × 2 × 2 × 2

CRITICAL THINKING
Tell whether the statement is always, sometimes, or never true. Explain your reasoning.

Question 50.
The GCF of two even numbers is 2.

Answer: Always

Explanation:
Example:
The factors of 14 are: 1, 2, 7, 14
The factors of 16 are: 1, 2, 4, 8, 16
Then the greatest common factor is 2.

Question 51.
The GCF of two prime numbers is 1.

Answer: Always

Explanation:
Example:
The factors of 3 are: 1, 3
The factors of 5 are: 1, 5
Then the greatest common factor is 1.

Question 52.
When one number is a multiple of another, the GCF of the numbers is the greater of the numbers.

Answer:
When one number is a multiple of another, the GCF of the numbers is the greater of the numbers. This is never true since the GCF is a factor of both numbers. So the GCF is the smaller of the two numbers.

Question 53.
PROBLEM SOLVING
A science museum makes gift bags for students using 168 magnets, 48 robot ﬁgurines, and 24 packs of freeze-dried ice cream. What is the greatest number of gift bags that can be made using all of the items? How many of each item are in each gift bag?
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 70$

Answer:
The greatest common factor of 24, 48, and 168 is 24, so 24 gift bags can be made. Each will have 1/24 of the number of gift items of each type that are available.
In each bag are
1/24 × 168 magnets = 7 magnets
1/24 × 48 robot figurines = 2 robot figurines
1/24 × 24 packs of ice cream = 1 pack of ice cream

Question 54.
VENN DIAGRAM
Consider the numbers 252, 270, and 300.
a. Create a Venn diagram using the prime factors of the numbers.

Answer:
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors img_6$
b. Use the Venn diagram to ﬁnd the GCF of 252, 270, and 300.

Answer:
The factors of 252 are: 1, 2, 3, 4, 6, 7, 9, 12, 14, 18, 21, 28, 36, 42, 63, 84, 126, 252
The factors of 270 are: 1, 2, 3, 5, 6, 9, 10, 15, 18, 27, 30, 45, 54, 90, 135, 270
The factors of 300 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 25, 30, 50, 60, 75, 100, 150, 300
Then the greatest common factor is 6.
c. What is the GCF of 252 and 270? 252 and 300? 270 and 300? Explain how you found your answers.

Answer:
The factors of 252 are: 1, 2, 3, 4, 6, 7, 9, 12, 14, 18, 21, 28, 36, 42, 63, 84, 126, 252
The factors of 270 are: 1, 2, 3, 5, 6, 9, 10, 15, 18, 27, 30, 45, 54, 90, 135, 270
The GCF of 252 and 270 is 18.
252 and 300:
The factors of 252 are: 1, 2, 3, 4, 6, 7, 9, 12, 14, 18, 21, 28, 36, 42, 63, 84, 126, 252
The factors of 300 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 25, 30, 50, 60, 75, 100, 150, 300
Then the greatest common factor is 12.
270 and 300:
The factors of 270 are: 1, 2, 3, 5, 6, 9, 10, 15, 18, 27, 30, 45, 54, 90, 135, 270
The factors of 300 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 25, 30, 50, 60, 75, 100, 150, 300
Then the greatest common factor is 30.

Question 55.
REASONING
You are making fruit baskets using 54 apples, 36 oranges, and 73 bananas.
a. Explain why you cannot make identical fruit baskets without leftover fruit.

Answer: 73 is a prime number. It can only be divided by 1 and by itself.
b. What is the greatest number of identical fruit baskets you can make with the least amount of fruit left over? Explain how you found your answer.

Answer:
The GCF of the three numbers:
54 36 73
1×54 1×36 1×73
2×27 2×18
3×18 3×12
6×9 4×9
6×6
GCF of 54, 36, and 73 is 1
GCF of 54 and 36 is 18
If we divide 54 apples into 18 baskets, we have 3 apples in each basket
If we divide 36 oranges into 18 baskets, we have 2 oranges in each basket
If we divide 73 bananas into 18 baskets, we have 4 bananas in each basket + one banana left over.
So the greatest number of identical fruit baskets we can make with the least amount of fruit left over is 18 baskets

Question 56.
DIG DEEPER!
Two rectangular, adjacent rooms share a wall. One-foot-by-one-foot tiles cover the ﬂoor of each room. Describe how the greatest possible length of the adjoining wall is related to the total number of tiles in each room. Draw a diagram that represents one possibility.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 71$

Answer:
Consider two adjacent rectangular rooms having Length=L, and, Breadth = B
Now Suppose the wall which is in between two rooms has a height or length =H.
The breadth of wall = B [ if the wall doesn’t exceed the breadth of the room]
Considering two rooms to be identical,
Area of each room= L × B square unit
Area of each tile = 1×1=1 square unit
Number of tiles required= L B ÷ 1= LB tiles( product of length and breadth of the room is the number of tiles required)
Suppose if,LB= N
B= N/LArea of the wall(W) = B×H= B H square unit
B =W/H
Equating (1) and (2)
N/L = W/ H
H = WL/N
H = WL/LB
H = W/B
H = Area of wall/Breadth of room or wall

Lesson 1.5 Least Common Multiple

EXPLORATION 1
Identifying Common Multiples

Work with a partner. In parts (a)–(d), create a Venn diagram that represents the ﬁrst several multiples of each number and identify any common multiples.
a. 8 and 12
b. 4 and 14
c. 10 and 15
d. 20 and 35
e. Look at the Venn diagrams in parts (a)–(d). Explain how to identify the least common multiple of each pair of numbers. Then circle it in each diagram.

Answer:
a. BIM Grade 6 Chapter 1 Answer Key img_7
b. $Big Ideas math 6th grade answers chapter 1 img_8$
c. $Big Ideas math 6th grade answers chapter 1 img_9$

d. $Big Ideas math 6th grade answers chapter 1 img_10$

EXPLORATION 2
Using Prime Factors
Work with a partner.
a. Create a Venn diagram that represents the prime factorizations of 8 and 12.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 72$

Answer:
BIM Grade 6 Chapter 1 Answer Key img_7
b. Repeat part (a) for the remaining number pairs in Exploration 1.

Answer:
$Big Ideas math 6th grade answers chapter 1 img_8$
c. STRUCTURE
Make a conjecture about the relationship between the least common multiples you found in Exploration 1 and the numbers in the Venn diagrams you just created.

Answer:
The numbers which are overlapped are the least common multiples of the numbers.
d. The Venn diagram shows the prime factors of two numbers.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 73$
Use the diagram to complete the following tasks.

Identify the two numbers.
Find the greatest common factor.
Find the least common multiple

Answer:

120, 180
The factors of 120 are: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120The factors of 180 are: 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180Then the greatest common factor is 60.
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 120:
120, 240, 360, 480, 600
Multiples of 180:
180, 360, 540, 720
Therefore,
LCM(120, 180) = 360

1.5 Lesson

Try It

Find the LCM of the numbers using lists of multiples.

Question 1.
3, 8

Answer: 24
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 3:
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Multiples of 8:
8, 16, 24, 32, 40
Therefore,
LCM(3, 8) = 24

Question 2.
9, 12

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 9:
9, 18, 27, 36, 45, 54
Multiples of 12:
12, 24, 36, 48, 60
Therefore,
LCM(9, 12) = 36

Question 3.
6, 10

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 6:
6, 12, 18, 24, 30, 36, 42
Multiples of 10:
10, 20, 30, 40, 50
Therefore,
LCM(6, 10) = 30

Try It
Find the LCM of the numbers using prime factorizations.

Question 4.
14, 18

Answer:
List all prime factors for each number.
Prime Factorization of 14 is:
2 x 7
Prime Factorization of 18 is:
2 x 3 x 3
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 3, 3, 7
Multiply these factors together to find the LCM.
LCM = 2 x 3 x 3 x 7 = 126
LCM(14, 18) = 126

Question 5.
28, 36

Answer:
List all prime factors for each number.
Prime Factorization of 28 is:
2 x 2 x 7
Prime Factorization of 36 is:
2 x 2 x 3 x 3
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 2, 3, 3, 7
Multiply these factors together to find the LCM.
LCM = 2 x 2 x 3 x 3 x 7 = 252
LCM = 252
Therefore,
LCM(28, 36) = 252

Question 6.
24, 90

Answer:
List all prime factors for each number.
Prime Factorization of 24 is:
2 x 2 x 2 x 3
Prime Factorization of 90 is:
2 x 3 x 3 x 5
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 2, 2, 3, 3, 5
Multiply these factors together to find the LCM.
LCM = 2 x 2 x 2 x 3 x 3 x 5 = 360
LCM = 360
Therefore,
LCM(24, 90) = 360

Try It

Find the LCM of the numbers.

Question 7.
2, 5, 8

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 2:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44
Multiples of 5:
5, 10, 15, 20, 25, 30, 35, 40, 45, 50
Multiples of 8:
8, 16, 24, 32, 40, 48, 56
Therefore,
LCM(2, 5, 8) = 40

Question 8.
6, 10, 12

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 6:
6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72
Multiples of 10:
10, 20, 30, 40, 50, 60, 70, 80
Multiples of 12:
12, 24, 36, 48, 60, 72, 84
Therefore,
LCM(6, 10, 12) = 60

Question 9.
Write three numbers that have a least common multiple of 100.

Answer: 1, 10, 100

Explanation:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 1:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102
Multiples of 10:
10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120
Multiples of 100:
100, 200, 300
Therefore,
LCM(1, 10, 100) = 100

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

FINDING THE LCM
Find the LCM of the numbers.

Question 10.
6, 9

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 6:
6, 12, 18, 24, 30
Multiples of 9:
9, 18, 27, 36
Therefore,
LCM(6, 9) = 18

Question 11.
30, 40

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 30:
30, 60, 90, 120, 150, 180
Multiples of 40:
40, 80, 120, 160, 200
Therefore,
LCM(30, 40) = 120

Question 12.
5, 11

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 5:
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65
Multiples of 11:
11, 22, 33, 44, 55, 66, 77
Therefore,
LCM(5, 11) = 55

Question 13.
Reasoning
Write two numbers such that 18 and 30 are multiples of the numbers. Justify your answer.

Answer: 3, 6

Explanation:
Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Multiples of 6: 6,12,18,24,30,36,42,48
Thus 18 and 30 are the multiples of 3 and 6.

Question 14.
REASONING
You need to ﬁnd the LCM of 13 and 14. Would you rather list their multiples or use their prime factorizations? Explain.

Answer:
List all prime factors for each number.
Prime Factorization of 13 shows:
13 is prime = 13
Prime Factorization of 14 is:
2 x 7
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 7, 13
Multiply these factors together to find the LCM.
LCM = 2 x 7 x 13 = 182
In exponential form:
LCM = 2 x 7 x 13 = 182
LCM = 182
Therefore,
LCM(13, 14) = 182

Question 15.
CHOOSE TOOLS
A student writes the prime factorizations of 8 and 12 in a table as shown. She claims she can use the table to ﬁnd the greatest common factor and the least common multiple of 8 and 12. How is this possible?
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 74$
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 75$

Answer:
$Big-Ideas-Math-Answer-Key-Grade-6-Chapter-1-Numerical-Numerical-Expressions-and-Factors-74$
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 2, 2, 3
Multiply these factors together to find the LCM.
LCM = 2 x 2 x 2 x 3 = 24
The least common multiple of 8 and 12 is 24.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 17.
A geyser erupts every fourth day. Another geyser erupts every sixth day. Today both geysers erupted. In how many days will both geysers erupt on the same day again?

Answer:
Given,
A geyser erupts every fourth day. Another geyser erupts every sixth day. Today both geysers erupted.
The geyser erupts on the same day after 12 days.

Question 18.
A water park has two large buckets that slowly ﬁll with water. One bucket dumps water every 12 minutes. The other bucket dumps water every 10 minutes. Five minutes ago, both buckets dumped water. When will both buckets dump water at the same time again?
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 76$

Answer:
Given,
A water park has two large buckets that slowly ﬁll with water.
One bucket dumps water every 12 minutes.
The other bucket dumps water every 10 minutes.
Five minutes ago, both buckets dumped water.
Both buckets will dump again at the same time in 60 minutes (1 hour.)

Question 19.
DIG DEEPER!
You purchase disposable plates, cups, and forks for a cookout. Plates are sold in packages of 24, cups in packages of 32, and forks in packages of 48. What are the least numbers of packages you should buy in order to have the same number of plates, cups, and forks?

Answer:
Given,
You purchase disposable plates, cups, and forks for a cookout. Plates are sold in packages of 24, cups in packages of 32, and forks in packages of 48.
We solve this question using the Lowest Common Multiple (LCM) method/
Step 1
We list multiples of each number until the first common multiple is found. This is referred to as the lowest common multiple.
Plates are sold in packages of 24
Cups in packages of 32
Forks in packages of 48
Multiples of 24:
24, 48, 72, 96, 120, 144
Multiples of 32:
32, 64, 96, 128, 160
Multiples of 48:
48, 96, 144, 192
Therefore,
LCM(24, 32, 48) = 96
Hence, the least numbers of packages you should buy in order to have the same number of plates, cups, and forks is 96 packages

Least Common Multiple Practice 1.5

Review & Refresh

Find the GCF of the numbers.

Question 1.
18, 42

Answer:
The factors of 18 are: 1, 2, 3, 6, 9, 18
The factors of 42 are: 1, 2, 3, 6, 7, 14, 21, 42
Then the greatest common factor is 6.

Question 2.
72, 96

Answer:
The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
The factors of 96 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96
Then the greatest common factor is 24.

Question 3.
38, 76, 114

Answer:
The factors of 38 are: 1, 2, 19, 38
The factors of 76 are: 1, 2, 4, 19, 38, 76
The factors of 114 are: 1, 2, 3, 6, 19, 38, 57, 114
Then the greatest common factor is 38.

Divide.

Question 4.
900 ÷ 6

Answer: 150

Explanation:
Divide the two numbers 900 and 6.
900/6 = 150
It means 6 divides 900 150 times.
Thus the quotient is 150.

Question 5.
1944 ÷ 9

Answer: 216

Explanation:
Divide the two numbers 1944 and 9
1944/9 = 216
It means that 9 divides 1944 216 times.
Thus the quotient is 216

Question 6.
672 ÷ 12

Answer: 56

Explanation:
Divide the two numbers 672 and 12.
672/12 = 56
It means 12 divides 672 56 times.
Thus the quotient is 56.

Write an ordered pair that corresponds to the point.

Question 7.
Point A

Answer: (2,4)
By seeing the below graph we can find the ordered pairs.
2 lies on the x-axis and 4 lies on the y-axis.
So, the ordered pairs to the point A is (2,4)

Question 8.
Point B

Answer: (3,1)
By seeing the below graph we can find the ordered pairs.
3 lies on the x-axis and 1 lies on the y-axis.
So, the ordered pairs to the point A is (3,1)

Question 9.
Point C

Answer: (4,7)
By seeing the below graph we can find the ordered pairs.
4 lies on the x-axis and 7 lies on the y-axis.
So, the ordered pairs to the point A is (4,7)

Question 10.
Point D

Answer: (9,6)
By seeing the below graph we can find the ordered pairs.
9 lies on the x-axis and 6 lies on the y-axis.
So, the ordered pairs to the point A is (9,6)

$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 77$

Concepts, Skills, & Problem Solving

USING A VENN DIAGRAM
Use a Venn diagram to ﬁnd the least common multiple of the numbers. (See Exploration 1, p. 27.)

Question 11.
3, 7

Answer: 21

$Big Ideas Math Grade 6 Chapter 1 img_10$

Question 12.
6, 8

Answer: 24

$Big Ideas Math 6th Grade Answers Chapter 1 img_11$

Question 13.
4, 5

Answer:
$Big Ideas Math 6th Grade Answers Chapter 1 img_12$

FINDING THE LCM
Find the LCM of the numbers using lists of multiples.

Question 14.
1, 5

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 1:
1, 2, 3, 4, 5, 6, 7
Multiples of 5:
5, 10, 15
Therefore,
LCM(1, 5) = 5

Question 15.
2, 6

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 2:
2, 4, 6, 8, 10
Multiples of 6:
6, 12, 18
Therefore,
LCM(2, 6) = 6

Question 16.
2, 3

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 2:
2, 4, 6, 8, 10
Multiples of 3:
3, 6, 9, 12
Therefore,
LCM(2, 3) = 6

Question 17.
2, 9

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 2:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22
Multiples of 9:
9, 18, 27, 36
Therefore,
LCM(2, 9) = 18

Question 18.
3, 4

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 3:
3, 6, 9, 12, 15, 18
Multiples of 4:
4, 8, 12, 16, 20
Therefore,
LCM(3, 4) = 12

Question 19.
8, 9

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 8:
8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88
Multiples of 9:
9, 18, 27, 36, 45, 54, 63, 72, 81, 90
Therefore,
LCM(8, 9) = 72

Question 20.
5, 8

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 5:
5, 10, 15, 20, 25, 30, 35, 40, 45, 50
Multiples of 8:
8, 16, 24, 32, 40, 48, 56
Therefore,
LCM(5, 8) = 40

Question 21.
11, 12

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 11:
11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, 132, 143, 154
Multiples of 12:
12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144, 156
Therefore,
LCM(11, 12) = 132

Question 22.
12, 18

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 12:
12, 24, 36, 48, 60
Multiples of 18:
18, 36, 54, 72
Therefore,
LCM(12, 18) = 36

FINDING THE LCM
Find the LCM of the numbers using prime factorizations.

Question 23.
7, 12

Answer:
List all prime factors for each number.
Prime Factorization of 7 shows:
7 is prime
Prime Factorization of 12 is:
2 x 2 x 3
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 2, 3, 7
Multiply these factors together to find the LCM.
LCM = 2 x 2 x 3 x 7 = 84
LCM = 84
Therefore,
LCM(7, 12) = 84

Question 24.
5, 9 4

Answer:
List all prime factors for each number.
Prime Factorization of 4 is:
2 x 2
Prime Factorization of 5 shows:
5 is prime
Prime Factorization of 9 is:
3 x 3
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 2, 3, 3, 5
Multiply these factors together to find the LCM.
LCM = 2 x 2 x 3 x 3 x 5 = 180
LCM = 180
Therefore,
LCM(4, 5, 9) = 180

Question 25.
4, 11

Answer:
List all prime factors for each number.
Prime Factorization of 4 is:
2 x 2
Prime Factorization of 11 shows:
11 is prime
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 2, 11
Multiply these factors together to find the LCM.
LCM = 2 x 2 x 11 = 44
LCM = 44
Therefore,
LCM(4, 11) = 44

Question 26.
9, 10

Answer:
List all prime factors for each number.
Prime Factorization of 9 is:
3 x 3
Prime Factorization of 10 is:
2 x 5
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 3, 3, 5
Multiply these factors together to find the LCM.
LCM = 2 x 3 x 3 x 5 = 90
LCM = 90
Therefore,
LCM(9, 10) = 90

Question 27.
12, 27

Answer:
List all prime factors for each number.
Prime Factorization of 12 is:
2 x 2 x 3
Prime Factorization of 27 is:
3 x 3 x 3
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 2, 3, 3, 3
Multiply these factors together to find the LCM.
LCM = 2 x 2 x 3 x 3 x 3 = 108
LCM = 108
Therefore,
LCM(12, 27) = 108

Question 28.
18, 45

Answer:
List all prime factors for each number.
Prime Factorization of 18 is:
2 x 3 x 3
Prime Factorization of 45 is:
3 x 3 x 5
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 3, 3, 5
Multiply these factors together to find the LCM.
LCM = 2 x 3 x 3 x 5 = 90
LCM = 90
Therefore,
LCM(18, 45) = 90

Question 29.
22, 23

Answer:
List all prime factors for each number.
Prime Factorization of 22 is:
2 x 11
Prime Factorization of 23 shows:
23 is prime
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 11, 23
Multiply these factors together to find the LCM.
LCM = 2 x 11 x 23 = 506
LCM = 506
Therefore,
LCM(22, 23) = 506

Question 30.
36, 60

Answer:
List all prime factors for each number.
Prime Factorization of 36 is:
2 x 2 x 3 x 3
Prime Factorization of 60 is:
2 x 2 x 3 x 5
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 2, 3, 3, 5
Multiply these factors together to find the LCM.
LCM = 2 x 2 x 3 x 3 x 5 = 180
LCM = 180
Therefore,
LCM(36, 60) = 180

Question 31.
35, 50

Answer:
List all prime factors for each number.
Prime Factorization of 35 is:
5 x 7
Prime Factorization of 50 is:
2 x 5 x 5
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 5, 5, 7
Multiply these factors together to find the LCM.
LCM = 2 x 5 x 5 x 7 = 350
LCM = 350
Therefore,
LCM(35, 50) = 350

Question 32.
YOU BE THE TEACHER
Your friend ﬁnds the LCM of 6 and 9. Is your friend correct? Explain your reasoning.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 78$

Answer: No friend is incorrect.
List all prime factors for each number.
Prime Factorization of 6 is:
2 x 3
Prime Factorization of 9 is:
3 x 3
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
The new superset list is
2, 3, 3
Multiply these factors together to find the LCM.
LCM = 2 x 3 x 3 = 18
LCM = 18
Therefore,
LCM(6, 9) = 18

Question 33.
MODELING REAL LIFE
You have diving lessons every ﬁfth day and swimming lessons every third day. Today you have both lessons. In how many days will you have both lessons on the same day again?

Answer:
After 15 days
diving is on 0 , 5 , 10 , 15 , 20 , … days
swimming is on 0 , 3, 6 , 9, 12 , 15 , 18 , … days
Thus the next will be in 15 days

Question 34.
REASONING
Which model represents an LCM that is different from the other three? Explain your reasoning.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 79$

Answer: The number line B is different from the other three number lines.

FINDING THE LCM
Find the LCM of the numbers.

Question 35.
2, 3, 7

Answer:
LCM(2,3) = (2 × 3) / GCF(2,3)
= (2 × 3) / 1
= 6 / 1
= 6
LCM(6,7) = (6 × 7) / GCF(6,7)
= (6 × 7) / 1
= 42 / 1
= 42
Therefore,
LCM(2, 3, 7) = 42

Question 36.
3, 5, 11

Answer:
LCM(3,5) = (3 × 5) / GCF(3,5)
= (3 × 5) / 1
= 15 / 1
= 15
LCM(15,11) = (15 × 11) / GCF(15,11)
= (15 × 11) / 1
= 165 / 1
= 165
Therefore,
LCM(3, 5, 11) = 165

Question 37.
4, 9, 12

Answer:
LCM(4,9) = (4 × 9) / GCF(4,9)
= (4 × 9) / 1
= 36 / 1
= 36
LCM(36,12) = (36 × 12) / GCF(36,12)
= (36 × 12) / 12
= 432 / 12
= 36
Therefore,
LCM(4, 9, 12) = 36

Question 38.
6, 8, 15

Answer:
LCM(6,8) = (6 × 8) / GCF(6,8)
= (6 × 8) / 2
= 48 / 2
= 24
LCM(24,15) = (24 × 15) / GCF(24,15)
= (24 × 15) / 3
= 360 / 3
= 120
Therefore,
LCM(6, 8, 15) = 120

Question 39.
7, 18, 21

Answer:
LCM(7,18) = (7 × 18) / GCF(7,18)
= (7 × 18) / 1
= 126 / 1
= 126
LCM(126,21) = (126 × 21) / GCF(126,21)
= (126 × 21) / 21
= 2646 / 21
= 126
Therefore,
LCM(7, 18, 21) = 126

Question 40.
9, 10, 28

Answer:
LCM(9,10) = (9 × 10) / GCF(9,10)
= (9 × 10) / 1
= 90 / 1
= 90
LCM(90,28) = (90 × 28) / GCF(90,28)
= (90 × 28) / 2
= 2520 / 2
= 1260
Therefore,
LCM(9, 10, 28) = 1260

Question 41.
PROBLEM SOLVING
At Union Station, you notice that three subway lines just arrived at the same time. How long must you wait until all three lines arrive at Union Station at the same time again?
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 80$

Answer: 60 minutes

Explanation:
The complete question in the attached figure
Step 1
Find the least common multiple (LCM) of the three numbers
List the prime factors of each number
10 = 2 × 5
12 = 2 × 2 × 3
15 = 3 × 5
Multiply each factor the greatest number of times it occurs in any of the numbers to find out the LCM
The LCM is equal to
4 × 3 × 5 = 60
Thus You must wait 60 minutes for all three lines to arrive at Union Station at the same time again.

Question 42.
DIG DEEPER!
A radio station gives away $15 to every 15th caller, $25 to every 25th caller, and a free concert ticket to every 100th caller. When will the station ﬁrst give away all three prizes to one caller? When this happens, how much money and how many tickets are given away?

Answer:
Given,
Radio Station gives :
1st prize: $15 to 15th caller
2nd prize: $25 to 25th caller
3rd prize: free concert tickets to 100th caller
So, in order to get all three prizes the caller must be 15th, 25th, and 100th caller at the same time. But to find when the radio station will give first all three prizes we calculate L.C.M. of ( 15, 25, 100 ) that is 300
Hence, the station first gives away all three prizes to the 300th caller.

Question 43.
LOGIC
You and a friend are running on treadmills. You run 0.5 mile every 3 minutes, and your friend runs 2 miles every 14 minutes. You both start and stop running at the same time and run a whole number of miles. What are the least possible numbers of miles you and your friend can run?

Answer:
If you run 0.5 miles every 3 minutes then you run 1 mile every 6 minutes.
If your friend runs 2 miles every 14 minutes then your friend runs 1 mile every 7 minutes.
You will both then both run a whole number of minutes for a time that is a multiple of 6 and 7.
The least common multiple of 6 and 7 is 42 so the least possible time you and your friend could run for and both run a whole number of miles is then 42 minutes.
Since you run 1 mile every 6 minutes, in 42 minutes you will run 42/6=7 miles.
Since your friend runs 1 mile every 7 minutes, in 42 minutes your friend will run 42/7=6 miles.

Question 44.
VENN DIAGRAM
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 81$
Refer to the Venn diagram.
a. Copy and complete the Venn diagram.
b. What is the LCM of 16, 24, and 40?
c. What is the LCM of 16 and 40? 24 and 40? 16 and 24? Explain how you found your answers.

Answer:
$Big-Ideas-Math-Answers-Grade-6-Chapter-1-Numerical-Numerical-Expressions-and-Factors-81$
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
LCM of 16, 40:
Multiples of 16:
16, 32, 48, 64, 80, 96, 112
Multiples of 40:
40, 80, 120, 160
Therefore,
LCM(16, 40) = 80
LCM of 24, 40:
Multiples of 24:
24, 48, 72, 96, 120, 144, 168
Multiples of 40:
40, 80, 120, 160, 200
Therefore,
LCM(24, 40) = 120
LCM of 16, 24:
Multiples of 16:
16, 32, 48, 64, 80
Multiples of 24:
24, 48, 72, 96
Therefore,
LCM(16, 24) = 48

CRITICAL THINKING
Tell whether the statement is always, sometimes, or never true. Explain your reasoning.

Question 45.
The LCM of two different prime numbers is their product.

Answer: Always true

Example:
3 × 5 = 15

Question 46.
The LCM of a set of numbers is equal to one of the numbers in the set.

Answer:
The LCM of a set of numbers is equal to one of the numbers in the set. Always Sometimes Never true. Question 909193: The LCM of a set of numbers is equal to one of the numbers in the set. This is sometimes true.

Question 47.
The GCF of two different numbers is the LCM of the numbers.

Answer:
Another way to find the LCM of two numbers is to divide their product by their greatest common factor ( GCF ). Example 2: Find the least common multiple of 18 and 20. The common factors are 2 and 3 .

Numerical Expressions and Factors Connecting Concepts

Getting Ready for Chapter Connecting Concepts

Using the Problem-Solving Plan

Question 1.
A sports team gives away shirts at the stadium. There are 60 large shirts, 1.6 times as many small shirts as large shirts, and 1.5 times as many medium shirts as small shirts. The team wants to divide the shirts into identical groups to be distributed throughout the stadium. What is the greatest number of groups that can be formed using every shirt?

Understand the Problem
You know the number of large shirts and two relationships among the numbers of small, medium, and large shirts. You are asked to ﬁnd the greatest number of identical groups that can be formed using every shirt.
Make a plan
Break the problem into parts. First use multiplication to ﬁnd the number of each size shirt. Then ﬁnd the GCF of these numbers.
Solve and Check
Use the plan to solve the problem. Then check your solution.

Answer:
Given,
A sports team gives away shirts at the stadium.
There are 60 large shirts, 1.6 times as many small shirts as large shirts, and 1.5 times as many medium shirts as small shirts. The team wants to divide the shirts into identical groups to be distributed throughout the stadium.
60 × 1.6 = 96
60 × 1.5 = 90
The factors of 90 are: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
The factors of 96 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96
Then the greatest common factor is 6.

Question 2.
An escape artist ﬁlls the tank shown with water. Find the number of cubic feet of water needed to ﬁll the tank. Then ﬁnd the number of cubic yards of water that are needed to ﬁll the tank. Justify your answer.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 75.1$

Answer:
Given,
An escape artist ﬁlls the tank shown with water.
side = 6 ft
We know that
The volume of the cube = s³
V = 6ft × 6ft × 6ft
V = 216 cubic ft.

Performance Task

Setting the Table

At the beginning of this chapter, you watched a STEAM video called “Filling Piñatas.” You are now ready to complete the performance task for this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 82$

Answer:
Factors of 50 – 1, 2, 5, 10, 25, and 50
Factors of 12 – 1, 2, 3, 4, 6, 12
Factors of 16 – 1, 2, 4, 8, 16
The factors of number 24 are 1, 2, 3, 4, 6, 8, 12, 24.
The factors of 100 are 1,2,4,5,10,20,25,50 and 100.

Numerical Expressions and Factors Chapter Review

Review Vocabulary

Write the deﬁnition and give an example of each vocabulary term.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 83$

Graphic Organizers

You can use an Information Frame to organize and remember concepts. Here is an example of an Information Frame for the vocabulary term power.

$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 84$

Choose and complete a graphic organizer to help you study the concept.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 85$

perfect square
numerical expression
order of operations
prime factorization
greatest common factor (GCF)
least common multiple (LCM)

Answer:
1. A perfect square is a number, from a given number system, that can be expressed as the square of a number from the same number system. Examples of Numbers that are Perfect Squares. 25 is a perfect square.
2. A numerical expression is a mathematical sentence involving only numbers and one or more operation symbols. Examples of operation symbols are the ones for addition, subtraction, multiplication, and division.
3. Order of operations refers to which operations should be performed in what order, but it’s just convention.
4. “Prime Factorization” is finding which prime numbers multiply together to make the original number.
5. Greatest Common Factor. The highest number that divides exactly into two or more numbers.
6. Least Common Multiple. The smallest positive number that is a multiple of two or more numbers.

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 86$

1.1 Powers and Exponents (pp. 3-8)

Write the product as a power.

Question 1.
3 × 3 × 3 × 3 × 3 × 3

Answer: The product of 3 × 3 × 3 × 3 × 3 × 3 is 3⁶

Question 2.
5 × 5 × 5

Answer: The product of 5 × 5 × 5 is 5³

Question 3.
17 . 17 . 17 . 17 . 17

Answer: The product of 17 . 17 . 17 . 17 . 17 is 17⁵

Question 4.
3³

Answer: 3 × 3 × 3

Question 5.
2⁶

Answer: 2 × 2 × 2 × 2 × 2 × 2

Question 6.
4⁴

Answer: 4 × 4 × 4 × 4

Question 7.
Write a power that has a value greater than 2³ and less than 3³.

Answer: The power that has a value greater than 2³ and less than 3³ is 4²

Question 8.
Without evaluating, determine whether 2⁵ or 4² is greater. Explain.

Answer: 2⁵ > 4²

Explanation:
The exponent with the highest number will be greater.

Question 9.
The bases on a softball ﬁeld are square. What is the area of each base?
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 87$
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 88$

Answer:
Given,
The bases on a softball ﬁeld are square.
s = 15 inches
We know that,
Area of the square = s × s
A = 15 × 15
A = 225 sq. in
Thus the area of each base is 225 sq. in.

1.2 Order of Operations (pp. 9–14)

Evaluate the expression.

Question 10.
3 × 6 – 12 ÷ 6

Answer: 16

Explanation:
You have to evaluate from left to right.
(3 × 6) – (12 ÷ 6)
18 – 2 = 16

Question 11.
30 ÷ (14 – 2²) × 5

Answer: 15

Explanation:
You have to evaluate from left to right.
30 ÷ (14 – 4) × 5
30 ÷ 10 × 5
3 × 5 = 15

Question 12.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 89$

Answer: 15

Explanation:
You have to evaluate from left to right.
2.3 + 3.7 = 6
5(6)/2 = 30/2 = 15

Question 13.
$Big Ideas Math Answers 6th Grade Chapter 1 Numerical Expressions and Factors 90$

Answer: 37

Explanation:
You have to evaluate from left to right.
7² + 5 = 49 + 5 = 54
1/2 × 54 = 27
4³ – 27 = 64 – 27 = 37

Question 14.
20 (3² – 4) ÷ 50

Answer: 2

Explanation:
You have to evaluate from left to right.
(3² – 4) = 9 – 4 = 5
20 × 5 ÷ 50
100 ÷ 50 = 2

Question 15.
5 + 3(4² – 2) ÷ 6

Answer: 12

Explanation:
You have to evaluate from left to right.
(4² – 2) = 16 – 2 = 14
5 + 3(14) ÷ 6
5 + 42 ÷ 6
5 + 7 = 12

Question 16.
Use grouping symbols and at least one exponent to write a numerical expression that has a value of 80.

Answer: 6 + (9² – 7) = 80

1.3 Prime Factorization (pp. 15–20)

List the factor pairs of the number.

Question 17.
28

Answer: The factor pairs of the number 28 are 1, 2, 4, 7, 14, 28

Explanation:
28 = 1 × 28
2 × 14
4 × 7
7 × 4
14 × 2
28 × 1

Question 18.
44

Answer: The factor pairs of the number 44 are 1, 2, 4, 11, 44.

Explanation:
44 = 1 × 44
2 × 22
4 × 11
11 × 4
44 × 1

Question 19.
96

Answer: The factor pairs of the number 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96

Explanation:
1 and 96 are a factor pair of 96 since 1 x 96= 96
2 and 48 are a factor pair of 96 since 2 x 48= 96
3 and 32 are a factor pair of 96 since 3 x 32= 96
4 and 24 are a factor pair of 96 since 4 x 24= 96
6 and 16 are a factor pair of 96 since 6 x 16= 96
8 and 12 are a factor pair of 96 since 8 x 12= 96
12 and 8 are a factor pair of 96 since 12 x 8= 96
16 and 6 are a factor pair of 96 since 16 x 6= 96
24 and 4 are a factor pair of 96 since 24 x 4= 96
32 and 3 are a factor pair of 96 since 32 x 3= 96
48 and 2 are a factor pair of 96 since 48 x 2= 96
96 and 1 are a factor pair of 96 since 96 x 1= 96

Question 20.
There are 36 graduated cylinders to put away on a shelf after science class. The shelf can ﬁt a maximum of 20 cylinders across and 4 cylinders deep. The teacher wants each row to have the same number of cylinders. List the possible arrangements of the graduated cylinders on the shelf.
$Big Ideas Math Answers Grade 6 Chapter 1 Numerical Expressions and Factors 91$

Answer:
Given,
There are 36 graduated cylinders to put away on a shelf after science class.
The shelf can ﬁt a maximum of 20 cylinders across and 4 cylinders deep.
The teacher wants each row to have the same number of cylinders.
There are three possible arrangements of the graduated cylinders on the shelf.
1. 4 rows of 9 graduated cylinders
2. 3 rows of 12 graduated cylinders.
3. 2 rows of 18 graduated cylinders.

Write the prime factorization of the number.

Question 21.
42

Answer: The prime factorization of the number 42 is 2 × 3 × 7

Explanation:
42 = 2 × 21
= 2 × 3 × 7

Question 22.
50

Answer: The prime factorization of the number 2 × 5 × 5

Explanation:
50 = 2 × 25
= 2 × 5 × 5

Question 23.
66

Answer: The prime factorization of the number 2 × 3 × 11

Explanation:
66 = 2 × 33
= 2 × 3 × 11

1.4 Greatest Common Factor (pp. 21–26)

Find the GCF of the numbers using lists of factors.

Question 24.
27, 45

Answer: GCF is 9

Explanation:
The factors of 27 are: 1, 3, 9, 27
The factors of 45 are: 1, 3, 5, 9, 15, 45
Then the greatest common factor is 9.

Question 25.
30, 48

Answer: GCF is 6

Explanation:
The factors of 30 are: 1, 2, 3, 5, 6, 10, 15, 30
The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Then the greatest common factor is 6.

Question 26.
28, 48

Answer: GCF is 4

Explanation:
The factors of 28 are: 1, 2, 4, 7, 14, 28
The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Then the greatest common factor is 4.

Find the GCF of the numbers using prime factorizations.

Question 27.
24, 80

Answer:
The prime factorization is the product of the circled primes. So the prime factorization of 24 is 24 = 2 · 2 · 2 · 3 = 2³ . 3
The prime factorization is the product of the circled primes. So the prime factorization of 80 is 80 = 2 x 2 x 2 x 2 x 5 = 2² . 2² . 5

Question 28.
52, 68

Answer:
The prime factorization is the product of the circled primes. So the prime factorization of 52 is 2 x 2 x 13 = 2² . 13
The prime factorization is the product of the circled primes. So the prime factorization of 68 is 2 × 2 × 17 = 2². 17

Question 29.
32, 56

Answer:
The prime factorization is the product of the circled primes. So the prime factorization of 32 is 2 x 2 x 2 x 2 x 2 = 2⁵
The prime factorization is the product of the circled primes. So the prime factorization of 56 is 2 x 2 x 2 x 7 = 2³ . 7

Question 30.
Write a pair of numbers that have a GCF of 20.

Answer: The prime factors of 20 are 2 x 2 x 5. The GCF of 20 is 5.

Question 31.
What is the greatest number of friends you can invite to an arcade using the coupon such that the tokens and slices of pizza are equally split between you and your friends with none left over? How many slices of pizza and tokens will each person receive?
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 92$

Answer: (n/4)-1

Explanation:
Total slices = n
Total number of people =4
Each people may be eat = n/4 slices
Here Harris eats 1 slice fewer
Then Harris eats (n/4)-1 slices

1.5 Least Common Multiple (pp. 27–32)

Find the LCM of the numbers using lists of multiples.

Question 32.
4, 14

Answer: 28

Explanation:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 4:
4, 8, 12, 16, 20, 24, 28, 32, 36
Multiples of 14:
14, 28, 42, 56
Therefore,
LCM(4, 14) = 28

Question 33.
6, 20

Answer: 60

Explanation:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 6:
6, 12, 18, 24, 30, 36, 42, 48, 54, 60.
Multiples of 20:
20, 40, 60, 80, 100
The LCM of 6, 20 is 60

Question 34.
12, 28

Answer: 84

Explanation:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 12:
12, 24, 36, 48, 60, 72, 84, 96
Multiples of 28:
28, 56, 84, 112, 140, 168, 196
LCM of 12, 28 is 84

Find the LCM of the numbers using prime factorizations.

Question 35.
6, 45

Answer:
Prime Factorisation of 6: 2 × 3
Prime Factorisation of 45: 3 × 3 × 5
LCM is 2 × 3 × 3 × 3 × 5 = 60

Question 36.
10, 12

Answer: 60
Prime factorization of 10: 2 × 5
Prime factorization of 12: 2 × 2 × 3
LCM is 5 × 2 × 2 × 3 = 60

Question 37.
18, 27

Answer:
Prime factorization of 18: 2 × 3 × 3
Prime factorization of 27: 3 × 3 × 3
LCM is 2 × 3 × 3 × 3 = 54

Question 38.
Find the LCM of 8, 12, and 18.

Answer: 72
Prime Factorisation of 8: 2 × 2 × 2
Prime Factorisation of 12: 2 × 2 × 3
Prime factorization of 18: 2 × 3 × 3
LCM = 72

Question 39.
Write a pair of numbers that have an LCM of 84.

Answer: 84 and 12

Explanation:
The LCM of 84 and 12 is 84.
Prime factorization of 12: 2 × 2 × 3
Prime factorization of 84: 2 × 2 × 3 × 7
The Least Common Multiple is 2 × 2 × 3 × 7 = 84

Question 40.
Write three numbers that have an LCM of 45.

Answer: 3, 15, 45

Explanation:
The prime factorization of 15: 3 × 5
The prime factorization of 45: 3 × 3 × 5
The LCM of 3, 15, 45 is 45.

Question 41.
You water your roses every sixth day and your hydrangeas every ﬁfth day. Today you water both plants. In how many days will you water both plants on the same day again?
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 93$

Answer: 30

Explanation:
Given,
You water your roses every sixth day and your hydrangeas every ﬁfth day. Today you water both plants.
6 × 5 = 30
You water both plants for 30 days on the same day again.

Question 42.
Hamburgers are sold in packages of 20, while buns are sold in packages of 12. What are the least numbers of packages you should buy in order to have the same number of hamburgers and buns?

Answer:
Given,
Hamburgers are sold in packages of 20, while buns are sold in packages of 12.
At least 5 packages of buns and 3 packages of hamburgers.
20×3=60
12×5=60
So that is how you get the answer by seeing if they have any integers in common.

Question 43.
A science museum is giving away a magnetic liquid kit to every 50th guest and a plasma ball to every 35th guest until someone receives both prizes.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 94$
a. Which numbered guest will receive both a magnetic liquid kit and a plasma ball?

Answer:
A magnetic liquid kit prize every 50 guests and a plasma ball every 35 guests.
1.Guest 50th
2.Guest 100th
3.Guest 150th
4.Guest 200th
5.Guest 250th
6.Guest 300th
7.Guest 350th
8.Guest 400th and so on, in case no coincidence would happen.

b. How many people will receive a plasma ball?

Answer:
1.Guest 35th
2.Guest 70th
3.Guest 105th
4.Guest 140th
5.Guest 175th
6.Guest 210th
7.Guest 245th
8.Guest 280th
9.Guest 315th
10.Guest 350th.
As you can see, Guest 350th will be the first one to receive both prizes, and including him or her, a total of ten guests will receive the plasma ball until that moment. There wasn’t any coincidence before Guest 350th.

Numerical Expressions and Factors Practice Test

Question 1.
Find the value of 2³.

Answer:
2³ can be written as 2 × 2 × 2 = 8
Thus the value of 2³is 8.

Question 2.
Evaluate $Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 95$

Answer:
5 + 4(12 – 2) = 5 + 4(10) = 5 + 40
(5 + 40)/3² = 45/9 = 5
Thus the value of $Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 95$ is 5.

Question 3.
Write 264.264.264 as a power

Answer: 264.264.264 can be written as 264³

Question 4.
List the factor pairs of 66.

Answer: The factor pairs of 66 are (1,66) (2, 33) (6, 11)
66 = 1 × 66
66 = 2 × 33
66 = 6 × 11

Question 5.
Write the prime factorization of 56.

Answer:
56 = 2 × 28
= 2 × 2 × 14
= 2 × 2 × 2 × 7
Thus the prime factorization of 56 is 2 × 2 × 2 × 7

Find the GCF of the numbers.

Question 6.
24, 54

Answer: GCF is 6

The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24
The factors of 54 are: 1, 2, 3, 6, 9, 18, 27, 54
Then the greatest common factor is 6.

Question 7.
16, 32, 72

Answer: GCF is 8

The factors of 16 are: 1, 2, 4, 8, 16
The factors of 32 are: 1, 2, 4, 8, 16, 32
The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Then the greatest common factor is 8.

Question 8.
52, 65

Answer: GCF is 13

The factors of 52 are: 1, 2, 4, 13, 26, 52
The factors of 65 are: 1, 5, 13, 65
Then the greatest common factor is 13.

Find the LCM of the numbers.

Question 9.
9, 24

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 9:
9, 18, 27, 36, 45, 54, 63, 72, 81, 90
Multiples of 24:
24, 48, 72, 96, 120
Therefore,
LCM(9, 24) = 72

Question 10.
26, 39

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 26:
26, 52, 78, 104, 130
Multiples of 39:
39, 78, 117, 156
Therefore,
LCM(26, 39) = 78

Question 11.
6, 12, 14

Answer:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 6:
6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96
Multiples of 12:
12, 24, 36, 48, 60, 72, 84, 96, 108
Multiples of 14:
14, 28, 42, 56, 70, 84, 98, 112
Therefore,
LCM(6, 12, 14) = 84

Question 12.
You have 16 yellow beads, 20 red beads, and 24 orange beads to make identical bracelets. What is the greatest number of bracelets that you can make using all of the beads?

Answer:
To find how many identical bracelets you can make, you need to find a common denominator.
In this case, all three numbers; 16, 20, and 24, can be divided by four.
So you now know you can have four bracelets.
Then you take your numbers of each color beads and divide them by four so you know how many of each color will be on the bracelets.
In the end, you have four bracelets, each with 4 yellow beads, 5 red beads and 6 orange beads

Question 13.
A bag contains equal numbers of green marbles and blue marbles. You can divide all of the green marbles into groups of 12 and all the blue marbles into groups of 16. What is the least number of each color of marble that can be in the bag?

Answer:
Given,
A bag contains equal numbers of green marbles and blue marbles.
You can divide all of the green marbles into groups of 12 and all the blue marbles into groups of 16.
To solve this problem, we need to find for the LCM of each number. That is:
12: 12, 24, 36, 48, 60
16: 16, 32, 48, 64, 80
So we can see that the LCM is 48.
Therefore the least number of each color of marble must be 48.

Question 14.
The ages of the members of a family are 65, 58, 27, 25, 5, and 2 years old. What is the total admission price for the family to visit the zoo?
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 96$

Answer:
The ages of the members of a family are 65, 58, 27, 25, 5, and 2 years old.
We can find the total admission price for the family to visit the zoo by following the above table.
$10 + $12 + $27 + $12 + $8 + $8 = $77

Question 15.
A competition awards prizes for fourth, third, second, and ﬁrst place. The fourth place winner receives $5. Each place above that receives a prize that is ﬁve times the amount of the previous prize. How much prize money is awarded?

Answer:
A competition awards prizes for fourth, third, second, and ﬁrst place. The fourth place winner receives $5. Each place above that receives a prize that is ﬁve times the amount of the previous prize.
Each place above that receives a prize that is five times the amount of the previous prize
So we can say that;
Pn = 5 × P(n+1)
Where n ⇒ Number Place
Pn = Price received by Number place
Substituting the values of n as 3,2,1 to find the price of third second first place winner.
P3 = 5 × P(3+1) = 5 × P4 = 5 × 5 = 25
P2 = 5 × P(2+1) = 5 × P3 = 5 × 25 = 125
P1 = 5 × P(1+1) = 5 × P2 = 5 × 125 = 625
Now We will find the Total Prize money awarded.
Total Prize money awarded = 625 + 125 + 25 + 5 = 780
Hence A total of $780 price money was awarded.

Question 16.
You buy tealight candles and mints as party favors for a baby shower. The tealight candles come in packs of 12 for $3.50. The mints come in packs of 50 for $6.25. What is the least amount of money you can spend to buy the same number of candles and mints?
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 96.1$

Answer: The least amount of money it can be spent is $125.

Explanation:
First, we write the prime factorization of each number:
12= 2·2·3
15= 2·5·5
Then, we search for each different factor which appears the greater number of times. The factor 2 appears in both factorizations so the least common multiple is:
LCM= 2·2·3·5·5=300
Hence, the total quantity of packs of each thing is:
Candles: 300÷12=25
Mints: 300÷50=6
The least amount of money it can be spent is:
T=25×$3.50 + 6×$6.25= $87.5 + $37.5= $125

Numerical Expressions and Factors Cumulative Practice

Question 1.
Find the value of 8 × 135?
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 96.2$
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 97$

Answer: Multiply the two numbers 8 and 135
8 × 135 = 1080

Question 2.
Which number is equivalent to the expression blow?
3.2³ – 8 ÷ 4

Answer: 22

Explanation:
Given the expression 3.2³ – 8 ÷ 4
3 × 8 -(8÷4)
24 – (2)
24 – 2 = 22

Question 3.
The top of an end table is a square with a side length of 16 inches. What is the area of the tabletop?
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 98$

Answer: I

Explanation:
Given that
The top of an end table is a square with a side length of 16 inches.
Area of the square = s × s
A = 16 × 16
A = 256 in²
Thus the correct answer is option I.

Question 4.
You are ﬁlling baskets using 18 green eggs, 36 red eggs, and 54 blue eggs. What is the greatest number of baskets that you can ﬁll so that the baskets are identical and there are no eggs left over?
A. 3
B. 6
C. 9
D. 18

Answer: D

Explanation:
Given,
You are ﬁlling baskets using 18 green eggs, 36 red eggs, and 54 blue eggs.
18/n = 36/n = 54/n
Factors of 18 are 2,3,6,9,18
Factors of 36 are 2,3,4,6,9,12,18.
Factors of 54 are 2,3,6,9,18,27,54.
The common multiples of 18,36 and 54 are 2,3,6,9,18.
Thus the greatest among them is 18.
Thus the correct answer is option D.

Question 5.
What is the value of 2³.3².5?
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 99$

Answer:
2³.3².5
2³ = 8
3² = 9
8 × 9 × 5 = 360

Question 6.
You hang the two strands of decorative lights shown below.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 100$
Both strands just changed color. After how many seconds will the strands change color at the same time again?
F. 3 seconds
G. 30 seconds
H. 90 seconds
I. 270 seconds

Answer: 3 seconds

Explanation:

Strand I: Changes between red and blue every 15 seconds
Strand II: Changes between green and gold every 18 seconds
18 – 15 = 3 seconds
Thus the correct answer is option F.

Question 7.
Point P is plotted in the coordinate plane below.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 101$
What are the coordinates of Point P ?
A. (5, 3)
B. (4, 3)
C. (3, 5)
D. (3, 4)

Answer: C
By seeing the above graph we can find the coordinates of point p.
The X-axis is on 3 and Y-axis is on 5.
Thus the correct answer is option c.

Question 8.
What is the prime factorization of 1100?
F. 2 × 5 × 11
G. 2² × 5² × 11
H. 4 × 5² × 11
I. 2² × 5 × 55

Answer: G
Prime factorization of 1100: 2 × 550
2 × 2 × 275
2 × 2 × 5 × 55
2 × 2 × 5 × 5 × 11
Thus the correct answer is option b.

Question 9.
What is the least common multiple of 3, 8, and 10?
A. 24
B. 30
C. 80
D. 120

Answer: D

Multiples of 3:
3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126
Multiples of 8:
8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 136
Multiples of 10:
10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 120, 140.
The common multiple among the three is 120.
Thus the correct answer is option D.

Question 10.
What is the area of the shaded region of the ﬁgure below?
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 102$

Answer:
The above figure is square.
s = 4 yd
Area of the square = s × s
A = 4 yd × 4 yd
A = 16 sq. yd
The area of the outer box.
s = 9 yd
Area of the square = s × s
A = 9 yd × 9 yd
A = 81 sq. yd
The area of the shaded region is 81 – 16 = 65 sq. yd
Thus the correct answer is option G.

Question 11.
Which expression represents a prime factorization?
A. 4 × 4 × 7
B. 2² × 21 × 23
C. 3⁴ × 5 × 7
D. 5 × 5 × 9 × 11

Answer: B

Prime factorization:
2² × 21 × 23
2, 21, 23 is a prime number.
Thus the correct answer is option B.

Question 12.
Find the greatest common factor for each pair of numbers.
$Big Ideas Math Answer Key Grade 6 Chapter 1 Numerical Expressions and Factors 103$
What can you conclude about the greatest common factor of 10, 15, and 21? Explain your reasoning.

Answer: 1

Explanation:
The factors of 10 are: 1, 2, 5, 10
The factors of 15 are: 1, 3, 5, 15
The factors of 21 are: 1, 3, 7, 21
Then the greatest common factor is 1.

Final Words:
I wish that the details prevailed in this article regarding Big Ideas Math Grade 6 Chapter 1 Numerical Expressions and Factors Answer Key is helpful for all the students and also teachers. Make use of the solutions and score good marks in the exams. Best Of Luck!!

Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions

January 4, 2021

Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions: Free step by step solutions to Big Ideas Math Grade 4 Chapter 8 Add and Subtract Fractions is available here. The Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions are given in the pdf format. So Download Big ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions pdf for free. By referring to the BIM 4th Grade Chapter 8 Add & Subtract Fractions you can finish your assignment and homework in time.

Big Ideas Math Book 4th Grade Answer Key Chapter 8 Add and Subtract Fractions

Practice Big Ideas Math Chapter 8 Add and Subtract Fractions Questions you can score good marks in the exams. We have provided the solutions according to the topics. This chapter includes Decompose Fractions, Add Fractions with Like Denominators, Subtract Fractions with Like Denominators, Add Mixed Numbers, Subtract Mixed Numbers, etc. You can get the solutions for all the topics in an easy manner in different methods. Tap the links and start practicing the problems and improve your math skills.

Lesson 1: Use Models to Add Fractions

Lesson 8.1 Use Models to Add Fractions
Use Models to Add Fractions Homework & Practice 8.1

Lesson 2: Decompose Fractions

Lesson 8.2 Decompose Fractions
Decompose Fractions Homework & Practice 8.2

Lesson 3: Add Fractions with Like Denominators

Lesson 8.3 Add Fractions with Like Denominators
Add Fractions with Like Denominators Homework & Practice 8.3

Lesson 4: Use Models to subtract Fractions

Lesson 5: Subtract Fractions with Like Denominators

Lesson 8.5 Subtract Fractions with Like Denominators
Subtract Fractions with Like Denominators Homework & Practice 8.5

Lesson 6: Model Fractions and Mixed Numbers

Lesson 8.6 Model Fractions and Mixed Numbers
Model Fractions and Mixed Numbers Homework & Practice 8.6

Lesson 7: Add Mixed Numbers

Lesson 8.7 Add Mixed Numbers
Add Mixed Numbers Homework & Practice 8.7

Lesson 8: Subtract Mixed Numbers

Lesson 8.8 Subtract Mixed Numbers
Subtract Mixed Numbers Homework & Practice 8.8

Lesson 9: Problem Solving: Fractions

Lesson 8.9 Problem Solving: Fractions
Problem Solving: Fractions Homework & Practice 8.9

Performance Task

Add and Subtract Fractions Performance Task 8
Add and Subtract Fractions Activity
Add and Subtract Fractions Chapter Practice 8

Lesson 8.1 Use Models to Add Fractions

Explore and Grow

Draw models to show $\frac{2}{8}$ and $\frac{5}{8}$.

Answer:
You can add fractions by joining parts that refer to the same whole.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_1$

Use your models to find $\frac{2}{8}$ + $\frac{5}{8}$. Explain your method.

Answer:
Combine the like terms
$\frac{2}{8}$ + $\frac{5}{8}$ = (2 + 5)/8 = 7/8

Repeated Reasoning
Write two fractions that have a sum of $\frac{6}{8}$. Explain your reasoning.

Think and Grow: Use Models to Add Fractions

You can add fractions by joining parts that refer to the same whole.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 1$
Answer:
The denominators of the fraction are the same so you have to add numerators.
$\frac{1}{5}$ + $\frac{3}{5}$ = $\frac{4}{5}$

Example
Use a number line to find $\frac{5}{4}$ + $\frac{2}{4}$.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 2$
Answer:
The denominators of the fraction are the same so you have to add numerators.
$\frac{5}{4}$ + $\frac{2}{4}$ = $\frac{7}{4}$

Show and Grow

Find the sum. Explain how you used the model to add.

Question 1.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 3$

Answer:
The denominators of the fraction are the same so you have to add numerators.
$Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-3$
(3+4)/10 = 7/10

Question 2.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 4$

Answer:
The denominators of the fraction are the same so you have to add numerators.
$Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-4$

Apply and Grow: Practice

Find the sum. Use a model or a number line to help.

Question 3.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 5$

Answer:
The denominators of the fraction are the same so you have to add numerators.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_2$

Question 4.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 6$

Answer:
The denominators of the fraction are the same so you have to add numerators.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_3$
$\frac{5}{12}$ + $\frac{4}{12}$ = $\frac{9}{12}$

Question 5.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 7$

Answer:
The denominators of the fraction are the same so you have to add numerators.
$Big Ideas Math Book 4th Grade Answer Key Chapter 8 Add and Subtract Fractions img_5$

Question 6.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 8$

Answer:
The denominators of the fraction are the same so you have to add numerators.
$Big Ideas Math Book 4th Grade Answer Key Chapter 8 Add and Subtract Fractions img_4$

Question 7.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 9$

Answer:
The denominators of the fraction are the same so you have to add numerators.
$Big Ideas Math Book 4th Grade Answer Key Chapter 8 Add and Subtract Fractions img_6$

Question 8.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 10$

Answer:
Add a fraction to the whole number.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_7$
5 + $\frac{6}{8}$ = $\frac{46}{8}$

Question 9.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 11$

Answer:
The denominators of the fraction are the same so you have to add numerators.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_8$

Question 10.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 12$

Answer:
The denominators of the fraction are the same so you have to add numerators.
$Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Fractions-img_10$

Question 11.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 13$

Answer:
The denominators of the fraction are the same so you have to add numerators.
$Big Ideas Math Book 4th Grade Answer Key Chapter 8 Add and Subtract Fractions img_11$

Question 12.
Structure
Write the addition equation represented by the models.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 14$

Answer:
By seeing the above model we can find the addition equation.
$\frac{4}{8}$ + $\frac{3}{8}$ = $\frac{7}{8}$

Question 13.
Open-Ended
Write three fractions with different numerators that have a sum of 1.

Answer:
$\frac{2}{8}$ + $\frac{5}{8}$ + $\frac{1}{8}$ = $\frac{8}{8}$ = 1

Question 14.
Writing
Explain why $\frac{1}{8}$ + $\frac{4}{8}$ does not equal $\frac{5}{16}$.

Answer:
In the above expressions, the denominators are the same but the numerators are different.
So, you have to add the numerators not denominators.
$\frac{1}{8}$ + $\frac{4}{8}$ = $\frac{5}{8}$

Think and Grow: Modeling Real Life

Example
You need $\frac{2}{3}$ cup of hot water and $\frac{4}{3}$ cups of cold water for a science experiment. How many cups of water do you need in all?
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 15$
Because each fraction represents a part of the same whole you can join the parts.
Use a model to find $\frac{2}{3}$ + $\frac{4}{3}$.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 16$

Answer:
Given that,
You need $\frac{2}{3}$ cup of hot water and $\frac{4}{3}$ cups of cold water for a science experiment.

$Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-16$
Thus you need 2 cups of water in all.

Show and Grow

Question 15.
You cut a foam noodle for a craft. You use $\frac{2}{4}$ of the noodle for one part of the craft and $\frac{1}{4}$ of the noodle for another part. What fraction of the foam noodle do you use altogether?
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 17$

Answer:
Given that,
You cut a foam noodle for a craft. You use $\frac{2}{4}$ of the noodle for one part of the craft and $\frac{1}{4}$ of the noodle for another part.
$\frac{2}{4}$ + $\frac{1}{4}$ = $\frac{3}{4}$
Thus $\frac{3}{4}$ of the foam noodle is used.

Question 16.
You make a fruit drink using $\frac{4}{8}$ gallon of orange juice, $\frac{2}{8}$ gallon of mango juice, and $\frac{4}{8}$ gallon of pineapple juice. How much juice do you use in all?

Answer:
Given that,
You make a fruit drink using $\frac{4}{8}$ gallon of orange juice, $\frac{2}{8}$ gallon of mango juice, and $\frac{4}{8}$ gallon of pineapple juice.
$\frac{4}{8}$ + $\frac{2}{8}$ = $\frac{6}{8}$
$\frac{6}{8}$ + $\frac{4}{8}$ = $\frac{10}{8}$
Thus you used $\frac{10}{8}$ fraction of juice.

Question 17.
DIG DEEPER!
A community plants cucumbers in $\frac{5}{12}$ of a garden, broccoli in $\frac{3}{12}$ of the garden, and carrots in $\frac{4}{12}$ of the garden. What fraction of the garden is planted with green vegetables?

Answer:
Given that,
A community plants cucumbers in $\frac{5}{12}$ of a garden, broccoli in $\frac{3}{12}$ of the garden, and carrots in $\frac{4}{12}$ of the garden.
$\frac{5}{12}$ + $\frac{3}{12}$ + $\frac{4}{12}$ = $\frac{12}{12}$ = 1
$\frac{12}{12}$ fraction of the garden is planted with green vegetables.

Use Models to Add Fractions Homework & Practice 8.1

Find the sum. Explain how you used the model to add.

Question 1.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 18$

Answer: $\frac{9}{6}$
$Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-18$

Question 2.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 19$

Answer: 1
$Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-19$

Find the sum. Use a model or a number line to help.

Question 3.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 20$

Answer: $\frac{7}{8}$
You can add fractions by joining parts that refer to the same whole.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_1$

Question 4.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 21$

Answer: 2
$Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Fractions-img_20$

Question 5.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 22$

Answer: 3 $\frac{1}{4}$

Explanation:
Add fraction to the whole number.
$\frac{1}{4}$ + 3
$\frac{1}{4}$ + 3 × $\frac{4}{4}$
$\frac{1}{4}$ + $\frac{13}{4}$ = $\frac{13}{4}$

Question 6.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 23$

Answer: $\frac{9}{12}$
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_21$

Question 7.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 24$

Answer: 2
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_22$
$\frac{12}{10}$ = $\frac{10}{10}$ + $\frac{2}{10}$
$\frac{10}{10}$ + $\frac{2}{10}$ + $\frac{8}{10}$ = $\frac{20}{10}$ = 2

Question 8.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 25$

Answer:
4/8 = 1/2
6 + 1/2 = (12 + 1)/2 = 13/2

Find the sum. Use a model or a number line to help.

Question 9.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 26$

Answer:
Add all the three unit fractions.
BIM Grade 4 Chapter 8 Add and Subtract Fractions img_23

Question 10.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 27$

Answer:
$Big Ideas Math Answers 4th Grade Chapter 8 img_24$

Question 11.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 28$

Answer:
The denominators of all the fractions are the same. So you have to add the numerators of the fraction.
$\frac{50}{100}$ + $\frac{25}{100}$ + $\frac{5}{100}$ = $\frac{80}{100}$

Question 12.
YOU BE THE TEACHER
Newton says $\frac{3}{5}$ + $\frac{1}{5}$ = $\frac{4}{10}$. Descartes says the sum is $\frac{4}{5}$. Who is correct? Explain.

Answer:
Newton says $\frac{3}{5}$ + $\frac{1}{5}$ = $\frac{4}{10}$. Descartes says the sum is $\frac{4}{5}$.
Descartes is correct.
$\frac{3}{5}$ + $\frac{1}{5}$ = $\frac{4}{5}$
You have to add numerators, not denominators.
So, Newton’s equation is not correct.

Question 13.
Make each statement true by writing two fractions whose denominators are the same and whose numerators are 3 and 2.
The sum of ___ and __ is greater than 1.
___________________________
The sum of ___ and ___ is less than 1.
___________________________
The sum of ___ and ___ is equal to 1.

Answer:
The sum of 3/2 and 2/2 is greater than 1.
3/2 + 2/2 = 5/2
5/2 > 1
The sum of 3/6 and 2/6 is less than 1.
3/6 + 2/6 = 5/6
5/6 < 1
The sum of 3/5 and 2/5 is equal to 1.
3/5 + 2/5 = 5/5 = 1

Question 14.
Modeling Real Life
Your teacher assigns 5 pages to read. You read $\frac{3}{5}$ of the pages in class and $\frac{1}{5}$ of the pages at home. What fraction of the reading assignment is complete?

Answer:
Given that,
Your teacher assigns 5 pages to read. You read $\frac{3}{5}$ of the pages in class and $\frac{1}{5}$ of the pages at home.
The denominators of all the fractions are the same. So you have to add the numerators of the fraction.
$\frac{3}{5}$ + $\frac{1}{5}$ = $\frac{4}{5}$

Question 15.
Modeling Real Life
In the Sahara Desert, it rains $\frac{2}{10}$ inch in September, $\frac{3}{10}$ inch in October, and $\frac{5}{10}$ inch in November. How much does it rain in the 3 months?

Answer:
Given that,
In the Sahara Desert, it rains $\frac{2}{10}$ inch in September, $\frac{3}{10}$ inch in October, and $\frac{5}{10}$ inch in November.
$\frac{2}{10}$ + $\frac{3}{10}$ + $\frac{5}{10}$
The denominators of all the fractions are the same. So you have to add the numerators of the fraction.
$\frac{2}{10}$ + $\frac{3}{10}$ + $\frac{5}{10}$ = (2 + 3 + 5)/10 = 10/10 = 1
It rains 10/10 in the 3 months.

Review & Refresh

Tell whether the number is prime or composite. Explain.

Question 16.
37

Answer: 37 is a prime number.
A prime number is a natural number greater than 1 that is not a product of two smaller natural numbers. A natural number greater than 1 that is not prime is called a composite number.

Question 17.
21

Answer: 21 is a composite number.
A composite number is a positive integer that can be formed by multiplying two smaller positive integers. Equivalently, it is a positive integer that has at least one divisor other than 1 and itself.

Question 18.
99

Answer: 99 is a composite number.
A composite number is a positive integer that can be formed by multiplying two smaller positive integers. Equivalently, it is a positive integer that has at least one divisor other than 1 and itself.

Lesson 8.2 Decompose Fractions

Explore and Grow

Use a model to find $Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 29$ .

How can you write $\frac{7}{10}$ as a sum of unit fractions? Explain your reasoning.

Answer: The sum of unit fraction of $\frac{7}{10}$ is $\frac{1}{10}$ + $\frac{1}{10}$ + $\frac{1}{10}$ + $\frac{1}{10}$ + $\frac{1}{10}$ + $\frac{1}{10}$ + $\frac{1}{10}$

Structure
Explain how you can write $\frac{7}{10}$ as a sum of two fractions. Draw a model to support your answer.

Answer: You can write sum of $\frac{7}{10}$ as $\frac{2}{10}$ + $\frac{5}{10}$
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_23$

Think and Grow: Decompose Fractions

A unit fraction represents one equal part of a whole. You can write a fraction as a sum of unit fractions.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 31$
Answer:
$Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-31$
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 31.1$

Answer:
$Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-31.1$

Show and Grow

Question 1.
Write $\frac{4}{5}$ as a sum of unit fractions.

Answer: The unit fraction of $\frac{4}{5}$ is $\frac{1}{5}$ + $\frac{1}{5}$ + $\frac{1}{5}$ + $\frac{1}{5}$
A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer.

Question 2.
Write $\frac{5}{6}$ as a sum of fractions in two different ways.

Answer: The unit fraction of $\frac{5}{6}$ is $\frac{1}{6}$ + $\frac{1}{6}$ + $\frac{1}{6}$ + $\frac{1}{6}$ + $\frac{1}{6}$
A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer.
You can also write $\frac{5}{6}$ as $\frac{2}{6}$ + $\frac{3}{6}$
That means $\frac{5}{6}$ can be written as 2 parts of $\frac{1}{6}$ and 3 parts of $\frac{1}{6}$

Apply and Grow: Practice

Question 3.
$\frac{4}{7}$

Answer: The unit fraction of $\frac{4}{7}$ is $\frac{1}{7}$ + $\frac{1}{7}$ + $\frac{1}{7}$ + $\frac{1}{7}$
A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer.

Question 4.
$\frac{7}{8}$

Answer: The unit fraction of $\frac{7}{8}$ is $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$
A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer.

Question 5.
$\frac{3}{10}$

Answer: The unit fraction of $\frac{3}{10}$ is $\frac{1}{10}$ + $\frac{1}{10}$ + $\frac{1}{10}$
A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer.

Question 6.
$\frac{10}{100}$

Answer: The unit fraction of $\frac{10}{100}$ is $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$
A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer.

Question 7.
$\frac{6}{2}$

Answer: 3
The unit fraction of $\frac{6}{2}$ is $\frac{1}{2}$ + $\frac{1}{2}$ + $\frac{1}{2}$ + $\frac{1}{2}$ + $\frac{1}{2}$ + $\frac{1}{2}$
A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer.

Question 8.
$\frac{9}{4}$

Answer:
Break apart 9 parts of $\frac{1}{4}$ into 5 parts of $\frac{1}{4}$ and 4 parts of $\frac{1}{4}$.

Question 9.
$\frac{8}{12}$

Answer: Break apart 8 parts of $\frac{1}{12}$ into 5 parts of $\frac{1}{12}$ and 3 parts of $\frac{1}{12}$.

Question 10.
$\frac{5}{3}$

Answer: The unit fraction of $\frac{5}{3}$ is $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$
A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer.

Question 11.
Writing
You write $\frac{4}{6}$ as a sum of unit fractions. Explain how the numerator of $\frac{4}{6}$ is related to the number of addends.

Answer: The unit fraction of $\frac{4}{6}$ is $\frac{1}{6}$ + $\frac{1}{6}$ + $\frac{1}{6}$ + $\frac{1}{6}$
A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer.
Also, you can write $\frac{4}{6}$ as 4 parts of $\frac{1}{6}$, 2 equal parts of $\frac{1}{6}$ and 2 equal parts of $\frac{1}{6}$.

Question 12.
DIG DEEPER!
Why is it important to be able to write a fraction as a sum of fractions in different ways?

Answer:
Asking students to write a fraction as a sum of unit fractions, or as a sum of other fractions, encourages students to make sense of quantities and their relationships. Students further develop their understandings about fractions and decomposing numbers through this process.

Question 13.
Precision
Match each fraction with an equivalent expression.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 32$

Answer:
$Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-32$

Think and Grow: Modeling Real Life

Example
A chef has $\frac{8}{10}$ liter of soup. How can the chef pour all of the soup into 2 bowls?
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 33$
Break apart $\frac{8}{10}$ into any two fractions that have a sum of $\frac{8}{10}$.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 34$

Answer:
$Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-34$

Show and Grow

Question 14.
You have $\frac{7}{3}$ pounds of almonds. What are two different ways you can put all of the almonds into 2 bags?

Answer:
Given that,
You have $\frac{7}{3}$ pounds of almonds.
Break apart 7 parts of $\frac{1}{3}$ into 5 parts of $\frac{1}{3}$ and 2 parts of $\frac{1}{3}$
Thus you can put 5 parts of $\frac{1}{3}$ and 2 parts of $\frac{1}{3}$ of the almonds into 2 bags.

Question 15.
A 3-person painting crew has $\frac{10}{12}$ of a fence left to paint. What is one way the crew can finish painting the fence when each person paints a fraction of the fence?
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 35$

Answer:
Given that,
A 3-person painting crew has $\frac{10}{12}$ of a fence left to paint.
$\frac{10}{12}$ can be written as $\frac{3}{12}$ + $\frac{3}{12}$ + $\frac{4}{12}$
Thus each person paints $\frac{3}{12}$ + $\frac{3}{12}$ + $\frac{4}{12}$ fraction of the fence.

Question 16.
DIG DEEPER!
Three teammates have to run a total of miles for a relay race. Can each team member run the same fraction of a mile, in fourths, to complete the race? Explain.

Answer:
Three teammates have to run a total of miles for a relay race.
No three members cannot run the same fraction of a mile, in fourths, to complete the race
$\frac{10}{12}$ can be written as $\frac{3}{12}$ + $\frac{3}{12}$ + $\frac{4}{12}$

Decompose Fractions Homework & Practice 8.2

write the fraction as a sum of unit fractions.

Question 1.
$\frac{2}{2}$

Answer: 1
The sum of unit fractions of $\frac{2}{2}$ is $\frac{1}{2}$ + $\frac{1}{2}$

Question 2.
$\frac{3}{5}$

Answer: The sum of unit fractions of $\frac{3}{5}$ is $\frac{1}{5}$ + $\frac{1}{5}$ + $\frac{1}{5}$

Question 3.
$\frac{4}{3}$

Answer: The sum of unit fractions of $\frac{4}{3}$ is $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$

Question 4.
$\frac{6}{4}$

Answer: The sum of unit fractions of $\frac{6}{4}$ is $\frac{1}{4}$ + $\frac{1}{4}$ + $\frac{1}{4}$ + $\frac{1}{4}$ + $\frac{1}{4}$ + $\frac{1}{4}$

write the fraction as a sum of fractions in two different ways.

Question 5.
$\frac{8}{12}$

Answer: The sum of unit fractions of $\frac{8}{12}$ is $\frac{1}{12}$ + $\frac{1}{12}$ + $\frac{1}{12}$ + $\frac{1}{12}$ + $\frac{1}{12}$ + $\frac{1}{12}$ + $\frac{1}{12}$ + $\frac{1}{12}$

Another Way:
Break apart $\frac{8}{12}$ as 4 parts of $\frac{1}{12}$ and 4 parts of $\frac{1}{12}$

Question 6.
$\frac{10}{6}$

Answer: The sum of unit fractions of $\frac{10}{6}$ is $\frac{1}{6}$ + $\frac{1}{6}$ + $\frac{1}{6}$ + $\frac{1}{6}$ + $\frac{1}{6}$ + $\frac{1}{6}$ + $\frac{1}{6}$ + $\frac{1}{6}$ + $\frac{1}{6}$ + $\frac{1}{6}$

Another way:
Break apart $\frac{10}{6}$ as 5 parts of $\frac{1}{6}$ and 5 parts of $\frac{11}{6}$

Question 7.
$\frac{11}{100}$

Answer: The sum of unit fractions of $\frac{11}{100}$ is $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$

Another way:
Break apart $\frac{11}{100}$ as 5 parts of $\frac{1}{100}$, 4 parts of $\frac{1}{100}$ and 2 parts of $\frac{1}{100}$

Question 8.
$\frac{14}{8}$

Answer: The sum of unit fractions of $\frac{14}{8}$ is $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$

Another way:
Break apart $\frac{14}{8}$ as 5 parts of $\frac{1}{8}$, 9 parts of $\frac{1}{8}$

Question 9.
Which One Doesn’t Belong? Which expression does belong with the other three?
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 36$

Answer: The expression $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ does not belong to the other three.

Question 10.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 37$

Answer: Yes your friend is correct.
$\frac{1}{10}$ + $\frac{3}{10}$ + $\frac{5}{10}$
Here the denominators are the same so you have to add the numerators.
$\frac{1}{10}$ + $\frac{3}{10}$ + $\frac{5}{10}$ = $\frac{9}{10}$
$\frac{2}{10}$ + $\frac{4}{10}$ + $\frac{3}{10}$
Here the denominators are the same so you have to add the numerators.
$\frac{2}{10}$ + $\frac{4}{10}$ + $\frac{3}{10}$ = $\frac{9}{10}$

Question 11.
Number Sense
Is it possible to write $\frac{7}{12}$ as the sum of three fractions with three different numerators and the same denominator? Explain.

Answer: Yes it is possible to write $\frac{7}{12}$ as the sum of three fractions with three different numerators and the same denominator.
$\frac{7}{12}$ = $\frac{3}{12}$ + $\frac{3}{12}$ + $\frac{1}{12}$

Question 12.
You have $\frac{8}{4}$ pounds of dried pineapple. What are two different ways you can put all of the pineapples into 2 bags?

Answer:
Given that,
You have $\frac{8}{4}$ pounds of dried pineapple.
Break apart $\frac{8}{4}$ as 4 parts of $\frac{1}{4}$ and 4 parts of $\frac{1}{4}$.
The two different ways you can put all of the pineapples into 2 bags are 4 parts of $\frac{1}{4}$.

Question 13.
DIG DEEPER!
A carpenter has 3 planks of wood. Each plank has a different thickness. When stacked, the thickness of the 3 planks is $\frac{6}{8}$ inch. What are the possible thickness of each plank?
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 38$

Answer:
Given that,
A carpenter has 3 planks of wood. Each plank has a different thickness.
When stacked, the thickness of the 3 planks is $\frac{6}{8}$ inch.
$\frac{6}{8}$ = $\frac{2}{8}$ + $\frac{3}{8}$ + $\frac{1}{8}$
The possible thickness of each plank are $\frac{2}{8}$, $\frac{3}{8}$, $\frac{1}{8}$

Review & Refresh

Find the product. Check whether your answer is reasonable.

Question 14.
Estimate: ___
608 × 5 = ___

Answer:
600 × 5 = 3000
The number close to 608 is 600.
Step 2:
608 × 5 = 3040
3040 is close to 3000. So, the answer is reasonable.

Question 15.
Estimate: ___
7 × 5,394 = ___

Answer:
7 × 5400 = 37,800
The number close to 5394 is 5400.
Step 2:
7 × 5394 = 37,758
37,758 is close to 37,800. So, the answer is reasonable.

Question 16.
Estimate: ___
927 × 3 = ___

Answer:
900 × 3 = 2700
The number close to 927 is 900.
Step 2:
927 × 3 = 2781
2781 is close to 2700. So, the answer is reasonable.

Lesson 8.3 Add Fractions with Like Denominators

Explore and Grow

Write each fraction as a sum of unit fractions. Use models to help.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 39$

How many unit fractions did you use in all to rewrite the fractions above? How does this relate to the sum $\frac{3}{6}+\frac{5}{6}$ ?

Answer:
$Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-39$
$\frac{3}{6}+\frac{5}{6}$ = $\frac{8}{6}$

Construct Arguments
How can you use the numerators and the denominators to add fractions with like denominators? Explain why your method makes sense.

Answer:
To add fractions with like denominators, add the numerators and keep the same denominator. Then simplify the sum. You know how to do this with numeric fractions.
$\frac{3}{6}+\frac{5}{6}$ = $\frac{8}{6}$

Think and Grow: Add Fractions

To add fractions with like denominators, add the numerators.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 40$
The denominator stays the same.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 41$

Answer:
$Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-41$
Add the numerators of the like denominators.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 42$

Answer:
Add the numerators of the like denominators.
$Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-42$

Show and Grow

Add.

Question 1.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 43$

Answer:
Add the numerators of the like denominators.
$Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-43$

Question 2.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 44$

Answer:
Add the numerators of the like denominators.
6 + 2 = 8
$\frac{6}{5}+\frac{2}{5}$ = $\frac{8}{5}$

Question 3.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 45$

Answer:
Add the numerators of the like denominators.
4 + 4 = 8
$\frac{4}{8}+\frac{4}{8}$ = $\frac{8}{8$ = 1

Apply and Grow: Practice

Add.

Question 4.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 46$

Answer:
Add the numerators of the like denominators.
3 + 2 = 5
$\frac{3}{6}+\frac{2}{6}$ = $\frac{5}{6}$

Question 5.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 47$

Answer:
Add the numerators of the like denominators.
8 + 4 = 12
$\frac{8}{2}+\frac{4}{2}$ = $\frac{12}{2}$ = 6

Question 6.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 48$

Answer:
Add the numerators of the like denominators.
4 + 1 = 5
$\frac{4}{5}+\frac{1}{5}$ = $\frac{5}{5}$ = 1

Question 7.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 49$

Answer:
Add the numerators of the like denominators.
60 + 35 = 95
$\frac{60}{100}+\frac{35}{100}$ = $\frac{95}{100}$

Question 8.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 50$

Answer:
The denominators are not the same. So first you have to make the common denominators and add the fraction with the number.
2 × 3/3 = 6/3
$\frac{6}{3}+\frac{5}{3}$ = $\frac{11}{3}$

Question 9.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 51$

Answer:
The denominators are not the same. So first you have to make the common denominators and add the fraction with the number.
6 × 12/12 = 72/12
$\frac{72}{12}+\frac{1}{12}$ = $\frac{73}{12}$

Question 10.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 52$

Answer:
Add the numerators of the like denominators.
3 + 1 + 1 = 5
3/4 + 1/4 + 1/4 = 5/4

Question 11.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 53$

Answer:
Add the numerators of the like denominators.
$\frac{6}{8}$ + $\frac{5}{8}$ + $\frac{4}{8}$
6 + 5 + 4 = 15
$\frac{6}{8}$ + $\frac{5}{8}$ + $\frac{4}{8}$ = $\frac{15}{8}$

Question 12.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 54$

Answer:
Add the numerators of the like denominators.
43 + 16 + 10 = 69
$\frac{43}{100}$ + $\frac{16}{100}$ + $\frac{10}{100}$ = $\frac{69}{100}$

Question 13.
You eat $\frac{2}{10}$ of a vegetable pizza. Your friend eats $\frac{3}{10}$ of the pizza. What fraction of the pizza do you and your friend eat together?

Answer:
Given that,
You eat $\frac{2}{10}$ of a vegetable pizza. Your friend eats $\frac{3}{10}$ of the pizza.
$\frac{2}{10}$ + $\frac{3}{10}$ = $\frac{5}{10}$ = $\frac{1}{2}$
$\frac{1}{2}$ fraction of the pizza do you and your friend eat together

Question 14.
Number Sense
A sum has 5 addends. Each addend is a unit fraction. The sum is 1. What are the addends?

Answer:
$\frac{1}{5}$ + $\frac{1}{5}$ + $\frac{1}{5}$ + $\frac{1}{5}$ + $\frac{1}{5}$ = $\frac{5}{5}$ = 1

Question 15.
Writing
Explain how to add $\frac{3}{4}$ and $\frac{1}{4}$. Use a model to support your answer.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 55$

Answer:
$Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-55$
$\frac{3}{4}$ + $\frac{1}{4}$ = $\frac{4}{4}$ = 1

Think and Grow: Modeling Real Life

Example
The table shows the natural hazards studied by 100 students for a science project. What fraction of the students studied a weather-based natural hazard?
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 56$
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 57$

Answer:
$Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-57$

Show and Grow

Question 16.
Use the graph above to find what fraction of the students studied an Earth-based natural hazard.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 57.1$

Answer:
We can find the fraction of the students who studied an Earth-based natural hazard
1 × 8 = 8
Half drop = 4
8 + 4 = 12
= Number of students/Total number of students surveyed
= 12/100
Thus $\frac{12}{100}$ fraction of the students who studied an Earth-based natural hazard.

Question 17.
DIG DEEPER!
A caterer needs at least 2 pounds of lunch meat to make a sandwich platter. She has $\frac{6}{4}$ pounds of turkey and $\frac{3}{4}$ pound of ham. Does the caterer have enough lunch meat to make a sandwich platter? Explain.

Answer:
Given that,
A caterer needs at least 2 pounds of lunch meat to make a sandwich platter. She has $\frac{6}{4}$ pounds of turkey and $\frac{3}{4}$ pound of ham.
$\frac{6}{4}$ + $\frac{3}{4}$ = $\frac{9}{4}$
Convert it into mixed fraction
$\frac{9}{4}$ = 1 $\frac{3}{4}$
Thus the caterer does not have enough lunch meat to make a sandwich platter.

Add Fractions with Like Denominators Homework & Practice 8.3

Add

Question 1.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 58$

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
$Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-58$

Question 2.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 59$

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
$\frac{2}{2}$ + $\frac{7}{2}$ = (2 + 7)/2 = $\frac{9}{2}$

Question 3.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 60$

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
$\frac{2}{5}$ + $\frac{2}{5}$ = (2 + 2)/5 = $\frac{4}{5}$

Question 4.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 61$

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
$\frac{4}{10}$ + $\frac{6}{10}$ = (4 + 6)/10 = $\frac{10}{10}$ = 1

Question 5.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 62$

Answer:
The denominators are not the same. So first you have to make the common denominators and add the fraction with the number.
Take the denominator as common and add the numerators.
4 × 3/3 = 12/3
$\frac{12}{3}$ + $\frac{1}{3}$ = (12 + 1)/3 = $\frac{13}{2}$

Question 6.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 63$

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
$\frac{27}{100}$ + $\frac{460}{100}$ = (27 + 460)/100 = $\frac{487}{100}$

Question 7.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 64$

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
$\frac{8}{4}$ + $\frac{5}{4}$ = (8 + 5)/4 = $\frac{13}{4}$

Question 8.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 65$

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
$\frac{4}{6}$ + $\frac{1}{6}$ = (4 + 1)/6 = $\frac{5}{6}$

Question 9.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 66$

Answer:
The denominators are not the same. So first you have to make the common denominators and add the fraction with the number.
Take the denominator as common and add the numerators.
10 × 12/12 = 120/12
$\frac{120}{12}$ + $\frac{7}{12}$ = (120 + 7)/3 = $\frac{127}{12}$

Question 10.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 67$

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
$\frac{1}{5}$ + $\frac{1}{5}$ + $\frac{2}{5}$ = (1 + 1 + 2)/5 = $\frac{4}{5}$

Question 11.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 68$

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
$\frac{38}{100}$ + $\frac{13}{100}$ + $\frac{21}{100}$ = (38+ 13 + 21)/100 = $\frac{72}{100}$

Question 12.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 69$

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
$\frac{8}{8}$ + $\frac{4}{8}$ + $\frac{2}{8}$ = (8 + 4 + 2)/8 = $\frac{14 }{8}$

Question 13.
You plant a sunflower seed. After 11 week, the plant is $\frac{1}{2}$ inch tall. The next week your plant grows $\frac{3}{2}$ inches. How tall is your plant after the second week?

Answer:
Given that,
You plant a sunflower seed. After 11 week, the plant is $\frac{1}{2}$ inch tall. The next week your plant grows $\frac{3}{2}$ inches.
$\frac{1}{2}$ + $\frac{3}{2}$ = $\frac{4}{2}$ = 2
The plant is 2 inches tall after the second week.

Question 14.
Writing
Explain how to find the unknown addend.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 70$

Answer:
1 can be written as $\frac{10}{10}$
$\frac{7}{10}$ + ? = $\frac{10}{10}$
? = $\frac{10}{10}$ – $\frac{7}{10}$
? = $\frac{3}{10}$
Thus the unknown addend is $\frac{3}{10}$

Question 15.
DIG DEEPER!
When you double me and add $\frac{1}{6}$, you get $\frac{5}{6}$. What fraction am I?

Answer: $\frac{2}{6}$

Explanation:
If you add $\frac{2}{6}$ twice and add $\frac{1}{6}$ to it you get $\frac{5}{6}$.

Question 16.
Reasoning
You eat $\frac{2}{8}$ of a large apple at lunch and another $\frac{4}{8}$ of it as a snack. Your friend eats $\frac{4}{8}$ of a small apple at lunch and another $\frac{2}{8}$ of it as a snack. Do you each eat the same amount? Explain.

Answer:
Given that,
You eat $\frac{2}{8}$ of a large apple at lunch and another $\frac{4}{8}$ of it as a snack. Your friend eats $\frac{4}{8}$ of a small apple at lunch and another $\frac{2}{8}$ of it as a snack.
$\frac{2}{8}$ + $\frac{4}{8}$ = $\frac{6}{8}$
$\frac{2}{8}$ + $\frac{4}{8}$ = $\frac{6}{8}$
Yes you and your friend eat same amount of food.

Question 17.
Modeling Real Life
The graph shows the classification of 100 species of birds in North America according to their extinction rate. What fraction of the species are classified as near threatened or vulnerable?
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 71$

Answer:
• = 4 species
Near threatened = 4 × 4 = 16 species
half • = 2 species
16 + 2 = 18 species
Fraction of near threatened = number of species/total number of species of birds in North America
= 18/100
Vulnerable = 5 × 4 = 20 species
half • = 2 species
20 + 2 = 22 species
Fraction of near threatened = number of species/total number of species of birds in North America
= 22/100

Question 18.
Modeling Real Life
Use the graph above to find what fraction of not the species are critically endangered.

Answer:
• = 4 species
critically endangered = 4 × 4 = 16 species
half • = 2 species
16 + 2 = 18 species
Fraction of near threatened = number of species/total number of species of birds in North America
= 18/100

Review & Refresh

Question 19.
A pet store has 25 tanks with 32 fish in each tank. A customer buys 7 fish. How many fish does the pet store have now?

Answer:
Given,
A pet store has 25 tanks with 32 fish in each tank. A customer buys 7 fish.
$\frac{25}{32}$ – $\frac{7}{32}$ = $\frac{18}{32}$
Thus there are 18 fishes in the pet store.

Lesson 8.4 Use Models to subtract Fractions

Explore and Grow

Draw a model to show $\frac{9}{12}$.
Answer:
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_25$

Use your model to find $\frac{9}{12}$ – $\frac{5}{12}$. Explain your method.
Answer:
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_26$

Repeated Reasoning
Write two fractions that have a difference of $\frac{7}{12}$. Explain your reasoning.

Answer: $\frac{9}{12}$ – $\frac{2}{12}$ = $\frac{7}{12}$

Think and Grow: Use Models to Subtract Fractions

You can subtract fractions by taking away parts that refer to the same whole.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 72$

Answer:
$Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-72$
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 73$

Answer:
$Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-73$

Show and Grow

Find the difference. Explain how you used the model to subtract.

Question 1.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 74$

Answer: 5/10 = 1/2
Take away a length of 4/10 from the length of 9/10.
$Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-74$

Question 2.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 75$

Answer:
Take away a length of 6/4 from the length of 2/4.
$Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-75$

Apply and Grow: Practice

Find the difference. Use a model or a number line to help.

Question 3.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 76$

Answer: 4/8
Take away a length of 8/8 from the length of 4/8.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_7$

Question 4.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 77$

Answer: 8/12
Take away a length of 10/12 from the length of 2/12.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_27$

Question 5.
$Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 78$

Answer: 3/5
Take away a length of 4/5 from the length of 1/5.
$Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-28$

Question 6.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 79$

Answer: 6/2 = 3
Take away a length of 9/2 from the length of 3/2.
$Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-29$

Question 7.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 80$

Answer: 10/6
Take away a length of 15/6 from the length of 5/6.
$Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-30$

Question 8.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 81$

Answer: $\frac{26}{100}$
The denominators of both the fractions are the same. So subtract the numerators.
$\frac{76}{100}$ – $\frac{50}{100}$ = $\frac{26}{100}$

Question 9.
You need to walk $\frac{3}{4}$ mile for your physical education class. So far, you have walked $\frac{2}{4}$ mile. How much farther do you need to walk?

Answer:
Given that,
You need to walk $\frac{3}{4}$ mile for your physical education class. So far, you have walked $\frac{2}{4}$ mile.
$\frac{3}{4}$ – $\frac{2}{4}$ = $\frac{1}{4}$
You need to walk $\frac{1}{4}$ miles more.

Question 10.
Number Sense
Which expressions have a difference of $\frac{4}{5}$ ?
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 82$

Answer:
5/5 – 1/5 = 4/5
10/5 – 6/5 = 4/5
6/5 – 3/5 = 3/5
9/5 – 5/5 = 4/5
i, ii, iv has the difference of $\frac{4}{5}$

Question 11.
Structure
Write the subtraction equation represented by the model.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 83$

Answer: $\frac{7}{8}$ – $\frac{4}{8}$ = $\frac{3}{8}$

Question 12.
Writing
Explain why the numerator changes when you subtract fractions with like denominators, but the denominator stays the same.

Answer:
The most simple fraction subtraction problems are those that have two proper fractions with a common denominator. That is, each denominator is the same. The process is just as it is for the addition of fractions with like denominators, except you subtract! You subtract the second numerator from the first and keep the denominator the same.

Think and Grow: Modeling Real Life

Example
A lizard’s tail is $\frac{10}{12}$ foot long. It sheds a $\frac{7}{12}$ foot long part of its tail to escape a predator. How long is the remaining part of the lizard’s tail?
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 84$
Because each fraction represents a part of the same whole, you can take away a part.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 85$

Answer:
Given that,
A lizard’s tail is $\frac{10}{12}$ foot long. It sheds a $\frac{7}{12}$ foot long part of its tail to escape a predator.
$Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-85$

Show and Grow

Question 13.
You have $\frac{9}{8}$ cups of raisins. You eat $\frac{2}{8}$ cup. What fraction of a cup of raisins do you have left?
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 86$

Answer:
Given that,
You have $\frac{9}{8}$ cups of raisins. You eat $\frac{2}{8}$ cup.
$\frac{9}{8}$ – $\frac{2}{8}$ = $\frac{7}{8}$
Thus $\frac{7}{8}$ fraction of a cup of raisins is left.

Question 14.
A large bottle has $\frac{7}{4}$ quarts of liquid soap. A small bottle has $\frac{3}{4}$ quart of liquid soap. How much more soap is in the large bottle than in the small bottle?

Answer:
Given that,
A large bottle has $\frac{7}{4}$ quarts of liquid soap. A small bottle has $\frac{3}{4}$ quart of liquid soap.
$\frac{7}{4}$ – $\frac{3}{4}$ = $\frac{4}{4}$ = 1
Thus 1 more soap is in the large bottle than in the small bottle.

Question 15.
DIG DEEPER!
You need 2 cups of milk for a recipe. You have cup of $\frac{1}{3}$ milk. How much more milk do you need? Explain.

Answer:
Given,
You need 2 cups of milk for a recipe. You have cup of $\frac{1}{3}$ milk.
2 × $\frac{1}{3}$ = $\frac{2}{3}$
Thus $\frac{2}{3}$ more milk you need.

Use Models to subtract Fractions Homework & Practice 8.4

Find the difference. Explain how you used the model to subtract.

Question 1.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 87$

Answer:
$Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-87$

Question 2.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 88$

Answer:
$Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-88$

Find the difference. Use a model or a number line to help.

Question 3.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 89$

Answer: 15/10
$Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-25$

Question 4.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 90$

Answer: 10/5

BIM 3rd Grade Answer Key Grade 4 Chapter 8 Add & Subtract Fractions img_31

Question 5.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 91$

Answer: 8/12

$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_32$

Question 6.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 92$

Answer:

$Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-226$

Question 7.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 93$

Answer: 7/4

$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-146$

Question 8.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 94$

Answer:
The denominators of both the fractions are the same. So subtract the numerators.
$\frac{70}{100}$ – $\frac{6}{100}$ = $\frac{64}{100}$

Question 9.
You have $\frac{2}{3}$ yard of ribbon. You cut off $\frac{1}{3}$ yard of the ribbon. How much ribbon do you have left?

Answer:
Given that,
You have $\frac{2}{3}$ yard of ribbon. You cut off $\frac{1}{3}$ yard of the ribbon.
The denominators of both the fractions are the same. So subtract the numerators.
$\frac{2}{3}$ – $\frac{1}{3}$ = $\frac{1}{3}$
$\frac{1}{3}$ ribbon has left.

Question 10.
Structure
When using circular models to find the difference of $\frac{4}{2}$ and $\frac{1}{2}$, why do you shade two circles to represent $\frac{4}{2}$?

Answer:
The denominators of both the fractions are the same. So subtract the numerators.
$\frac{4}{2}$ – $\frac{1}{2}$ = $\frac{3}{2}$

Question 11.
YOU BE THE TEACHER
In a box of pens, $\frac{3}{4}$ of the pens are blue. Your friend takes $\frac{1}{4}$ of the blue pens and says that now $\frac{2}{4}$ of the pens in the box are blue. Is your friend correct? Explain.

Answer:
Given,
In a box of pens, $\frac{3}{4}$ of the pens are blue. Your friend takes $\frac{1}{4}$ of the blue pens and says that now $\frac{2}{4}$ of the pens in the box are blue.
The denominators of both the fractions are the same. So subtract the numerators.
$\frac{3}{4}$ – $\frac{1}{4}$ = $\frac{2}{4}$
Yes, your friend is correct.

Question 12.
DIG DEEPER!
Using numerators that even number, write two different subtraction equations that each have a difference of 1.

Answer: $\frac{6}{4}$ – $\frac{2}{4}$ = $\frac{4}{4}$ = 1

Question 13.
Modeling Real Life
In our solar system, $\frac{6}{8}$ of the planets have moons, and $\frac{4}{8}$ of the planets have moons and rings. What fraction of the planets in our solar system have moons, but do not have rings?

Answer:
Given,
In our solar system, $\frac{6}{8}$ of the planets have moons, and $\frac{4}{8}$ of the planets have moons and rings.
The denominators of both the fractions are the same. So subtract the numerators.
$\frac{6}{8}$ – $\frac{4}{8}$ = $\frac{2}{8}$

Question 14.
Modeling Real Life
A professional pumpkin carver carves a pumpkin that weighs $\frac{7}{10}$ ton. He carves a second pumpkin that weighs $\frac{6}{10}$ ton. How much heavier is the first pumpkin than the second pumpkin?
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 95$

Answer:
Given that,
A professional pumpkin carver carves a pumpkin that weighs $\frac{7}{10}$ ton. He carves a second pumpkin that weighs $\frac{6}{10}$ ton.
The denominators of both the fractions are the same. So subtract the numerators.
$\frac{7}{10}$ – $\frac{6}{10}$ = $\frac{1}{10}$
The first pumpkin is $\frac{1}{10}$ heavier than the second pumpkin.

Review & Refresh

Find an equivalent fraction.

Question 15.
$\frac{7}{4}$

Answer:
The equivalent fraction of $\frac{7}{4}$ is given below,
$\frac{7}{4}$ × $\frac{2}{2}$ = $\frac{14}{8}$

Question 16.
$\frac{3}{5}$

Answer:
The equivalent fraction of $\frac{3}{5}$ is given below,
$\frac{3}{5}$ × $\frac{3}{3}$ = $\frac{9}{15}$

Question 17.
$\frac{2}{3}$

Answer:
The equivalent fraction of $\frac{2}{3}$ is given below,
$\frac{2}{3}$ × $\frac{2}{2}$ = $\frac{4}{6}$

Lesson 8.5 Subtract Fractions with Like Denominators

Explore and Grow

Write each fraction as a sum of unit fractions. Use models to help.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 96$
How many more unit fractions did you use to rewrite $\frac{4}{5}$ than $\frac{3}{5}$?
How does this relate to the difference $\frac{4}{5}$ – $\frac{3}{5}$ ?

Answer:
$Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-96$
$\frac{4}{5}$ – $\frac{3}{5}$ = $\frac{1}{5}$

Construct Arguments
How can you use the numerators and the denominators to subtract fractions with like denominators? Explain why your method makes sense.

Answer: Steps on How to Add and Subtract Fractions with the Same Denominator. To add fractions with like or the same denominator, simply add the numerators then copy the common denominator. Always reduce your final answer to its lowest term.

Think and Grow: Subtract Fractions

To subtract fractions with like denominators, subtract the numerators. The denominator stays the same.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 97$
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 98$

Answer:
$Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-98$

Show and Grow

Subtract.

Question 1.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 100$

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.

$Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-100$

Question 2.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 101$

Answer:
First, make the denominators common and then subtract the numerators
1 can be written as $\frac{12}{12}$
$\frac{12}{12}$ – $\frac{8}{12}$ = (12 – 8)/12
= $\frac{4}{12}$ or $\frac{1}{3}$

Question 3.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 102$

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
$\frac{50}{100}$ – $\frac{30}{100}$ = (50 – 30)/100
= $\frac{20}{100}$ or $\frac{1}{5}$

Apply and Grow: Practice

Subtract.

Question 4.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 103$

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
$Big Ideas Math Book 4th Grade Answer Key Chapter 8 Add and Subtract Fractions img_37$

Question 5.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 104$

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_35$

Question 6.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 105$

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
$\frac{12}{6}$ – $\frac{7}{6}$ = (12- 7)/6
$\frac{5}{6}$

Question 7.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 106$

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
$\frac{4}{5}$ – $\frac{3}{5}$ = (4- 3)/5
$\frac{1}{5}$

Question 8.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 107$

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
$\frac{60}{100}$ – $\frac{43}{100}$ = (60 – 43)/100
$\frac{17}{100}$

Question 9.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 108$

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
$\frac{10}{2}$ – $\frac{2}{2}$ = (10 – 2)/2
$\frac{8}{4}$ = 2

Question 10.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 109$

Answer:
First, make the denominators common and then subtract the numerators.
$\frac{12}{12}$ – $\frac{7}{12}$ = (12 – 7)/12
= $\frac{5}{12}$
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_34$

Question 11.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 110$

Answer:
First, make the denominators common and then subtract the numerators
1 can be written as $\frac{8}{8}$
$\frac{8}{8}$ – $\frac{5}{8}$ = $\frac{3}{8}$

Question 12.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 111$

Answer:
First, make the denominators common and then subtract the numerators
2 can be written as $\frac{8}{4}$
$\frac{8}{4}$ – $\frac{1}{4}$ = $\frac{7}{4}$

Question 13.
You have 1 gallon of paint. You use $\frac{2}{3}$ gallon to paint a wall. How much paint do you have left?
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 112$

Answer:
Given that,
You have 1 gallon of paint. You use $\frac{2}{3}$ gallon to paint a wall.
First, make the denominators common and then subtract the numerators
1 – $\frac{2}{3}$
1 can be written as $\frac{3}{3}$
$\frac{3}{3}$ – $\frac{2}{3}$ = $\frac{1}{3}$

Question 14.
Reasoning
Why is it unreasonable to get a difference of $\frac{7}{8}$ when subtracting $\frac{1}{8}$ from $\frac{7}{8}$? Use a model to support your answer.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 113$

Answer:
The difference of $\frac{7}{8}$ when subtracting $\frac{1}{8}$ from $\frac{7}{8}$ is,
$Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-113$

Question 15.
Your friend says each difference is $\frac{3}{10}$. Is your friend correct? Explain.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 114$

Answer:
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_36$
Your friend is correct.
10/10 – 7/10 = 3/10
100/100 = 70/100 = 30/100 = 3/10

Think and Grow: Modeling Real Life

Example
A flock of geese has completed $\frac{5}{12}$ of its total migration. What fraction of its migration does the flock of geese have left to complete?
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 115$
Because the total migration is 1 whole, find 1 − $\frac{5}{12}$.
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 116$

Answer:
Given,
A flock of geese has completed $\frac{5}{12}$ of its total migration.
Because the total migration is 1 whole, find 1 − $\frac{5}{12}$.
First, make the denominators common and then subtract the numerators.
$Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-116$

Show and Grow

Question 16.
A runner has completed $\frac{6}{10}$ of a race. What fraction of the race does the runner have left to complete?

Answer:
Given that,
A runner has completed $\frac{6}{10}$ of a race.
1 – $\frac{6}{10}$
1 can be written as $\frac{10}{10}$
$\frac{10}{10}$ – $\frac{6}{10}$ = $\frac{4}{10}$
The runner has left $\frac{4}{10}$ fraction of the race to complete.

Question 17.
A pizza buffet serves pizzas of the same size with different toppings. There is $\frac{7}{8}$ of a vegetable pizza and $\frac{2}{8}$ of a pineapple pizza left. How much more vegetable pizza is left than pineapple pizza?

Answer:
Given,
A pizza buffet serves pizzas of the same size with different toppings.
There is $\frac{7}{8}$ of a vegetable pizza and $\frac{2}{8}$ of a pineapple pizza left.
$\frac{7}{8}$ – $\frac{2}{8}$ = $\frac{5}{8}$
$\frac{5}{8}$ more vegetable pizza is left than pineapple pizza.

Question 18.
DIG DEEPER!
Baseball practice is 1 hour long. You stretch for 7 minutes and play catch for 8 minutes. What fraction of an hour do you have left to practice?
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 117$

Answer:
Given,
Baseball practice is 1 hour long. You stretch for 7 minutes and play catch for 8 minutes.
7 minutes + 8 minutes = 15 minutes
15 minutes = $\frac{1}{4}$ hour
1 – $\frac{1}{4}$ = $\frac{3}{4}$
Thus $\frac{3}{4}$ fraction of an hour is left to practice.

Subtract Fractions with Like Denominators Homework & Practice 8.5

Subtract

Question 1.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 118$

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
$\frac{7}{8}$ – $\frac{3}{8}$ = $\frac{4}{8}$
$Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-118$

Question 2.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 119$

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
$\frac{5}{4}$ – $\frac{3}{4}$ = $\frac{2}{4}$

Question 3.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 120$

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
$\frac{13}{5}$ – $\frac{6}{5}$ = $\frac{7}{6}$

Question 4.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 121$

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
$\frac{5}{12}$ – $\frac{1}{12}$ = $\frac{4}{12}$

Question 5.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 122$

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
$\frac{9}{6}$ – $\frac{4}{6}$ = $\frac{5}{6}$

Question 6.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 123$

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
$\frac{11}{3}$ – $\frac{7}{3}$ = $\frac{4}{3}$

Question 7.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 124$

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
$\frac{10}{10}$ – $\frac{4}{10}$ = $\frac{6}{10}$

Question 8.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 125$

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
$\frac{20}{2}$ – $\frac{8}{2}$ = $\frac{12}{2}$

Question 9.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 126$

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
$\frac{36}{100}$ – $\frac{21}{100}$ = $\frac{15}{100}$

Question 10.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 127$

Answer:
First, make the denominators common and then subtract the numerators.
1 can be written as 5/5.
$\frac{5}{5}$ – $\frac{3}{5}$ = $\frac{2}{5}$

Question 11.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 128$

Answer:
First, make the denominators common and then subtract the numerators.
2 can be written as 8/4
$\frac{8}{4}$ – $\frac{2}{4}$ = $\frac{6}{4}$

Question 12.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 129$

Answer:
First, make the denominators common and then subtract the numerators
3 can be written as 24/8.
$\frac{24}{8}$ – $\frac{15}{8}$ = $\frac{9}{8}$

Question 13.
A family eats $\frac{2}{3}$ of a tray of lasagna. What fraction of the tray of lasagna is left?

Answer:
Given,
A family eats $\frac{2}{3}$ of a tray of lasagna.
1 – $\frac{2}{3}$
1 can be written as $\frac{3}{3}$
$\frac{3}{3}$ – $\frac{2}{3}$ = $\frac{1}{3}$
Therefore $\frac{1}{3}$ fraction of the tray of lasagna is left.

Question 14.
Writing
Explain how ﬁnding is $\frac{7}{10}-\frac{4}{10}$ similar to ﬁnding 7 – 4.

Answer:
Yes $\frac{7}{10}-\frac{4}{10}$ similar to ﬁnding 7 – 4. Because the denominators of the fractions are the same.
$\frac{7}{10}-\frac{4}{10}$ = $\frac{3}{10}$

Question 15.
Open-Ended
The model shows equal parts of a 1 whole. Write a subtraction problem whose answer is shown.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 130$

Answer:
By seeing the above figure we can write the subtraction problem.
1 – $\frac{3}{8}$

Question 16.
Modeling Real Life
you fill $\frac{2}{4}$ of your plate with vegetables. What fraction of your plate does not contain vegetables?

Answer:
Given,
you fill $\frac{2}{4}$ of your plate with vegetables.
1 – $\frac{2}{4}$
1 can be written as $\frac{4}{4}$
$\frac{4}{4}$ – $\frac{2}{4}$ = $\frac{2}{4}$
Thus $\frac{2}{4}$ fraction of your plate does not contain vegetables.

Question 17.
Modeling Real Life
A group of students designs a rectangular playground. They use $\frac{2}{8}$ of the playground for a basketball court and $\frac{3}{8}$ of the playground for a soccer field. How much space is left?

Answer:
Given,
A group of students designs a rectangular playground.
They use $\frac{2}{8}$ of the playground for a basketball court and $\frac{3}{8}$ of the playground for a soccer field.
$\frac{2}{8}$ + $\frac{3}{8}$ = $\frac{5}{8}$
1 – $\frac{5}{8}$ = $\frac{3}{8}$
Thus $\frac{3}{8}$ space is left.

Review & Refresh

Find the quotient and the remainder

Question 18.
34 ÷ 7 = ___R___

Answer: 4R6

Explanation:
34 ÷ 7 = $\frac{34}{7}$
$\frac{34}{7}$ = 4R6
Thus the quotient is 4 and the remainder is 6.

Question 19.
28 ÷ 3 = ___R___

Answer: 9 R1

Explanation:
28 ÷ 3 = $\frac{28}{3}$
$\frac{28}{3}$ = 9 R1
Thus the quotient is 9 and the remainder is 1.

Lesson 8.6 Model Fractions and Mixed Numbers

Explore and Grow

Draw a model to show 1 + 1 + $\frac{2}{3}$.

Use your model to write the sum as a fraction.

Answer:
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_37$

Repeated Reasoning
How can you write a fraction greater than 1 as the sum of a whole number and a fraction less than 1? Explain.

Answer:
$\frac{3}{2}$ = 1 $\frac{1}{2}$
The fraction 1 $\frac{1}{2}$ the whole fraction is greater than 1 and the fraction is less than 1.

Think and Grow: Write Fractions and Mixed Numbers

A mixed number represents the sum of a whole number and a fraction less than 1.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 132$
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 133$

Answer:
$Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-133$

Example
Write $\frac{5}{2}$ as a mixed number.

Find how many wholes are in $\frac{5}{2}$ and how many halves are left over.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 134$

Answer:
$Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-134$

Show and Grow

Question 1.
Write 3$\frac{1}{4}$ as a fraction. Use a model or a number line to help.

Answer:

$Big Ideas math 4th grade answers chapter 8 img_38$

Question 2.
Write $\frac{9}{6}$ as a mixed number. Use a model or a number line to help.

Answer:
$\frac{9}{6}$ can be written as $\frac{3}{2}$
Now convert $\frac{3}{2}$ into the mixed fraction
$\frac{3}{2}$ = 1 $\frac{1}{2}$
$BIM 4th Grade Answer key Chapter 8 Add and subtract fractions img_39$

Apply and Grow: Practice

Write the mixed number as a fraction.

Question 3.
3$\frac{4}{5}$

Answer: $\frac{19}{5}$

Explanation:
Step 1
Multiply the denominator by the whole number
5 × 3 = 15
Step 2
Add the answer from Step 1 to the numerator
15 + 4 = 19
Step 3
Write an answer from Step 2 over the denominator
19/5

Question 4.
2$\frac{1}{3}$

Answer: $\frac{7}{3}$

Explanation:
Step 1
Multiply the denominator by the whole number
3 × 2 = 6
Step 2
Add the answer from Step 1 to the numerator
6 + 1 = 7
Step 3
Write an answer from Step 2 over the denominator
7/3

Question 5.
6$\frac{7}{12}$

Answer: $\frac{79}{12}$

Explanation:
Step 1
Multiply the denominator by the whole number
12 × 6 = 72
Step 2
Add the answer from Step 1 to the numerator
72 + 7 = 79
Step 3
Write an answer from Step 2 over the denominator
79/12

Question 6.
1$\frac{82}{100}$

Answer: $\frac{182}{100}$

Explanation:
Step 1
Multiply the denominator by the whole number
100 × 1 = 100
Step 2
Add the answer from Step 1 to the numerator
100 + 82 = 182
Step 3
Write an answer from Step 2 over the denominator
$\frac{182}{100}$

Question 7.
11$\frac{3}{8}$

Answer: $\frac{91}{8}$

Explanation:
Step 1
Multiply the denominator by the whole number
8 × 11 = 88
Step 2
Add the answer from Step 1 to the numerator
88 + 3 = 91
Step 3
Write an answer from Step 2 over the denominator
91/8

Question 8.
9$\frac{5}{10}$

Answer: $\frac{95}{10}$

Explanation:
Step 1
Multiply the denominator by the whole number
10 × 9 = 90
Step 2
Add the answer from Step 1 to the numerator
90 + 5 = 95
Step 3
Write an answer from Step 2 over the denominator
95/10

Write the fraction as a mixed number or a whole number.

Question 9.
$\frac{9}{8}$

Answer: 1 $\frac{1}{8}$

Explanation:
9÷8=1R1
$\frac{9}{8}$ = 1 $\frac{1}{8}$

Question 10.
$\frac{19}{3}$

Answer: 6 $\frac{1}{3}$

Explanation:
19÷3=6R1
$\frac{19}{3}$ = 6 $\frac{1}{3}$

Question 11.
$\frac{38}{5}$

Answer: 7 $\frac{3}{5}$

Explanation:
38÷5=7R3
$\frac{38}{5}$ = 7 $\frac{3}{5}$

Question 12.
$\frac{22}{10}$

Answer: 2 $\frac{1}{5}$

Explanation:
11÷5=2R1
$\frac{22}{10}$ = 2 $\frac{1}{5}$

Question 13.
$\frac{460}{100}$

Answer: 4 $\frac{3}{5}$

Explanation:
23÷5=4R3
$\frac{460}{100}$ = 4 $\frac{3}{5}$

Question 14.
$\frac{20}{4}$

Answer: 5

Explanation:
4 divides 20 five times.
$\frac{20}{4}$ = 5

Compare

Question 15.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 135$

Answer: =

Explanation:
$\frac{3}{2}$ can be written as 1 $\frac{1}{2}$
1 × 2 + 1 = 3
So, 1 $\frac{1}{2}$ = $\frac{3}{2}$

Question 16.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 136$

Answer: >

Explanation:
3$\frac{3}{12}$ can be written as $\frac{39}{12}$
$\frac{39}{12}$ > $\frac{15}{12}$
So, 3$\frac{3}{12}$ > $\frac{15}{12}$

Question 17.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 137$

Answer: <

Explanation:
$\frac{21}{6}$ can be written as 3 $\frac{3}{6}$ or 3 $\frac{1}{2}$
So, 3 $\frac{1}{2}$ < 4
$\frac{21}{6}$ < 4

Question 18.
Which One Doesn’t Belong? Which expression does not Belong to the other three?
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 138$

Answer:
3 $\frac{2}{3}$ = $\frac{11}{3}$
$\frac{9}{3}$ + $\frac{3}{3}$ = $\frac{12}{3}$
$\frac{3}{3}$ + $\frac{3}{3}$ +$\frac{3}{3}$ + $\frac{2}{3}$ = $\frac{11}{3}$
$\frac{11}{3}$
So, the second expression does not belong to the other three expressions.

DIG DEEPER!
Find the unknown Number

Question 19.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 139$

Answer: 2

Explanation:
$\frac{8}{6}$ is 4÷3=1R1
$\frac{8}{6}$ = 1 $\frac{2}{6}$
So, the unknown number is 2.

Question 20.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 140$

Answer: 3

Explanation:
$\frac{35}{4}$ = 8 R 3
8 $\frac{3}{4}$ = $\frac{35}{4}$
So, the unknown number is 3.

Question 21.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 141$

Answer: 10

Explanation:
12 × 10 + 9 = 129
$\frac{129}{12}$ = 10 $\frac{9}{12}$
So, the unknown number is 10.

Think and Grow: Modeling Real Life

Example
A construction worker needs nails that are $\frac{9}{4}$ inches long. Which size of nails should the worker use?
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 142$
Write $\frac{9}{4}$ as a mixed number.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 143$

Answer:
Given,
A construction worker needs nails that are $\frac{9}{4}$ inches long.
Convert from improper fraction to the mixed fraction.
$Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-143$

Show and Grow

Question 22.
You need screws that are $\frac{13}{8}$ inches long to build a birdhouse. Which size of screws should you use?
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 144$

Answer:
Given,
You need screws that are $\frac{13}{8}$ inches long to build a birdhouse.
Convert from improper fraction to the mixed fraction.
$\frac{13}{8}$ = 1 $\frac{5}{8}$
8 × 1 + 5 = 13
So, you should use 1 $\frac{5}{8}$ inches of screws.

Question 23.
You and your friend each measure the distance between two bean bag toss boards. You record the distance as 3$\frac{3}{5}$ meters. Your friend records the distance as $\frac{18}{5}$ meters. Did you and your friend record the same distance? Explain.

Answer:
Given that,
You and your friend each measure the distance between two bean bag toss boards.
You record the distance as 3$\frac{3}{5}$ meters. Your friend records the distance as $\frac{18}{5}$ meters.
3$\frac{3}{5}$
5 × 3 + 3 = 18
3$\frac{3}{5}$ = $\frac{18}{5}$
Yes you and your friend record the same distance.

Question 24.
DIG DEEPER!
You use a $\frac{1}{3}$-cup scoop to measure 3$\frac{1}{3}$ cups of rice. How many times do you fill the scoop?

Answer:
Given,
You use a $\frac{1}{3}$-cup scoop to measure 3$\frac{1}{3}$ cups of rice.
$\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ = $\frac{10}{3}$
You need to measure 10 times to fill the scoop.

Question 25.
DIG DEEPER!
A sunflower plant is $\frac{127}{10}$ centimeters tall. A snapdragon plant is 8$\frac{9}{10}$ centimeters tall. Which plant is taller? Explain.

Answer:
Given that,
A sunflower plant is $\frac{127}{10}$ centimeters tall. A snapdragon plant is 8$\frac{9}{10}$ centimeters tall.
8$\frac{9}{10}$ = $\frac{89}{10}$
$\frac{127}{10}$ is greater than $\frac{89}{10}$
So, sunflower plant is taller.

Model Fractions and Mixed Numbers Homework & Practice 8.6

Write a mixed number as a fraction.

Question 1.
1$\frac{7}{10}$

Answer: $\frac{17}{10}$

Explanation:
Step 1
Multiply the denominator by the whole number
10 × 1 = 10
Step 2
Add the answer from Step 1 to the numerator
10 + 7 = 17
Step 3
Write an answer from Step 2 over the denominator
=17/10

Question 2.
1$\frac{5}{6}$

Answer: $\frac{11}{6}$

Explanation:
Step 1
Multiply the denominator by the whole number
6 × 1 = 6
Step 2
Add the answer from Step 1 to the numerator
6 + 5 = 11
Step 3
Write an answer from Step 2 over the denominator
11/6

Question 3.
2$\frac{2}{3}$

Answer: $\frac{8}{3}$

Explanation:
Step 1
Multiply the denominator by the whole number
3 × 2 = 6
Step 2
Add the answer from Step 1 to the numerator
6 + 2 = 8
Step 3
Write an answer from Step 2 over the denominator
8/3

Question 4.
4$\frac{1}{2}$

Answer: $\frac{9}{2}$

Explanation:
Step 1
Multiply the denominator by the whole number
2 × 4 = 8
Step 2
Add the answer from Step 1 to the numerator
8 + 1 = 9
Step 3
Write an answer from Step 2 over the denominator
9/2

Question 5.
3$\frac{2}{8}$

Answer: $\frac{13}{4}$

Explanation:
Step 1
Multiply the denominator by the whole number
8 × 3 = 24
Step 2
Add the answer from Step 1 to the numerator
24 + 2 = 26
Step 3
Write an answer from Step 2 over the denominator
26/8

Question 6.
9$\frac{8}{12}$.

Answer: $\frac{29}{3}$

Explanation:
Step 1
Multiply the denominator by the whole number
12 × 9 = 108
Step 2
Add the answer from Step 1 to the numerator
108 + 8 = 116
Step 3
Write an answer from Step 2 over the denominator
116/12 = $\frac{29}{3}$

write the fraction as a mixed number or a whole number.

Question 7.
$\frac{7}{5}$

Answer: 1 $\frac{2}{5}$

Explanation:
Given the expression $\frac{7}{5}$
We have to convert the improper fraction to the mixed fraction.
7 ÷ 5=1R2
$\frac{7}{5}$ = 1 $\frac{2}{5}$

Question 8.
$\frac{10}{3}$

Answer: 3 $\frac{1}{3}$

Explanation:
Given the expression $\frac{10}{3}$
We have to convert the improper fraction to the mixed fraction.
10÷3=3R1
$\frac{10}{3}$ = 3 $\frac{1}{3}$

Question 9.
$\frac{15}{4}$

Answer: 3 $\frac{3}{4}$

Explanation:
Given the expression $\frac{15}{4}$
We have to convert the improper fraction to the mixed fraction.
15÷4=3 R 3
$\frac{15}{4}$ = 3 $\frac{3}{4}$

Question 10.
$\frac{32}{6}$

Answer: 5 $\frac{1}{3}$

Explanation:
Given the expression $\frac{32}{6}$
We have to convert the improper fraction to the mixed fraction.
$\frac{32}{6}$ = $\frac{16}{3}$
16÷3=5R1
$\frac{32}{6}$ = 5 $\frac{1}{3}$

Question 11.
$\frac{75}{8}$

Answer: 9 $\frac{3}{8}$

Explanation:
Given the expression $\frac{75}{8}$
We have to convert the improper fraction to the mixed fraction.
75÷8=9R3
$\frac{75}{8}$ = 9 $\frac{3}{8}$

Question 12.
$\frac{40}{10}$

Answer: 4

Explanation:
Given the expression $\frac{40}{10}$
We have to convert the improper fraction to the mixed fraction.
$\frac{40}{10}$ = $\frac{4}{1}$ = 4
Thus $\frac{40}{10}$ = 4

Compare

Question 13.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 145$

Answer: <

Explanation:
We have to convert the improper fraction to the mixed fraction.
5 $\frac{1}{2}$ = $\frac{11}{2}$
$\frac{11}{2}$ < $\frac{15}{2}$

Question 14.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 146$

Answer: =

Explanation:
We have to convert the improper fraction to the mixed fraction.
$\frac{27}{12}$ = $\frac{27}{12}$

Question 15.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 147$

Answer:

Explanation:
We have to convert the improper fraction to the mixed fraction.
6 $\frac{7}{8}$ = $\frac{55}{8}$
$\frac{55}{8}$ > $\frac{50}{8}$

Question 16.
Number Sense
Complete the number line.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 148$

Answer:
$Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-148$

Question 17.
Modeling Real Life
You need pencil lead that is $\frac{12}{10}$ millimeters thick to complete an art project. Which size of pencil lead should you use?
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 149$

Answer:
Given,
You need pencil lead that is $\frac{12}{10}$ millimeters thick to complete an art project.
1 $\frac{1}{10}$ = $\frac{11}{10}$
1 $\frac{2}{10}$ = $\frac{12}{10}$
1 $\frac{4}{10}$ = $\frac{14}{10}$
You should use 2nd pencil lead.

Question 18.
DIG DEEPER!
You have a $\frac{1}{4}$-cup measuring cup and a $\frac{1}{2}$-cup measuring cup. What are two ways you can 2$\frac{3}{4}$ cups of water?

Answer:
Given,
You have a $\frac{1}{4}$-cup measuring cup and a $\frac{1}{2}$-cup measuring cup.
$\frac{1}{4}$ + $\frac{1}{2}$ = $\frac{3}{4}$
2$\frac{3}{4}$ – $\frac{3}{4}$ = 2

Review & Refresh

Question 19.
67 × 31 = ___

Answer:
Multiply the two numbers 67 and 31.
67 × 31 = 2077

Question 20.
83 × 47 = ___

Answer:
Multiply the two numbers 83 and 47.
83 × 47 = 3901

Lesson 8.7 Add Mixed Numbers

Use model to find 2$\frac{3}{8}$ + 1$\frac{1}{8}$.

Answer: 3 $\frac{1}{2}$

Explanation:
Add the fractional parts and then the whole numbers.
Rewriting our equation with parts separated
=2+3/8+1+1/8
Solving the whole number parts
2+1=3
Solving the fraction parts
3/8+1/8=4/8
Reducing the fraction part, 4/8,
4/8=1/2
Combining the whole and fraction parts
3+1/2=3 1/2

Construct Arguments
How can you use the whole number parts and the fractional parts to add mixed numbers with like denominators? Explain why your method makes sense.

Answer: To add mixed numbers, we first add the whole numbers together, and then the fractions. If the sum of the fractions is an improper fraction, then we change it to a mixed number.

Think and Grow: Add Mixed Numbers

To add mixed numbers, add the fractional parts and add the whole number parts. Another way to add mixed numbers is to rewrite each number as a fraction, then add.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 150$
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 151$

Answer:
$Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-151$

Example
Find 4$\frac{2}{8}$ + 2$\frac{7}{8}$.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 152$

Answer:
$Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-152$

Apply and Grow: Practice

Add.

Question 3.
5$\frac{1}{3}$ + 3$\frac{2}{3}$ = ___

Answer: 9

Explanation:
Add the fractional parts and then the whole numbers.
Rewriting our equation with parts separated
=5+1/3+3+2/3
Solving the whole number parts
5+3=8
Solving the fraction parts
1/3+2/3=33
Reducing the fraction part, 3/3,
3/3=1/1
Simplifying the fraction part, 1/1,
1/1=1
Combining the whole and fraction parts
8+1=9

Question 4.
2$\frac{8}{12}$ + 7$\frac{5}{12}$ = ___

Answer: 10 $\frac{1}{12}$

Explanation:
Add the fractional parts and then the whole numbers.
Rewriting our equation with parts separated
=2+8/12+7+5/12
Solving the whole number parts
2+7=9
Solving the fraction parts
8/12+5/12=13/12
Simplifying the fraction part, 13/12,
13/12=1 1/12
Combining the whole and fraction parts
9+1+1/12=10 1/12

Question 5.
4 + 1$\frac{1}{2}$ = ___

Answer: 5 $\frac{1}{2}$

Explanation:
Add the fractional parts and then the whole numbers.
Rewriting our equation with parts separated
=4+1+1/2
Solving the whole number parts
4+1=5
Combining the whole and fraction parts
5+1/2=5 1/2

Question 6.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 153$

Answer: 5 $\frac{2}{100}$

Explanation:
Add the fractional parts and then the whole numbers.
Rewriting our equation with parts separated
=78/100+124/100 + 3
Solving the fraction parts
78/100+124/100=202/100
3 + 202/100 = 5 $\frac{2}{100}$

Question 7.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 154$

Answer: 16 $\frac{3}{8}$

Explanation:
Add the fractional parts and then the whole numbers.
8 + 5 + 2 = 15
$\frac{4}{8}$ + $\frac{3}{8}$ + $\frac{4}{8}$ = (4 + 4 + 3)/8 = $\frac{11}{8}$
$\frac{11}{8}$ = 1 $\frac{3}{8}$
15 + 1$\frac{3}{8}$ = 16 $\frac{3}{8}$

Question 8.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 155$

Answer: 24 $\frac{2}{5}$

Explanation:
Add the fractional parts and then the whole numbers.
10 + 9 + 4 = 23
$\frac{4}{5}$ + $\frac{2}{5}$ + $\frac{1}{5}$ = $\frac{7}{5}$
Convert it into the mixed fraction.
$\frac{7}{5}$ = 1 $\frac{2}{5}$
23 + 1 $\frac{2}{5}$ = 24 $\frac{2}{5}$

Question 9.
Number Sense
Explain how to use the addition properties to find $Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 156$ mentally. Then find the sum.

Answer: 16 $\frac{3}{4}$

Explanation:
Add the fractional parts and then the whole numbers.
6 + 8 + 1 = 15
$\frac{3}{4}$ + $\frac{2}{4}$ + $\frac{1}{4}$ = $\frac{6}{4}$
Convert it into the mixed fraction.
$\frac{6}{4}$ = 1 $\frac{2}{4}$
15 + 1 $\frac{2}{4}$ = 16 $\frac{2}{4}$

Question 10.
DIG DEEPER!
When adding mixed numbers, is it always necessary to write the sum as a mixed number? Explain.

Answer: To add mixed numbers, we first add the whole numbers together, and then the fractions. If the sum of the fractions is an improper fraction, then we change it to a mixed number.

Question 11.
DIG DEEPER!
Find the unknown number.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 156.1$

Answer:
Let the unknown number be x.
4 $\frac{5}{6}$ + x = 8 $\frac{3}{6}$
x = 8 $\frac{3}{6}$ – 4 $\frac{5}{6}$
x = 3 $\frac{2}{3}$

Think and Grow: Modeling Real Life

Example
You pick 2$\frac{3}{4}$ pounds of cherries. Your friend picks 1$\frac{2}{4}$ pounds of cherries. How many pounds of cherries do you and your friend pick in all?
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 157$
Add the amounts of cherries you and your friend each pick.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 158$

Answer:
You pick 2$\frac{3}{4}$ pounds of cherries. Your friend picks 1$\frac{2}{4}$ pounds of cherries.
$Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-158$

Show and Grow

Question 12.
Before noon, 2$\frac{3}{8}$ inches of snow falls in a city. Afternoon, 4$\frac{6}{8}$ inches of snow falls. How many inches of snow falls in the city that day?

Answer:
Given that,
Before noon, 2$\frac{3}{8}$ inches of snow falls in a city. Afternoon, 4$\frac{6}{8}$ inches of snow falls.
2$\frac{3}{8}$ + 4$\frac{6}{8}$
2 + 4 = 6
$\frac{3}{8}$ + $\frac{6}{8}$ = $\frac{9}{8}$
Convert it into the mixed fraction.
$\frac{9}{8}$ = 1 $\frac{1}{8}$
1 $\frac{1}{8}$ inches of snow falls in the city that day.

Question 13.
DIG DEEPER!
A student driver must practice driving at night for a total of at least 10 hours. Has the student met the nighttime driving requirement yet?
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 159$

Answer:
2 $\frac{1}{2}$ + 3 $\frac{1}{2}$ + 2 $\frac{1}{2}$
First add the whole numbers
2 + 3 + 2 = 7
$\frac{1}{2}$ + $\frac{1}{2}$ + $\frac{1}{2}$ = 1 $\frac{1}{2}$
7 + 1 $\frac{1}{2}$ = 8 $\frac{1}{2}$

Add Mixed Numbers Homework & Practice 8.7

Add.

Question 1.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 160$

Answer: 12 $\frac{4}{5}$

Explanation:
Add the fractional parts and then the whole numbers.
4 + 8 = 12
Add the fractions
$\frac{1}{5}$ + $\frac{3}{5}$ = $\frac{4}{5}$
Now add the fractions and thw whole numbers
12 + $\frac{4}{5}$ = 12 $\frac{4}{5}$

Question 2.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 161$

Answer: 20 $\frac{3}{8}$

Explanation:
Add the fractional parts and then the whole numbers.
10 + 9 = 19
Add the fractions
$\frac{5}{8}$ + $\frac{6}{8}$ = $\frac{11}{8}$
Convert it into the mixed fraction.
$\frac{11}{8}$ = 1 $\frac{3}{8}$
Now add the fractions and the whole numbers
19 + 1 $\frac{3}{8}$ = 20 $\frac{3}{8}$

Question 3.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 162$

Answer: 8 $\frac{1}{3}$

Explanation:
Add the fractional parts and then the whole numbers.
2 + 6 = 8
Now add the fractions and the whole numbers
$\frac{1}{3}$ + 8 = 8 $\frac{1}{3}$

Question 4.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 163$

Answer:

Explanation:
Add the fractional parts and then the whole numbers.
3 + 4 = 7
$\frac{10}{12}$ + $\frac{10}{12}$ = $\frac{20}{12}$
Convert it into the mixed fraction.
$\frac{20}{12}$ = 1 $\frac{8}{12}$
Now add the fractions and the whole numbers
7 + 1 $\frac{8}{12}$ = 8 $\frac{8}{12}$

Question 5.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 164$

Answer: 9 $\frac{2}{6}$

Explanation:
Add the fractional parts and then the whole numbers.
Convert it into the mixed fraction.
$\frac{11}{6}$ = 1 $\frac{5}{6}$
7 + 1 = 8
$\frac{3}{6}$ + $\frac{5}{6}$ = $\frac{8}{6}$
Now add the fractions and the whole numbers
8 + $\frac{8}{6}$
$\frac{8}{6}$ = 1 $\frac{2}{6}$
8 + 1 $\frac{2}{6}$ = 9 $\frac{2}{6}$

Question 6.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 165$

Answer: 15 $\frac{1}{4}$

Explanation:
Add the fractional parts and then the whole numbers.
Rewriting our equation with parts separated
=8+70/100+6+55/100
Solving the whole number parts
8+6=14
Solving the fraction parts
70/100+55/100=125/100
Reducing the fraction part, 125/100,
125/100=5/4
Simplifying the fraction part, 5/4,
5/4=1 1/4
Combining the whole and fraction parts
14+1+1/4= 15 $\frac{1}{4}$

Add

Question 7.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 166$

Answer: 11

Explanation:
Add the fractional parts and then the whole numbers.
5 + 3 + 2 = 10
Now add the fractional part,
3/4 + 1/4 = 1
10 + 1 = 11

Question 8.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 167$

Answer: 12 $\frac{1}{2}$

Explanation:
Add the fractional parts and then the whole numbers.
Add the whole numbers
1 + 1 + 9 = 11
$\frac{1}{2}$ + $\frac{1}{2}$ + $\frac{1}{2}$ = 1 $\frac{1}{2}$
Combining the whole and fraction parts
11 + 1 $\frac{1}{2}$ = 12 $\frac{1}{2}$

Question 9.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 168$

Answer: 9 $\frac{2}{10}$

Explanation:
Add the fractional parts and then the whole numbers.
3 + 5 = 8
$\frac{4}{10}$ + $\frac{2}{10}$ + $\frac{6}{10}$ = $\frac{12}{10}$
Convert it into the mixed fraction.
$\frac{12}{10}$ = 1 $\frac{2}{10}$
Combining the whole and fraction parts
8 + 1 $\frac{2}{10}$ = 9 $\frac{2}{10}$

Question 10.
Structure
Find 7$\frac{4}{5}$ + 8$\frac{2}{5}$ two different ways. Which way do you prefer? Why?

Answer:

Explanation:
Add the fractional parts and then the whole numbers.
7$\frac{4}{5}$ + 8$\frac{2}{5}$
Add the fractional parts and then the whole numbers.
7 + 8 = 15
$\frac{4}{5}$ + $\frac{2}{5}$ = $\frac{6}{5}$
Convert it into the mixed fraction.
$\frac{6}{5}$ = 1 $\frac{1}{5}$
15 + 1 $\frac{1}{5}$ = 16 $\frac{1}{5}$

Question 11.
Modeling Real Life
A homeowner has two strings of lights. One is 8$\frac{1}{3}$ yards long. The other is 16$\frac{2}{3}$ yards long. He connects the strings of lights. How long will the string of lights be in all?
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 169$

Answer:
Given,
A homeowner has two strings of lights. One is 8 $\frac{1}{3}$ yards long. The other is 16 $\frac{2}{3}$ yards long. He connects the strings of lights.
8 $\frac{1}{3}$ + 16 $\frac{2}{3}$ = 24 $\frac{3}{3}$
$\frac{3}{3}$ = 1
24 + 1 = 25

Question 12.
DIG DEEPER!
You can play the song “Mary Had a Little Lamb” by striking three glasses filled with water to make the tone. The first glass needs 1$\frac{3}{4}$ cups, the second glass needs 1$\frac{1}{2}$ cups, and the third glass needs 1$\frac{1}{4}$ cups of water. How much water do you need in all?

Answer:
Given that,
You can play the song “Mary Had a Little Lamb” by striking three glasses filled with water to make the tone. The first glass needs 1$\frac{3}{4}$cups, the second glass needs 1$\frac{1}{2}$ cups, and the third glass needs 1$\frac{1}{4}$ cups of water.
1$\frac{1}{2}$ + 1$\frac{1}{4}$ + 1$\frac{3}{4}$
1 + 1 + 1 = 3
$\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{3}{4}$ = 1 $\frac{1}{2}$
3 + 1 $\frac{1}{2}$ = 4 $\frac{1}{2}$
Therefore you need 4 $\frac{1}{2}$ cups of water.

Review & Refresh

Write the first six numbers in the pattern. Then describe another feature of the pattern.

Question 13.
Rule: Add 11.
First number: 22

Answer:
The first number is 22 you need to add 11 to it.
22 + 11 = 33
33 + 11 = 44
44 + 11 = 55
55 + 11 = 66
66 + 11 = 77
77 + 11 = 88

Question 14.
Rule: Multiply by 4.
First number: 7

Answer:
The first number is 7. You need to multiply by 4.
7 × 4 = 28
28 × 4 = 112
112 × 4 = 448
448 × 4 = 1792
1792 × 4 = 7168
7168 × 4 = 28672

Lesson 8.8 Subtract Mixed Numbers

Explore and Grow

Use a model to find 2$\frac{3}{8}$ – 1$\frac{1}{8}$.

Answer:
2$\frac{3}{8}$ – 1$\frac{1}{8}$.
First subtract the whole numbers
2 – 1 = 1
$\frac{3}{8}$ – $\frac{1}{8}$ = $\frac{2}{8}$
Combine the whole numbers and fractions.
1 $\frac{2}{8}$ = 1 $\frac{1}{4}$

Construct Arguments
How can you use the whole number parts and the fractional parts to subtract mixed numbers with like denominators? Explain why your method makes sense.

Answer:
First, you have to subtract the whole number parts and then subtract the fraction parts with like denominators.
You can subtract the mixed fractions by using the number line or model.

Think and Grow: subtract Mixed Numbers

To subtract mixed numbers, subtract the fractional parts and subtract the whole number parts. Another way to subtract mixed numbers is to rewrite each number as a fraction, then subtract.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 170$

Answer:
$Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-170$

Example
Find 5$\frac{3}{6}$ – 4$\frac{5}{6}$.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 171$

Answer:
$Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-171$

Show and Grow

Subtract

Question 1.
5$\frac{4}{5}$ – 1$\frac{2}{5}$ = ____

Answer:
There are enough fifths.
Subtract the whole numbers
5 – 1 = 4
$\frac{4}{5}$ – $\frac{2}{5}$ =$\frac{2}{5}$
4 + $\frac{2}{5}$ = 4 $\frac{2}{5}$
Thus, 5$\frac{4}{5}$ – 1$\frac{2}{5}$ = 4 $\frac{2}{5}$

Question 2.
7$\frac{1}{3}$ – 2$\frac{2}{3}$ = _____

Answer:
There are not enough thirds.
7$\frac{1}{3}$ = 6 $\frac{3}{3}$ + $\frac{1}{3}$ = 6$\frac{4}{3}$
Subtract the whole numbers
6 – 2 = 4
$\frac{4}{3}$ – $\frac{2}{3}$ = $\frac{2}{3}$
4 + $\frac{2}{3}$ = 4 $\frac{2}{3}$

Question 3.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 172$

Answer:
There are enough twelfths.
Subtract the whole numbers
15 – 4 = 11
$\frac{10}{12}$ – $\frac{8}{12}$ = $\frac{2}{12}$
11 + $\frac{2}{12}$ = 11 $\frac{2}{12}$

Question 4.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 173$

Answer:
There are enough eighths.
$\frac{6}{8}$ – $\frac{6}{8}$ = 0
Subtract the whole numbers
6 – 3 = 3
3 + 0 = 3

Question 5.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 174$

Answer:
There are not enough tenths.
5 $\frac{7}{10}$ can be written as 4 $\frac{17}{10}$
4 $\frac{17}{10}$ – 1 $\frac{9}{10}$
Subtract the whole numbers
4 – 1 = 3
Subtract the fractional parts
$\frac{17}{10}$ – $\frac{9}{10}$ = $\frac{8}{10}$
3 + $\frac{8}{10}$ = 3 $\frac{8}{10}$

Question 6.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 175$

Answer:
There are not enough hundreds.
11 $\frac{50}{100}$ can be written as 10 $\frac{150}{100}$
Subtract the whole numbers
10 – 7 = 3
Subtract the fractional parts
$\frac{150}{100}$ – $\frac{85}{100}$ = $\frac{65}{100}$
3 + $\frac{65}{100}$ = 3 $\frac{65}{100}$

Question 7.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 176$

Answer:
There are not enough sixths.
8 can be written as 7 $\frac{6}{6}$
Subtract the whole numbers
7 – 1 = 6
Subtract the fractional parts
$\frac{6}{6}$ – $\frac{3}{6}$ = $\frac{3}{6}$
6 + $\frac{3}{6}$ = 6 $\frac{3}{6}$

Question 8.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 177$

Answer:
There are not enough fourths.
10 can be written as 9 $\frac{4}{4}$
Subtract the whole numbers
9 – 9 = 0
Subtract the fractional parts
$\frac{4}{4}$ – $\frac{3}{4}$ = $\frac{1}{4}$
0 + $\frac{1}{4}$ = $\frac{1}{4}$

Question 9.
YOU BE THE TEACHER
Your friend says the difference of 9 and 2$\frac{3}{5}$ is 7$\frac{3}{5}$. Is your friend correct? Explain.

Answer:
9 – 2$\frac{3}{5}$ = 7 $\frac{3}{5}$
Thus by this we can say that your friend is correct.

Question 10.
Writing
Explain how adding and subtracting mixed numbers are similar and different.

Answer:
Any mixed number can also be written as an improper fraction, in which the numerator is larger than the denominator.
Subtracting mixed numbers is very similar to adding them.
Write both fractions as equivalent fractions with a denominator. Then subtract the fractions.

Question 11.
DIG DEEPER!
Write two mixed numbers with like denominators that have a sum of 5$\frac{2}{3}$ and a difference of 1.

Answer:
5$\frac{2}{3}$ = 3 $\frac{1}{3}$ + 2$\frac{1}{3}$
Now if you subtract the same fraction you need to get the difference as 1.
3 $\frac{1}{3}$ – 2$\frac{1}{3}$
3 – 2 = 1
$\frac{1}{3}$ – $\frac{1}{3}$ = 0
So, 3 $\frac{1}{3}$ – 2$\frac{1}{3}$ = 1

Think and Grow: Modeling Real Life

Example
A replica of the Eiffel Tower is 6 inches tall. It is 2$\frac{2}{5}$ inches taller than a replica of the Space Needle. How tall is the replica of the Space Needle?
Find the difference between the height of the Eiffel Tower replica, 6 inches, and 2$\frac{2}{5}$ inches.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 178$
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 179$

Answer:
Given,
A replica of the Eiffel Tower is 6 inches tall. It is 2$\frac{2}{5}$ inches taller than a replica of the Space Needle.
$Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-179$

Show and Grow

Question 12.
A cook has a 5-pound bag of potatoes. He uses 2$\frac{1}{3}$ pounds of potatoes to make a casserole. How many pounds of potatoes are left?

Answer:
Given,
A cook has a 5-pound bag of potatoes. He uses 2$\frac{1}{3}$ pounds of potatoes to make a casserole.
5 – 2$\frac{1}{3}$
4 $\frac{3}{3}$ – 2$\frac{1}{3}$
Subtract the whole numbers
4 – 2 = 2
Subtract the fractional parts
$\frac{3}{3}$ – $\frac{1}{3}$ = $\frac{2}{3}$
2 $\frac{2}{3}$
Thus 2 $\frac{2}{3}$ pounds of potatoes are left.

Question 13.
A half-marathon is 13$\frac{1}{10}$ miles long. A competitor runs 9$\frac{6}{10}$ miles. How many miles does the competitor have left to run?

Answer:
Given,
A half-marathon is 13$\frac{1}{10}$ miles long. A competitor runs 9$\frac{6}{10}$ miles.
13$\frac{1}{10}$ – 9$\frac{6}{10}$
12 $\frac{11}{10}$ – 9$\frac{6}{10}$
Subtract the whole numbers
12 – 9 = 3
$\frac{11}{10}$ – $\frac{6}{10}$ = $\frac{5}{10}$
3 + $\frac{5}{10}$ = 3 $\frac{1}{2}$
The competitor has left 3 $\frac{1}{2}$ miles to run.

Question 14.
DIG DEEPER!
You want to mail a package that weighs 18$\frac{2}{4}$ ounces. The weight limit is 13 ounces, so you remove 4$\frac{3}{4}$ ounces of items from the package. Does the lighter package meet the weight requirement? If not, how much more weight do you need to remove?

Answer:
Given that,
You want to mail a package that weighs 18$\frac{2}{4}$ ounces.
The weight limit is 13 ounces, so you remove 4$\frac{3}{4}$ ounces of items from the package.
18$\frac{2}{4}$ – 4$\frac{3}{4}$ = 13 $\frac{3}{4}$
13 $\frac{3}{4}$ – 13 = $\frac{3}{4}$
Thus you need to remove $\frac{3}{4}$ ounces more.

Subtract Mixed Numbers Homework & Practice 8.8

Subtract

Question 1.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 180$

Answer: 5 $\frac{1}{2}$

Explanation:
Rewriting our equation with parts separated
=10+3/4−5−1/4
Solving the whole number parts
10−5=5
Solving the fraction parts
3/4−1/4=2/4
Reducing the fraction part, 2/4,
2/4=1/2
Combining the whole and fraction parts
5+1/2=5 1/2

Question 2.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 181$

Answer: 6

Explanation:
Rewriting our equation with parts separated
9 + 1/3 – 3 – 1/3
9 – 3 = 6
So, 9 $\frac{1}{3}$ – 3 $\frac{1}{3}$ = 6

Question 3.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 182$

Answer: 4 $\frac{2}{3}$

Explanation:
Rewriting our equation with parts separated
=6+7/12−1−11/12
Solving the whole number parts
6−1=5
Solving the fraction parts
7/12−11/12=−4/12
Reducing the fraction part, 4/12,
−4/12=−1/3
Combining the whole and fraction parts
5−1/3=4 2/3

Question 4.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 183$

Answer: 6 $\frac{43}{50}$

Explanation:
Rewriting our equation with parts separated
=15+6/100−8−20/100
Solving the whole number parts
15−8=7
Solving the fraction parts
6/100−20/100=−14/100
Reducing the fraction part, 14/100,
−14/100=−7/50
Combining the whole and fraction parts
7−7/50=6 43/50

Question 5.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 184$

Answer: 1 $\frac{2}{3}$

Explanation:
Rewriting our equation with parts separated
=4+3/6−2−5/6
Solving the whole number parts
4−2=2
Solving the fraction parts
3/6−5/6=−2/6
Reducing the fraction part, 2/6,
−2/6=−1/3
Combining the whole and fraction parts
2−1/3=1 2/3

Question 6.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 185$

Answer: 1 $\frac{3}{5}$

Explanation:
20 – 19 = 1
$\frac{4}{5}$ – $\frac{1}{5}$ = $\frac{3}{5}$
1 + $\frac{3}{5}$ = 1 $\frac{3}{5}$

Subtract.

Question 7.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 186$

Answer: 2 $\frac{3}{5}$

Explanation:
Rewriting our equation with parts separated
=5+6/10−3
Solving the whole number parts
5−3=2
Combining the whole and fraction parts
2+6/10=2 6/10

Question 8.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 187$

Answer: 10 $\frac{1}{2}$

Explanation:
Rewriting our equation with parts separated
=13−2−1/2
Solving the whole number parts
13−2=11
Combining the whole and fraction parts
11−1/2=10 1/2

Question 9.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 188$

Answer: 3 $\frac{1}{4}$

Explanation:
Rewriting our equation with parts separated
=18−14−6/8
Solving the whole number parts
18−14=4
Combining the whole and fraction parts
4−6/8=3 2/8

Question 10.
Reasoning
Explain why you rename 4$\frac{1}{3}$ when finding 4$\frac{1}{3}$ – $\frac{2}{3}$ .

Answer:
4$\frac{1}{3}$ – $\frac{2}{3}$
4 can be written as 3 $\frac{3}{3}$
3 $\frac{3}{3}$ – $\frac{2}{3}$
3 + $\frac{3}{3}$ – $\frac{2}{3}$
3 + $\frac{1}{3}$ = 3 $\frac{1}{3}$
So, 4$\frac{1}{3}$ – $\frac{2}{3}$ = 3 $\frac{1}{3}$

Question 11.
DIG DEEPER!
Find the unknown number.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 189$

Answer:
Let the unknown number be x.
10 $\frac{3}{12}$ – x = $\frac{4}{12}$
10 $\frac{3}{12}$ – $\frac{4}{12}$ = x
9 $\frac{15}{12}$ – $\frac{4}{12}$ = x
x = 9 $\frac{11}{12}$
Thus the unknown number is 9 $\frac{11}{12}$.

Question 12.
Modeling Real Life
A rare flower found in Indonesian rain forests can grow wider than a car tire. How much wider is the flower than a car tire that is 1$\frac{11}{12}$ feet wide?
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 189.1$

Answer:
Given,
A rare flower found in Indonesian rain forests can grow wider than a car tire.
3 – 1$\frac{11}{12}$
2 $\frac{12}{12}$ – 1$\frac{11}{12}$
= 1 $\frac{1}{12}$

Question 13.
Modeling Real Life
Your tablet battery is fully charged. You use $\frac{32}{100}$ of the charge listening to music, and $\frac{13}{100}$ of the charge playing games. What fraction of the charge remains on your tablet battery?

Answer:
Given,
Your tablet battery is fully charged. You use $\frac{32}{100}$ of the charge listening to music, and $\frac{13}{100}$ of the charge playing games.
$\frac{32}{100}$ – $\frac{13}{100}$ = $\frac{19}{100}$
Thus $\frac{19}{100}$ fraction of the charge remains on your tablet battery.

Review & Refresh

Divide. Then check your answer.

Question 14.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 190$

Answer:
Divide 84 by 5
84/5 = 16.8

Question 15.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 191$

Answer:
Divide 51 by 4.
51/4 = 12.75

Question 16.
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 192$

Answer:
Divide 89 by 8.
89/8 = 11.125

Lesson 8.9 Problem Solving: Fractions

Explore and Grow

Make a plan to solve the problem.

The table shows the tusk lengths of two elephants. Which elephant’s tusks have a greater total length? How much greater?
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 193$
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 194$

Answer:
Male Elephant = 4 1/12 + 4 3/12 = 8 4/12
Female Elephant = 4 + 3 7/12 = 7 7/12
The Right Tusk of a Male Elephant is greater than Female Elephant.
The left tusk of a Male Elephant is greater than Female Elephant.
Thus the total length of the Male Elephant is greater than Female Elephant.

Make Sense of Problems
A $\frac{7}{12}$-foot long piece of one of the male elephant’s tusks breaks off. Does this change your plan to solve the problem? Will this change the answer? Explain.

Answer:
A $\frac{7}{12}$-foot long piece of one of the male elephant’s tusks breaks off.
8 4/12 – 7 7/12 = 3/4
No, if $\frac{7}{12}$-foot long piece of one of the male elephant’s tusks breaks off it will not change the answer. Still, the Male Elephant is greater than Female Elephant.

Think and Grow: Problem Solving: Fractions

Example
A family spends 2$\frac{2}{4}$ hours traveling to a theme park, 7$\frac{1}{4}$ hours at the theme park, and 2$\frac{3}{4}$ hours traveling home. How much more time does the family spend at the theme park than traveling?
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 195$

Understand the Problem

What do you know?

The family spends 2$\frac{2}{4}$ hours traveling to the theme park, 7$\frac{1}{4}$ hours at the theme park, 2$\frac{3}{4}$ hours traveling home.
What do you need to find?
You need to find how much more time the family spends at the theme park than the traveling.

Make a plan

How will you solve it?

Add 2$\frac{2}{4}$ and 2$\frac{3}{4}$ to find how much time the family spends traveling.
Then subtract the sum from 7$\frac{1}{4}$ to find how much more time they spend at the theme park.

Solve
So, the family spends ___ more hours at the theme park than traveling.

Show and Grow

Question 1.
Explain how you can check your answer in each step of the example above.

Answer:
$Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-196$
So, the family spends 2 more hours at the theme park than traveling.

Apply any and Grow: Practice

Understand the problem. What do you know? What do you need to find? Explain.

Answer:

The family spends 2$\frac{2}{4}$ hours traveling to the theme park, 7$\frac{1}{4}$ hours at the theme park, 2$\frac{3}{4}$ hours traveling home.
What do you need to find?
You need to find how much more time the family spends at the theme park than the traveling.

Question 2.
You are making a sand art bottle. You fill $\frac{1}{6}$ of the bottle with pink sand, $\frac{3}{6}$ with red sand, and $\frac{2}{6}$ with white sand. How much of the bottle is filled?

Answer:
Given that,
You are making a sand art bottle. You fill $\frac{1}{6}$ of the bottle with pink sand, $\frac{3}{6}$ with red sand, and $\frac{2}{6}$ with white sand.
$\frac{1}{6}$ + $\frac{3}{6}$ + $\frac{2}{6}$ = $\frac{1}{6}$
Thus $\frac{1}{6}$ of the bottle is filled.

Question 3.
Your friend has $\frac{1}{8}$ of a photo album filled with beach photographs and $\frac{4}{8}$ of the album filled with photos of friends. What fraction of the photo album is left?

Answer:
Given that,
Your friend has $\frac{1}{8}$ of a photo album filled with beach photographs and $\frac{4}{8}$ of the album filled with photos of friends.
$\frac{1}{8}$ + $\frac{4}{8}$ = $\frac{5}{8}$
$\frac{8}{8}$ – $\frac{5}{8}$ = $\frac{3}{8}$
Thus $\frac{3}{8}$ fraction of the photo album is left.

Understand the problem. Then make a plan. How will you solve? Explain.

Question 4.
In Race A, an Olympic swimmer swims 100 meters in 62$\frac{25}{100}$ seconds. In Race B, she cuts 2$\frac{38}{100}$ seconds off her Race A time. How many seconds does she need to cut off her Race B time to swim 100 meters in 58$\frac{45}{100}$ seconds?

Answer:
Given,
In Race A, an Olympic swimmer swims 100 meters in 62$\frac{25}{100}$ seconds. In Race B, she cuts 2$\frac{38}{100}$ seconds off her Race A time.
62$\frac{25}{100}$ – 2$\frac{38}{100}$ = 59 $\frac{87}{100}$
59 $\frac{87}{100}$ – 58$\frac{45}{100}$ = 1 $\frac{42}{100}$
She need 1 $\frac{42}{100}$ to cut off her Race B time to swim 100 meters in 58$\frac{45}{100}$ seconds.

Question 5.
A semi-truck has 2 fuel tanks that each hold the same amount of fuel. A truck driver fills up both tanks and uses $\frac{3}{4}$ tank of gasoline driving to his first stop. He uses $\frac{2}{4}$ tank of gasoline driving to his second stop. How much gasoline does he have left?

Answer:
Given that,
A semi-truck has 2 fuel tanks that each hold the same amount of fuel. A truck driver fills up both tanks and uses $\frac{3}{4}$ tank of gasoline driving to his first stop. He uses $\frac{2}{4}$ tank of gasoline driving to his second stop.
$\frac{3}{4}$ + $\frac{2}{4}$ = $\frac{5}{4}$
2 – $\frac{5}{4}$ = $\frac{3}{4}$
Thus $\frac{3}{4}$ gasoline has left.

Question 6.
A bootlace worm holds the record as the longest animal at 180 feet long. How much longer is it than 2 blue whales combined?
$Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 197$

Answer:
Given,
A bootlace worm holds the record as the longest animal at 180 feet long.
1 blue whale = 85 $\frac{8}{12}$
2 blue whales = 85 $\frac{8}{12}$ + 85 $\frac{8}{12}$ = 171 $\frac{1}{3}$
180 – 171 $\frac{1}{3}$
179 $\frac{3}{3}$ – 171 $\frac{1}{3}$ = 8 $\frac{2}{3}$

Think and Grow: Modeling Real Life

Example
You walk $\frac{1}{10}$ kilometer on Monday, $\frac{3}{10}$ kilometer on Tuesday, and $\frac{5}{10}$ kilometer on Wednesday. You continue the pattern on Thursday and Friday. How many kilometers do you walk in all?
Think: What do you know? What do you need to find? How will you solve?
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 198$
Step 1: Identify the pattern.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 199$
Step 2: Use the pattern to find the distances you walk on Thursday and Friday.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 200$
Step 3: Add all of the distances.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 201$
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 202$ .

Answer:
Step 1: Identify the pattern.
$Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-199$
Step 2: Use the pattern to find the distances you walk on Thursday and Friday.
$Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-200$
Step 3: Add all of the distances.
$Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-201$
So, you walk 2 $\frac{5}{10}$ kilometers in all.

Show and Grow

Question 7.
You save $\frac{1}{4}$ dollar the first week, $\frac{2}{4}$ dollar the next week, and dollar $\frac{3}{4}$ dollar the following week. You continue the pattern for 3 more weeks. How much money do you save after 6 weeks?
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 203$

Answer:
You save $\frac{1}{4}$ dollar the first week, $\frac{2}{4}$ dollar the next week, and dollar $\frac{3}{4}$ dollar the following week. You continue the pattern for 3 more weeks.
$\frac{1}{4}$, $\frac{2}{4}$, $\frac{3}{4}$, $\frac{4}{4}$, $\frac{5}{4}$, $\frac{6}{4}$
You save $\frac{6}{4}$ dollar after 6 weeks.

Problem Solving: Fractions Homework & Practice 8.9

Question 1.
An older washing machine uses 170$\frac{3}{10}$ liters of water per load. A new, high-efficiency, washing machine uses 75$\frac{7}{10}$ fewer liters than the older washing machine. How many liters of water will the high-efficiency washing machine use for 2 loads of laundry?

Answer:
Given,
An older washing machine uses 170$\frac{3}{10}$ liters of water per load. A new, high-efficiency, washing machine uses 75$\frac{7}{10}$ fewer liters than the older washing machine.
75$\frac{7}{10}$ + 75$\frac{7}{10}$ = 151$\frac{2}{5}$
170$\frac{3}{10}$ – 151$\frac{2}{5}$ = 18 $\frac{9}{10}$

Question 2.
A student jumps 40 $\frac{5}{12}$ inches for the high jump. On his second try, he jumps 1$\frac{8}{12}$ inches higher. He can tie the school record if he raises the bar another 3$\frac{10}{12}$ inches and successfully jumps over it. What is the school record for the high jump?

Answer:
Given,
A student jumps 40 $\frac{5}{12}$ inches for the high jump.
On his second try, he jumps 1$\frac{8}{12}$ inches higher.
He can tie the school record if he raises the bar another 3$\frac{10}{12}$ inches and successfully jumps over it.
40 $\frac{5}{12}$ + 1 $\frac{8}{12}$ = 42 $\frac{1}{12}$
40 $\frac{5}{12}$ + 3$\frac{10}{12}$ = 44 $\frac{3}{12}$
44 $\frac{3}{12}$ is the school record for the high jump.

Question 3.
You are shipping three care packages. The first package weighs 10$\frac{1}{10}$ pounds. The second weighs 5$\frac{7}{10}$ pounds, and the third weighs 25$\frac{8}{10}$ pounds. What is the total weight of the packages?

Answer:
Given,
You are shipping three care packages. The first package weighs 10$\frac{1}{10}$ pounds.
The second weighs 5$\frac{7}{10}$ pounds, and the third weighs 25$\frac{8}{10}$ pounds.
10$\frac{1}{10}$ + 5$\frac{7}{10}$ + 25$\frac{8}{10}$ = 41 $\frac{6}{10}$
The total weight of the packages is 41 $\frac{6}{10}$ pounds.

Question 4.
A person’s arm span is approximately equal to the person’s height. How tall is this fourth grader according to his arm span?
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 204$

Answer:
Given,
A person’s arm span is approximately equal to the person’s height.
By using the pattern we can find the arm span of the fourth-grader i.e., 1 $\frac{7}{12}$

Question 5.
Writing
Write and solve a two-step word problem with mixed numbers that can be solved using addition or subtraction.

Answer:
I have 5 $\frac{8}{12}$ episodes of my favorite series download onto my computer. I Downloaded some yesterday and $\frac{7}{12}$ of the episodes this morning. The download speed was really slow. What fraction of the episodes did I download yesterday?
5 $\frac{8}{12}$ – $\frac{7}{12}$ = 5 $\frac{1}{12}$ = $\frac{61}{12}$

Question 6.
Modeling Real Life
Your friend walks $\frac{2}{10}$ mile to school each day. She walks the same distance home. How many miles does she walk to and from school in one 5-day school week?

Answer:
Given,
Your friend walks $\frac{2}{10}$ mile to school each day. She walks the same distance home.
$\frac{2}{10}$ + $\frac{2}{10}$ = $\frac{4}{10}$
5 × $\frac{4}{10}$ = $\frac{20}{10}$ = 2
Thus she walk to and from school in one 5-day school week is 2 miles.

Question 7.
DIG DEEPER!
A store sells cashews in $\frac{2}{3}$-pound bags. You buy some bags and repackage the cashews into 1-pound bags. What is the least number of bags you should buy so that you do not have any cashews left over?

Answer:
Given,
A store sells cashews in $\frac{2}{3}$-pound bags. You buy some bags and repackage the cashews into 1-pound bags.
1 – $\frac{2}{3}$ = $\frac{1}{3}$
Thus $\frac{1}{3}$ pound of cashews left over.

Review & Refresh

Compare.

Question 8.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 205$

Answer: >

Explanation:
$\frac{8}{12}$ = $\frac{4}{6}$
$\frac{4}{6}$ > $\frac{1}{6}$

Question 9.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 206$

Answer: <

Explanation:
First, make the denominators common.
$\frac{9}{10}$ = $\frac{18}{20}$
$\frac{14}{8}$ = $\frac{35}{20}$
$\frac{18}{20}$ < $\frac{35}{20}$

Question 10.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 207$

Answer: >

Explanation:
First, make the denominators common.
$\frac{3}{4}$
$\frac{1}{2}$ × 2/2 = $\frac{2}{4}$
$\frac{3}{4}$ > $\frac{2}{4}$

Add and Subtract Fractions Performance Task 8

The notes on sheet music tell you what note to play and how long to hold each note. The table shows how long you hold some notes compared to the length of one whole note.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 208$
1. a. Complete the table by writing equivalent fractions.

Answer:
$Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-208$
A whole note is nothing but 1 so the fraction is 8/8.
1/2 note is nothing but 4/8.
1/4 note is nothing but 2/8.

b. Each group of notes represents one measure. What is the sum of the values of the notes in each measure?
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 209$

Answer:
$Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-209$

c. Draw the missing note to complete each measure.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 210$

Answer:
$Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-210$

d. Draw one measure of notes where the sum of the values is 1. Show your work.
___________

Answer:
$Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-209 (1)$

e. Write the fraction represented by the sum of the notes. Then write the fraction as a sum of fractions in two different ways.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 211$

Answer:
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 211$ = $\frac{1}{8}$ + $\frac{4}{8}$ + $\frac{2}{8}$ = $\frac{7}{8}$

Add and Subtract Fractions Activity

Three In a Row: Fraction Add or Subtract

Directions:

Players take turns.
On your turn, spin both spinners. Choose whether to add or subtract.
Add or subtract the mixed number and fraction. Cover the sum or difference.
If the sum or difference is already covered, you lose your turn.
The ﬁrst player to get three in a row wins!

$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 212$

Answer:
1 $\frac{1}{8}$ + $\frac{3}{8}$ = 1 + $\frac{1}{8}$ + $\frac{3}{8}$ = 1 $\frac{4}{8}$
3 $\frac{7}{8}$ + $\frac{8}{8}$ = 3 + $\frac{7}{8}$ + 1 = 4 $\frac{7}{8}$
2 $\frac{5}{8}$ + $\frac{4}{8}$ = 2 + $\frac{5}{8}$ + $\frac{4}{8}$ = 3 $\frac{1}{8}$

Add and Subtract Fractions Chapter Practice 8

8.1 Use Models to Add Fractions

Find the sum. Explain how you used the model to add.

Question 1.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 214$

Answer: 5/6
$Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-214$

Question 2.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 215$

Answer:
$Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-215$

Find the sum. Use a model or a number line to help.

Question 3.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 216$

Answer:
$BIM Grade 4 Chapter 8 add & subtract fractions img_25$

Question 4.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 217$

Answer:
$Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_216$

Question 5.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 218$

Answer:
Denominators are the same so add the numerators.
$\frac{45}{100}$ + $\frac{10}{100}$ + $\frac{9}{100}$ = $\frac{64}{100}$

8.2 Decompose Fractions

Write the fraction as a sum of unit fractions.

Question 6.
$\frac{2}{12}$

Answer:
The unit fraction for $\frac{2}{12}$ is $\frac{1}{12}$ + $\frac{1}{12}$

Question 7.
$\frac{3}{3}$

Answer: The unit fraction for $\frac{3}{3}$ is $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$

Write the fraction as a sum of fractions in two different ways.

Question 8.
$\frac{5}{8}$

Answer:
The unit fraction for $\frac{5}{8}$ is $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$

Question 9.
$\frac{6}{100}$

Answer:
The unit fraction for $\frac{6}{100}$ is $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$ + $\frac{1}{100}$

Question 10.
$\frac{90}{100}$

Answer:
$\frac{90}{100}$ = 9/10
The unit fraction for $\frac{9}{10}$ is $\frac{1}{10}$ + $\frac{1}{10}$ + $\frac{1}{10}$ + $\frac{1}{10}$ + $\frac{1}{10}$ + $\frac{1}{10}$ + $\frac{1}{10}$ + $\frac{1}{10}$ + $\frac{1}{10}$

Question 11.
$\frac{4}{5}$

Answer:
The unit fraction for $\frac{4}{5}$ is $\frac{1}{5}$ + $\frac{1}{5}$ + $\frac{1}{5}$ + $\frac{1}{5}$

8.3 Add Fractions with Like Denominators

Add

Question 12.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 219$

Answer:
Denominators are the same so add the numerators.
$\frac{5}{10}$ + $\frac{10}{10}$ = $\frac{15}{10}$

Question 13.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 220$

Answer:
$BIM Grade 4 Chapter 8 add & subtract fractions img_24$
Denominators are the same so add the numerators.
$\frac{1}{3}$ + $\frac{1}{3}$ = $\frac{2}{3}$

Question 14.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 221$

Answer:
Denominators are the same so add the numerators.
$\frac{1}{8}$ + $\frac{6}{8}$ = $\frac{7}{8}$

Question 15.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 222$

Answer:
Denominators are the same so add the numerators.
$\frac{7}{4}$ + $\frac{3}{4}$ = $\frac{11}{4}$

Question 16.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 223$

Answer:
Denominators are the same so add the numerators.
$\frac{2}{6}$ + $\frac{2}{6}$ = $\frac{4}{6}$

Question 17.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 224$

Answer:
Denominators are the same so add the numerators.
$\frac{8}{12}$ + $\frac{4}{12}$ = $\frac{12}{12}$ = 1

Question 18.
Logic
When you add two of me you get $\frac{100}{100}$. What fraction am I?

Answer: $\frac{50}{100}$
If you add $\frac{50}{100}$ two times you get $\frac{100}{100}$
$\frac{50}{100}$ + $\frac{50}{100}$ = $\frac{100}{100}$

8.4 Use Models to Subtract Fractions

Find the difference. Explain how you used the model to subtract.

Question 19.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 225$

Answer:
$Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-225$

Question 20.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 226$

Answer:
$Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-226$

Find the difference. Use a model or a number line to help.

Question 21.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 227$

Answer: 3/2
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-139$

Question 22.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 228$

Answer:
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-139$

Question 23.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 229$

Answer:
Denominators are the same so subtract the numerators.
$\frac{30}{100}$ – $\frac{21}{100}$ = (30 – 21)/100 = $\frac{9}{100}$

Question 24.
Modeling Real Life
A football team wins $\frac{7}{10}$ of their games this season. They lose $\frac{3}{10}$ of their games. How many more games does the team win than lose?

Answer:
Given that,
A football team wins $\frac{7}{10}$ of their games this season. They lose $\frac{3}{10}$ of their games.
$\frac{7}{10}$ – $\frac{3}{10}$ = $\frac{4}{10}$
Thus $\frac{4}{10}$ more games the team win than lose.

8.5 Subtract Fractions with Like Denominators

Subtract.

Question 25.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 230$

Answer: $\frac{5}{10}$

Explanation:
Denominators are the same so subtract the numerators.
$\frac{9}{10}$ – $\frac{4}{10}$ = $\frac{5}{10}$

Question 26.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 231$

Answer: $\frac{7}{12}$

Explanation:
Denominators are the same so subtract the numerators.
$\frac{14}{12}$ – $\frac{7}{12}$ = $\frac{7}{12}$

Question 27.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 232$

Answer: $\frac{24}{100}$

Explanation:
Denominators are the same so subtract the numerators.
$\frac{80}{100}$ – $\frac{56}{100}$ = $\frac{24}{100}$

Question 28.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 233$

Answer: 3/8$\frac{3}{8}$

Explanation:
Denominators are the same so subtract the numerators.
1 can be written as $\frac{8}{8}$
$\frac{8}{8}$ – $\frac{5}{8}$ = $\frac{3}{8}$

Question 29
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 234$

Answer: $\frac{2}{3}$

Explanation:
Denominators are the same so subtract the numerators.
1 can be written as $\frac{3}{3}$
$\frac{3}{3}$ – $\frac{1}{3}$ = $\frac{2}{3}$

Question 30.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 235$

Answer: $\frac{2}{6}$

Explanation:
Denominators are the same so subtract the numerators.
2 can be written as $\frac{12}{6}$
$\frac{12}{6}$ – $\frac{10}{6}$ = $\frac{2}{6}$

8.6 Model Fractions and Mixed Numbers

Write the mixed number as a fraction.

Question 31.
1 $\frac{6}{8}$

Answer: $\frac{7}{4}$

Explanation:
Step 1
Multiply the denominator by the whole number
8 × 1 = 8
Step 2
Add the answer from Step 1 to the numerator
8 + 6 = 14
Step 3
Write an answer from Step 2 over the denominator
14/8 = $\frac{7}{4}$

Question 32.
4 $\frac{1}{2}$

Answer: $\frac{9}{2}$

Explanation:
Step 1
Multiply the denominator by the whole number
2 × 4 = 8
Step 2
Add the answer from Step 1 to the numerator
8 + 1 = 9
Step 3
Write an answer from Step 2 over the denominator
$\frac{9}{2}$

Question 33.
5 $\frac{10}{12}$

Answer: $\frac{35}{6}$

Explanation:
Step 1
Multiply the denominator by the whole number
12 × 5 = 60
Step 2
Add the answer from Step 1 to the numerator
60 + 10 = 70
Step 3
Write an answer from Step 2 over the denominator
70/12 = $\frac{35}{6}$

Write the fraction as a mixed number or a whole number.

Question 34.
$\frac{17}{4}$

Answer: 4 $\frac{1}{4}$

Explanation:
Converting from improper fraction to the mixed fraction.
$\frac{17}{4}$ = 4 $\frac{1}{4}$

Question 35.
$\frac{30}{6}$

Answer: 5

Explanation:
Converting from improper fraction to the mixed fraction.
6 divides 30 five times.
So, $\frac{30}{6}$ = 5

Question 36.
$\frac{63}{10}$

Answer: 6 $\frac{3}{10}$

Explanation:
Converting from improper fraction to the mixed fraction.
$\frac{63}{10}$ = 63 ÷ 10
= 6 $\frac{3}{10}$

Compare.

Question 37.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 236$

Answer: <

Explanation:
2 4/100
Step 1
Multiply the denominator by the whole number
100 × 2 = 200
Step 2
Add the answer from Step 1 to the numerator
200 + 4 = 204
Step 3
Write an answer from Step 2 over the denominator
204/100
240/100
Step 1
Multiply the denominator by the whole number
100 × 2 = 200
Step 2
Add the answer from Step 1 to the numerator
200 + 40 = 240
Step 3
Write an answer from Step 2 over the denominator
240/100
204/100 < 240/100

Question 38.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 237$

Answer: >

Explanation:
Step 1
Multiply the denominator by the whole number
3 × 8 = 24
Step 2
Add the answer from Step 1 to the numerator
24 + 2 = 26
Step 3
Write an answer from Step 2 over the denominator
26/3
26/3 > 25/3

Question 39.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 238$

Answer: =

Explanation:
25/5 = 5
5 = 5

Question 40.
Which One Doesn’t Belong? Which expression does not belong with the other three?
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 239$

Answer: 20/8 does not belong with the other three.

8.7 Add Mixed Numbers

Add.

Question 41.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 240$

Answer: 9

Explanation:
Rewriting our equation with parts separated
=5+1/2+3+1/2
Solving the whole number parts
5+3=8
Solving the fraction parts
1/2+1/2=2/2
Reducing the fraction part, 2/2,
2/2=1/1
Simplifying the fraction part, 1/1,
1/1=1
Combining the whole and fraction parts
8+1=9

Question 42.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 241$

Answer: 4 1/3

Explanation:
Rewriting our equation with parts separated
=2+5/6+1+3/6
Solving the whole number parts
2+1=3
Solving the fraction parts
5/6+3/6 = 8/6
8/6 = 4/3
4/3 = 4 1/3

Question 43.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 242$

Answer: 5 5/6

Explanation:
Rewriting our equation with parts separated
=4+1+10/12
Solving the whole number parts
4+1=5
Combining the whole and fraction parts
5+10/12= 5 10/12 = 5 5/6

Question 44.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 243$

Answer: 19

Explanation:
Rewriting our equation with parts separated
=8+3/5+10+2/5
Solving the whole number parts
8+10=18
Solving the fraction parts
3/5+2/5=5/5
Simplifying the fraction part, 1/1,
1/1 = 1
Combining the whole and fraction parts
18+1=19

Question 45.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 244$

Answer: 14 1/4

Explanation:
Rewriting our equation with parts separated
=7+2/4+1+2/4
Solving the whole number parts
7+1=8
Solving the fraction parts
2/4+2/4=4/4
Reducing the fraction part, 4/4,
4/4=1/1
Simplifying the fraction part, 1/1,
1/1=1
Combining the whole and fraction parts
8+1=9
9 + 5 1/4
Rewriting our equation with parts separated
=9+5+1/4
Solving the whole number parts
9+5=14
Combining the whole and fraction parts
14+1/4=14 1/4

Question 46.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 245$

Answer: 19 5/100

Explanation:
Rewriting our equation with parts separated
=4+25/100+11+75/100
Solving the whole number parts
4+11=15
Solving the fraction parts
25/100+75/100=100/100
Reducing the fraction part, 100/100,
100/100=11
Simplifying the fraction part, 1/1,
1/1=1
Combining the whole and fraction parts
15+1=16
Rewriting our equation with parts separated
=16+3+5/100
Solving the whole number parts
16+3=19
Combining the whole and fraction parts
19+5/100=19 5/100

8.8 Subtract Mixed Numbers

Subtract

Question 47.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 246$

Answer: 3

Explanation:
Rewriting our equation with parts separated
=9+2/3-6-2/3
Solving the whole number parts
9−6=3
Solving the fraction parts
2/3−2/3=0/3
Simplifying the fraction part, 0/3,
0/3=0
Combining the whole and fraction parts
3+0=3

Question 48.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 247$

Answer: 5 2/5

Explanation:
Rewriting our equation with parts separated
=13+9/10−8−5/10
Solving the whole number parts
13−8=5
Solving the fraction parts
9/10−5/10=4/10
Reducing the fraction part, 4/10,
4/10=2/5
Combining the whole and fraction parts
5+2/5=5 2/5

Question 49.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 248$

Answer: 1/2

Explanation:
Rewriting our equation with parts separated
=3+2/8−2−6/8
Solving the whole number parts
3−2=1
Solving the fraction parts
2/8−6/8=−4/8
Reducing the fraction part, 4/8,
−4/8=−1/2
Combining the whole and fraction parts
1−1/2=1/2

Question 50.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 249$

Answer: 5 1/2

Explanation:
6 + 1/2 – 1 = 5 1/2

Question 51.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 250$

Answer: 2 3/4

Explanation:
Rewriting our equation with parts separated
=7−4−1/4
Solving the whole number parts
7−4=3
Combining the whole and fraction parts
3−1/4=2 3/4

Question 52.
$Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 251$

Answer: 1/6

Explanation:
Rewriting our equation with parts separated
=20−19−5/6
Solving the whole number parts
20−19=1
Combining the whole and fraction parts
1−5/6=1/6

8.9 Problem Solving: Fractions

Question 53.
You give $\frac{3}{12}$ of your bag of grapes to one friend and $\frac{5}{12}$ of your bag to another friend. What fraction of the bag of grapes do you have left?

Answer:
Given that,
You give $\frac{3}{12}$ of your bag of grapes to one friend and $\frac{5}{12}$ of your bag to another friend
$\frac{3}{12}$ + $\frac{5}{12}$ = $\frac{8}{12}$
$\frac{12}{12}$ – $\frac{8}{12}$ = $\frac{4}{12}$
Thus $\frac{4}{12}$ fraction of the bag of grapes are left.

Final Words:

Hope you are all satisfied with the solutions provided in the BIM Grade 4 Chapter 8 Add and Subtract Fractions pdf. If you have any doubts regarding the problems you can ask your doubts in the below comment box. We are ready to clarify your doubts at any time. Stay with us to get the solutions of all 4th-grade chapters.

Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers

December 31, 2020

Hey Guys!!! Are you looking for solutions for BIM Grade 3 Chapters? If your answer is yes, then you are on the right page. Here the students can get a clear cut explanation for all the Big Ideas Math Answers Chapter 8 Add and Subtract Multi-Digit Numbers. It is necessary for you to learn the basics from the primary level itself. The Big Ideas Math Book 3rd Grade Answer Key Chapter 8 Add and Subtract Multi-Digit Numbers helps you overcome the issues you face while solving the problems.

Big Ideas Math Book 3rd Grade Answer Key Chapter 8 Add and Subtract Multi-Digit Numbers

First, go through the topics covered in this chapter to know which topic is easy and which is difficult. Practice the questions a number of times to understand the concepts in depth. To make your task easy we have provided the links in the below section. Just hit the links and Download Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers pdf.

Lesson 1 – Identify Addition Properties

Lesson 8.1 Identify Addition Properties
Identify Addition Properties Homework & Practice 8.1

Lesson 2 – Use Number Lines to Addition

Lesson 8.2 Use Number Lines to Addition
Use Number Lines to Addition Homework & Practice 8.2

Lesson 3 – Use Mental Math to Add

Lesson 8.3 Use Mental Math to Add
Use Mental Math to Add Homework & Practice 8.3

Lesson 4 – Use Partial Sums to Add

Lesson 8.4 Use Partial Sums to Add
Use Partial Sums to Add Homework & Practice 8.4

Lesson 5 – Add Three-Digit Numbers

Lesson 8.5 Add Three-Digit Numbers
Add Three-Digit Numbers Homework & Practice 8.5

Lesson 6 – Add Three or More Numbers

Lesson 8.6 Add Three or More Numbers
Add Three or More Numbers Homework & Practice 8.6

Lesson 7 – Use Number Lines to Subtract

Lesson 8.7 Use Number Lines to Subtract
Use Number Lines to Subtract Homework & Practice 8.7

Lesson 8 – Use Mental Math to Subtract

Lesson 8.8 Use Mental Math to Subtract
Use Mental Math to Subtract Homework & Practice 8.8

Lesson 9 – Subtract Three-Digit Numbers

Lesson 8.9 Subtract Three-Digit Numbers
Subtract Three-Digit Numbers Homework & Practice 8.9

Lesson 10 – Relate Addition and Subtraction

Lesson 8.10 Relate Addition and Subtraction
Relate Addition and Subtraction Homework & Practice 8.10

Lesson 11 – Problem Solving: Addition and Subtraction

Lesson 8.11 Problem Solving: Addition and Subtraction
Problem Solving: Addition and Subtraction Homework & Practice 8.11

Performance Task

Add and Subtract Multi-Digit Numbers Performance Task
Add and Subtract Multi-Digit Numbers Activity
Add and Subtract Multi-Digit Numbers Chapter Practice
Add and Subtract Multi-Digit Numbers Cumulative Practice 1 – 8
Add and Subtract Multi-Digit Numbers Steam Performance Task 1-8

Lesson 8.1 Identify Addition Properties

Explore and Grow

Use the addition table to write all of the addition equations that have a sum of 13.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 1$
What do you notice now?
Answer:
By adding a vertical number and the horizontal number we can find the sum of 13.
0 + 0 = 0
0 + 1 = 0
Add blue line and yellow line.
Add 10 in the blue column and 3 in the yellow column.
10 + 3 = 13
Thus the addition equation has the sum of 13.

Structure
Use the addition table to write all of the equations that have a sum of 12. What do you notice?

Answer:
Add blue line and yellow line.
Add 10 in the blue column and 2 in the yellow column.
10 + 2 = 12
Thus the addition equation has the sum of 12.

Think and Grow: Addition Properties
Changing the order of addends does not change the sum.
3 + 5 = 5 + 3
Associative Property of Addition
Changing the grouping of addends does not change the sum.
(7 + 6) + 4 = 7 + (6 + 4)
Addition Property of Zero
The sum of any number and 0 is that number.
9 + 0 = 9
Example
Identify the property.
56 + 0 = ___
12 + 29 = 29 + 12 _____
(24 + 17) + 23 = 24 + (17 + 23) ____

Answer:
56 + 0 = 56
It shows Addition Property of Zero. The Addition Property of Zero defines the sum of any number and 0 is that number.
12 + 29 = 29 + 12
Changing the order of addends does not change the sum.
(24 + 17) + 23 = 24 + (17 + 23)
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Show and Grow

Identify the property.

Question 1.
16 + (14 + 19) = (16 + 14) + 19

Answer:
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 2.
11 + 54 = 54 + 11

Answer:
Associative Property of Addition defines the changing the order of addends does not change the sum.

Question 3.
0 + 43 = 43

Answer: Addition Property of Zero
The Addition Property of Zero defines the sum of any number and 0 is that number.

Question 4.
(27 + 18) + 22 = 27 + (18 + 22)

Answer:
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Apply and Grow: Practice

Identify the property.

Question 5.
(28 + 16) + 14 = 28 + (16 + 14)

Answer:
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 6.
12 + 35 = 35 + 12

Answer:
Associative Property of Addition defines the changing the order of addends does not change the sum.

Question 7.
36 + 0 = 36

Answer:
The Addition Property of Zero defines the sum of any number and 0 is that number.

Question 8.
11 + (9 + 57) = (11 + 9) + 57

Answer:
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Find the missing number.

Question 9.
23 + 45 = 45 + ___

Answer: 23

Explanation:
Let the missing number be x.
By using the Associative Property of Addition property we can find the missing number.
23 + 45 = 45 + x
23 + 45 = 45 + 23
Thus the missing number is 23.

Question 10.
(13 + 12) + __ = 13 + (12 + 45)

Answer: 45

Explanation:
Let the missing number be x.
By using the Associative Property of Addition property we can find the missing number.
(13 + 12) + x = 13 + (12 + 45)
Thus the missing number is 45.

Question 11.
4 + (76 + 10) = (___ + 76) + 10

Answer: 4

Explanation:
Let the missing number be x.
By using the Associative Property of Addition property we can find the missing number.
4 + (76 + 10) = (x + 76) + 10
4 + (76 + 10) = (4 + 76) + 10
Thus the missing number is 4.

Question 12.
98 + ___ = 98

Answer: 0

Explanation:
Let the missing number be x.
The Addition Property of Zero defines the sum of any number and 0 is that number.
98 + x = 98
x = 98 – 98
x = 0
Thus the missing number is 0.

Question 13.
(___ + 0) + 32 = 6 + 32

Answer: 6

Explanation:
Let the missing number be x.
By using the Associative Property of Addition property we can find the missing number.
(x + 0) + 32 = 6 + 32
x + 32 = 38
x = 38 – 32
x = 6
Thus the missing number is 6.

Question 14.
64 + (5 + 23) = (23 + ___) + 64

Answer: 5

Explanation:
Let the missing number be x.
By using the Associative Property of Addition property we can find the missing number.
64 + (5 + 23) = (23 + x) + 64
64 + 28 = 23 + x + 64
82 = x + 87
x = 87 – 82
x = 5
Thus the missing number is 5.

Question 15.
DIG DEEPER!
Use the numbers 24, 54, and 11 to write an equation that shows the Associative Property of Addition.

Answer:
We can write the equation by using the Associative Property of Addition.
24 + (54 + 11) = (24 + 54) + 11

Writing
Use a property to find the sum. Which property did you use? Why?

Question 16.
54 + 0 = __

Answer: 54

Explanation:
We can find the sum of the 54 + 0 by using the addition property of zero.
The Addition Property of Zero defines the sum of any number and 0 is that number.
54 + 0 = 54

Question 17.
(46 + 17) + 33 = ___

Answer: 96

Explanation:
We can find the sum of the (46 + 17) + 33 by using the Associative Property of Addition.
(46 + 17) + 33 = 96

Question 8.
20 + 63 = ___

Answer: 83

Explanation:
We can find the sum of the Associative Property of Addition.
20 + 63 = 83

Think and Grow: Modeling Real Life

$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 2$
A tourist visits 13 museums, 19 memorials, and 11 monuments. Explain how to use a property to find the total number of sites the tourist visits.
(13 + 19) + 11 = ?
Explain:
The tourist visits ___ sites.

Answer:
Given,
A tourist visits 13 museums, 19 memorials, and 11 monuments.
We can find the total number of sites the tourist visits.
(13 + 19) + 11 = 13 + (19 + 11) = 43
Thus the tourist visits 43 sites.

Show and Grow

Question 19.
A farmer sells 34 cucumbers, 48 ears of corn, and 26 bell peppers at a farmer’s market. Explain how to use properties to find the total number of vegetables the farmer sells.
(34 + 48) + 26 = ?

Answer:
Given,
A farmer sells 34 cucumbers, 48 ears of corn, and 26 bell peppers at a farmer’s market.
We can find the sum of the Associative Property of Addition.
(34 + 48) + 26 = 34 + (48 + 26)
34 + 48 + 26 = 108
Thus the total number of vegetables the farmer sells is 108.

Question 20.
How many people go on the field trip?
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 3$

Answer:
The number of adults = 20
The number of grade 2 students = 47
The number of grade 3 students = 53
20 + 47 + 53 = 120
Thus 120 people go on the field trip.

DIG DEEPER!
The Grade 2 and Grade 3 students are divided into 10 equal groups. How many students are in each group? Explain.

Answer:
Given that,
The Grade 2 and Grade 3 students are divided into 10 equal groups.
The number of grade 2 students = 47
The number of grade 3 students = 53
47 + 53 = 100
100/10 = 10
Thus there are 10 students in each group.

Identify Addition Properties Homework & Practice 8.1

Identify the property.

Question 1.
(79 + 12) + 13 = 79 + (12 + 13)

Answer: Associative property of addition
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 2.
24 + 63 = 63 + 24

Answer: Associative property of addition
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 3.
0 + 64 = 64

Answer: Associative property of zero
The Addition Property of Zero defines the sum of any number and 0 is that number.

Question 4.
37 + (43 + 19) = (37 + 43) + 19

Answer: Associative property of addition
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 5.
17 + 38 = 38 + 17

Answer: Associative property of addition
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 6.
18 + 48 = 48 + 18

Answer: Associative property of addition
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Find the missing number.

Question 7.
36 + __ = 36

Answer: 0

Explanation:
Let the missing number be x.
36 + x = 36
x = 36 – 36
x = 0
Thus the missing number is 0.

Question 8.
25 + __ + 11 = 25 + 11

Answer: 0

Explanation:
Let the missing number be y.
25 + y + 11 = 25 + 11
y + 36 = 36
y = 36 – 36
y = 0
Thus the missing number is 0.

Question 9.
0 + 43 = __ + 0

Answer: 43

Explanation:
Let the missing number be t.
0 + 43 = t + 0
43 = t
Thus the missing number is 43.

Question 10.
(22 + 19) + 28 = 19 + (___ + 28)

Answer: 22

Explanation:
Let the missing number be p.
22 + 19 + 28 = 19 + p + 28
69 = 47 + p
p = 69 – 47
p = 22
Thus the missing number is 22.

Question 11.
Number Sense
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 4$
Newton uses two properties. Identify the properties he uses.

Answer:
Given the expression (18 + 27) + 12 = 27 + (18 + 12)
By seeing the above expression we can say that Newton used Associative Property of Addition and Commutative Property of Addition.

Question 12.
Open-Ended
Write an equation that shows the Commutative Property of Addition.

Answer: 4 + 2 = 2 + 4
Commutative property of addition: Changing the order of addends does not change the sum.

Question 13.
Structure
Explain how the Associative Property of Addition and the Associative Property of Multiplication are alike and how they are different.

Answer:
Associative property explains that addition and multiplication of numbers are possible regardless of how they are grouped.
Example:
2 × (3 × 5) = (2 × 3) × 5
2 + (3 + 5) = (2 + 3) + 5
The method is same but the solution for both the equations are different.

Question 14.
Modeling Real Life
A florist uses 11 roses, 12 lilies, and 19 daisies to make bouquets. How many flowers does he use?

Answer:
Given that,
A florist uses 11 roses, 12 lilies, and 19 daisies to make bouquets.
Add all the flowers to find the total number of flowers he used.
11 + 12 + 19 = 42
Thus he used 42 flowers.

DIG DEEPER!
The florist uses 6 flowers for each bouquet. How many bouquets does he make? Explain.

Answer:
Given,
The florist uses 6 flowers for each bouquet.
Total flowers = 42
Divide the total number of flowers by the number of flowers in each bouquet
42/6 = 7
Thus he made 7 bouquets.

Review & Refresh

Find the product.

Question 15.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 400$

Answer:
Multiply the two numbers 0 and 3.
0 × 3 = 0

Question 16.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 5$

Answer:
Multiply the two numbers 7 and 7.
7 × 7 = 49

Question 17.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 6$

Answer:
Multiply the two numbers 10 and 4.
10 × 4 = 40

Question 18
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 7$

Answer:
Multiply the two numbers 7 and 6.
7 × 6 = 42

Lesson 8.2 Use Number Lines to Addition

Explore and Grow

Color to find 33 + 25. Then model your jumps on the number line.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 8$

Answer:
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-8$
First, you have to color the two numbers 33 and 25.
Now add 33 and 25
33 + 25 = 58
Now try to solve the problem by using the number line.
Start at 33. Count by twenties, then by 5s.
33 + 20 + 5 = 58

Reasoning

How can finding 33 + 25 help you find 533 + 25?

Answer:
We can find the sum of 533 and 25 with the help of the above problem.
33 + 25 = 58
533+ 25 = 538
Just add 5 on the left side to get the sum.

Think and Grow: Adding on a Number Line

Example
Find 245 + 38
One Way: Use the count on strategy. Start at 245. Count on by tens, then by ones.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 9$
Answer:
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-9$

Another Way: Use the make a ten strategy. Start at 245. Count on to the nearest ten. Then count on by tens and by ones.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 10$
Answer:
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-10$

Show and Grow

Question 1.
Use the count on strategy to find 368 + 24.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 11$

Answer: 392
Use the count on strategy. Start at 368. Count on by tens, then by ones.
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-11$
368 + 10 + 10 + 2 + 2 = 392

Question 2.
Use the count on strategy to find 57 + 179.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 12$

Answer: 236
Use the count on strategy. Start at 57. Count on by hundreds, tens, then by ones.
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-12$

Apply and Grow: Practice

Question 3.
Use the count on strategy to find 47 + 216.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 13$

Answer: 263
Use the count on strategy. Start at 47. Count on by hundreds, tens, then by ones.
47 + 100 + 100 + 10 + 6 = 263
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-13$

Question 4.
Use the make a ten strategy to find 478 + 64.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 14$

Answer: 542
Use the count on strategy. Start at 478. Count on by tens, then by ones.
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-14$

Find the sum

Question 5.
395 + 85 = ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 15$

Answer: 480
Use the count on strategy. Start at 395. Count on by tens, then by ones.
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-15$

Question 6.
653 + 109 = ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 16$

Answer: 762
Use the count on strategy. Start at 353. Count on by tens, then by ones.
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-15$

Question 7.
Humans have 24 rib bones and 182 other types of bones. How many bones do humans have in all?
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 17$

Answer:
Given,
Humans have 24 rib bones and 182 other types of bones.
182 + 24 = 206
Thus Humans have 206 bones.

Question 8.
Structure
Write the equation shown by the number line.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 18$

Answer:
436 + 4 + 20 + 4
440 + 20 + 4 = 464

Question 9.
Structure
Show two different ways to find 225 + 39 using a number line.

Answer:
Use the count on strategy. Start at 225. Count on by tens, then by ones.
225 + 30 + 9 = 264
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-15$

Think and Grow: Modeling Real Life

$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 19$
The Leaning Tower of Pisa has 294 steps. A visitor climbs 156 steps, takes a break, and then climbs 78 more steps. Does the visitor reach the top of the tower?
Addition equation:
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 20$
The visitor ___ reach the top of the tower.

Answer:
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-16$
156 + 78 = 234
294 – 234 = 60
Thus the visitor does not reach the top of the tower.

Show and Grow

Question 10.
A book has 216 pages. You have already read 167 pages. You read 49 more pages. Do you finish reading the book?

Answer:
Given that,
A book has 216 pages. You have already read 167 pages. You read 49 more pages.
167 + 49 = 216 pages
216 – 216 = 0
Thus you finish reading the book.

Question 11.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 21$
DIG DEEPER!
A puzzle has 350 pieces. You put 95 pieces together. Your friend puts 185 pieces together. Do you and your friend complete the puzzle? If not, how many pieces are left?

Answer:
Given that,
A puzzle has 350 pieces. You put 95 pieces together. Your friend puts 185 pieces together.
95 + 185 = 280 pages
350 – 280 = 70 pages
No you and your friend do not complete the puzzle.
70 pieces are left to complete the puzzle.

Question 12.
A music library has 483 songs. You listen to162 different songs one week and 171 more songs the next week. How many songs are left?

Answer:
Given that,
A music library has 483 songs.
You listen to162 different songs one week and 171 more songs the next week.
162 + 171 = 333 songs
483 – 333 = 150 songs
Thus 150 songs are left.

Use Number Lines to Addition Homework & Practice 8.2

Question 1.
Use the count on strategy to find 402 + 39.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 22$

Answer:
Use the count on strategy. Start at 402. Count on by tens, then by ones.
$Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-22$

Question 2.
Use the make a ten strategy to find 81 + 647.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 23$

Answer:
Use the count on strategy. Start at 81. Count on by hundreds, tens, then by ones.
$Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-23$

Question 3.
Find the sum.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 24$

Answer:

Question 4.
718 + 226 = __
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 25$

Answer:
Use the count on strategy. Start at 718. Count on by hundreds, tens, then by ones.
$Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-25$
718 + 226 = 944

Question 5.
YOU BE THE TEACHER
Your friend uses a number line to find 435 + 27. Is your friend correct? Explain.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 26$

Answer:
Your friend uses a number line to find 435 + 27.
435 + 27 = 462
Your friend is not correct.

Question 6.
YOU BE THE TEACHER
Your friend says she can find 64 + 691 by starting at 691 on a number line because of the Associative Property of Addition. Is your friend correct? Explain.

Answer:
Your friend says she can find 64 + 691 by starting at 691 on a number line because of the Associative Property of Addition.
$Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-25$
Yes, your friend is correct.

Question 7.
Modeling Real Life
Your cousin needs to write a 400-word essay. She writes 318 words during class. She finishes her essay by writing 94 words at home. Does your cousin’s essay have enough words?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 401$

Answer:
Given,
Your cousin needs to write a 400-word essay.
She writes 318 words during class. She finishes her essay by writing 94 words at home.
318 + 94 = 412
412 – 400 = 12
Yes your cousin’s essay have enough words.

Question 8.
DIG DEEPER!
There are 275 apartments in an apartment building. There are 203 two-bedroom apartments rented, and 56 one-bedroom not apartments rented. How many apartments are rented?

Answer:
Given that,
There are 275 apartments in an apartment building.
There are 203 two-bedroom apartments rented, and 56 one-bedroom not apartments rented.
203 + 56 = 259
Thus 259 apartments are rented.

Find the quotient

Question 9.
18 ÷ 6 = ___

Answer: 3

Explanation:
Divide the two numbers 18 and 6.
18/6 = 3
Thus the quotient is 3.

Question 10.
32 ÷ 8 = ___

Answer: 4

Explanation:
Divide the two numbers 32 and 8.
32/4 = 4
Thus the quotient is 4.

Question 11.
72 ÷ 9 = __

Answer: 8

Explanation:
Divide the two numbers 72 and 9.
72/9 = 8
Thus the quotient is 8.

Lesson 8.3 Use Mental Math to Add

Explore and Grow

What addition problem is shown? How can you find the sum?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 27$

Answer: 366 + 504 = 870

Explanation:
By seeing the above figure we can find the sum.
First count the number of blocks
There are 100 blocks in each figure
There are three 100 blocks, six 10 blocks, and 6 blocks.
300 + 60 + 6 = 366
There are five 100 blocks and 4 blocks.
500 + 4 = 504
Now add both
366 + 504 = 870

Reasoning
Show how to find 402 + 248.

Answer:
You can find the sum of 402 and 248 by using the above arrays.
Take 10 × 10 block
402 – Four 10 × 10 blocks, 2 blocks
248 – Two 10 × 10 blocks, Four 10 blocks, and eight blocks
402 + 248 = 650

Think and Grow: Mental Math Strategies for Addition

Example
Find 247 + 328.
Use compensation to add.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 28$

Answer:
You can find the sum of 247 and 328 by using mental math strategies.
247 + 3 = 250
328 – 3 = 325
250
+325
575
Thus the sum of 247 and 328 is 575.

Example
Find 119 + 163.
Make a ten and count on to add.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 29$

Answer:
119 + 163 = 119 + (1 + 100 + 60 + 2)
(119 + 1) + 100 + 60 + 2
120 + 100 + 60 + 2
382

Show and Grow

Question 1.
Use compensation to find 322 + 158.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 30$

Answer:
322 – 2 = 320
158 + 2 = 160
320
+160
480
So, 322 + 158 = 480

Question 2.
Make a ten and count on to find 695 + 187.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 31$

Answer:
695 + 187 = 695 + (5 + 100 + 80 + 2)
(695 + 5) + (100 + 80 + 2)
700 + 100 + 80 + 2
882
So, 695 + 187 = 882

Apply and Grow: Practice

Question 3.
Use compensation to find the sum.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 32$

Answer:
604 – 4 = 600
275 + 4 = 279
600
+279
879

Question 4.
Make a ten and count on to find the sum.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 33$

Answer:
359 + 1 = 360
360 + 300 = 660
660 + 17 = 677
So, 359 + 318 = 677

Use mental math to find the sum.

Question 5.
436 + 248 = __

Answer:
436 + 248 = 436 + (8 + 40 + 200)
436 + (4 + 4 + 40 + 200)
(436 + 4) + 4 + 40 + 200
440 + 244
684
So, 436 + 248 = 684

Question 6.
795 + 102 = ___

Answer:
102 – 2 = 100
795 + 2 = 797
797
+100
897
So, 795 + 102 = 897

Question 7.
503 + 71 = ___

Answer:
503 – 3 = 500
71 + 3 = 74
500
+74
574
So, 503 + 71 = 574

Question 8.
589 + 407 = ___

Answer:
589 + 1 = 590
407 – 1 = 406
590
+406
996

Question 9.
734 + 97 = ___

Answer:
734 – 3 = 731
97 + 3 = 100
731
+100
831

Question 10.
352 + 164 = ___

Answer:
352 – 2 = 350
164 + 2 = 166
350
+166
516

Question 11.
297 + 211 = ___

Answer:
297 + 3 = 300
211 – 3 = 208
300
+208
508

Question 12.
426 + 364 = ___

Answer:
426 + 4 = 430
364 – 4 = 360
430
+360
790

Question 13.
159 + 104 = ___

Answer:
159 + 1 = 160
104 – 1 = 103
160
+103
263

Question 14.
A community shelter collects 101 shirts and 109 pairs of pants from a clothing drive. How many clothing items does the community shelter collect?

Answer:
Given,
A community shelter collects 101 shirts and 109 pairs of pants from a clothing drive.
101 – 1 = 100
109 + 1 = 110
100
+110
210
Thus the community shelter collect 210 clothing items.

Question 15.
Number Sense
Descartes wants to use compensation to find 238 + 127. Which numbers could he use?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 34$

Answer:
Descartes wants to use compensation to find 238 + 127.
The number near 238 is 240, 127 is 130.
Thus he could use 240 and 130

Think and Grow: Modeling Real Life

A company manager has $900. Does he have enough money to buy the laptop and the cell phone?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 35$
Addition equation:
Compare:
The manager ___ have enough money.

Answer:
Given,
A company manager has $900.
The cost of the laptop is $595
The cost of the mobile is $249.
595 + 5 = 600
249 – 5 = 244
600
+244
844
900 – 844 = 56
Thus the manager does not have enough money.

Show and Grow

Question 16.
A USB drive holds 600 photos. You have 279 photos on a digital camera and 337 photos on a cell phone. Can the USB drive hold all of your photos?

Answer:
Given,
A USB drive holds 600 photos. You have 279 photos on a digital camera and 337 photos on a cell phone.
279 + 1 = 280
337 – 1 = 336
280
+336
616
Yes the USB drive can hold all of your photos

Question 17.
A school board president has $1,000. Which two items can she buy?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 36$

DIG DEEPER!
The president buys the two items. How much money does she have left?

Answer:
A school board president has $1,000.
The cost of a swing set is $648
The cost of the seesaw is $372
The cost of a Dome Climber is $498
498
+372
870
1000 – 870 = 130
Thus she has left $130.

Use Mental Math to Add Homework & Practice 8.3

Use compensation to find the sum.

Question 1.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 37$

Answer:
248 + 2 = 250
137 – 2 = 135
250
+135
385
So, 248 + 137 = 385

Question 2.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 38$

Answer:
401 – 1 = 400
165 + 1 = 166
400
+166
566

Make a ten and count on to find the sum

Question 3.
506 + 369 = ?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 39$

Answer:
506 + 369 = 506 + (4 + 300 + 60 + 5)
(506 + 4) + 300 + 60 + 5
510 + 365
875
So, 506 + 369 = 875

Question 4.
617 + 348 = ?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 40$

Answer:
617 + 348 = 617 + (3 + 300 + 40 + 5)
(617 + 3) + 300 + 40 + 5
620 + 345
965
617 + 348 = 965

Use mental math to find the sum.

Question 5.
478 + 219 = ___

Answer:
478 + 2 = 480
219 – 2 = 217
480
+217
697

Question 6.
503 + 64 = ___

Answer:
503 – 3 = 500
64 + 3 = 67
500
+67
567

Question 7.
288 + 242 = ___

Answer:
288 + 2 = 290
242 – 2 = 240
290
+240
530

Question 8.
Structure
Explain how to make a ten to find the sum.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 41$

Answer:
738 + 126 = 738 + (6 + 20 + 100)
(738 + 2) + 4 + 20 + 100
740 + 124
864

Question 9.
Writing
How is compensation make a ten similar to the strategy? How is it different?

Answer:
Compensation is a mental math strategy for multi-digit addition that involves adjusting one of the addends to make the equation easier to solve. The methods are different but the solutions are the same.

Question 10.
Modeling Real Life
A binder holds 250 sheets of paper. You have 107 science papers and 142 math papers. Can the binder hold all of your papers?

Answer:
Given,
A binder holds 250 sheets of paper. You have 107 science papers and 142 math papers.
107 – 2 = 105
142 + 2 = 144
105 + 144 = 249
Yes, the binder can hold all of your papers.

Question 11.
Modeling Real Life
A school nurse has $450. Which two items can she buy?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 42$

DIG DEEPER!
The school nurse buys the two items. How much money does she have left?

Answer:
A school nurse has $450.
The cost of a Stethoscope = $119
The cost of thermometer = $ 308
119 + 1 = 120
308 – 1 = 307
120
+307
427
450 – 427 = 23
She has left $23

Review & Refresh

Question 12.
It costs $1 to get into each football game. Newton buys a chicken wrap for $2 and a drink for $1 each game. How much money does Newton spend in 4 games?

Answer:
Given,
It costs $1 to get into each football game. Newton buys a chicken wrap for $2 and a drink for $1 each game.
1 + 2 + 1 = 4
4 × $4 = $8
Thus Newton spent $8 for 4 games.

Lesson 8.4 Use Partial Sums to Add

Explore and Grow

Model each number. Draw to show each model.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 43$
How can you use the models to find 341 + 227?

Answer:
$Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-43$
You can find the sum of 341 and 227 is

300 + 40 + 1
200 + 20 + 7
500 + 60 + 8 = 568

Reasoning
How can breaking apart addends help you add three-digit numbers?

Answer: Breaking apart addends helps you add three-digit numbers easily. Mind math is possible with this breaking apart addends.

Think and Grow: Use Partials Sums to Add

Example
Find 356 + 408.
Step 1: Write each number in expanded form.
Step 2: Find the partial sums.
Step 3: Add the partial sums.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 44$

Answer: 764

Show and Grow

Use partial sums to add.

Question 1.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 45$

Answer:
319 = 300 + 10 + 9
234 = 200 + 30 + 4
553 = 500 + 40 + 13

Question 2.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 46$

Answer:
568 = 500 + 60 + 8
173 = 100 + 70 + 3
741 = 600 + 130 + 11

Question 3.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 47$

Answer:
424 = 400 + 20 + 4
450 = 400 + 50 + 0
874 = 800 + 70 + 4

Question 4.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 48$

Answer:
281 = 200 + 80 + 1
365 = 300 + 60 + 5
646 = 500 + 140 + 6

Question 5.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 49$

Answer:
127 = 100 + 20 + 7
609 = 600 + 0 + 9
736 = 700 + 20 + 16

Question 6.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 50$

Answer:
276 = 200 + 70 + 6
39 = 000 + 30 + 9
315 = 200 + 100 + 15

Apply and Grow: Practice

Question 7.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 51$

Answer:
759 = 700 + 50 + 9
202 = 200 + 00 + 2
961 = 900 + 50 + 11

Question 8.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 52$

Answer:
864 = 800 + 60 + 4
131 = 100 + 30 + 1
995 = 900 + 90 + 5

Question 9.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 53$

Answer:
387 = 300 + 80 + 7
94 = 000 + 90 + 4
481 = 300 + 170 + 11

Question 10.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 54$

Answer:
560 = 500 + 60 + 0
273 = 200 + 70 + 3
833 = 700 + 130 + 3

Question 11.
498 + 375 = ___

Answer:
498 = 400 + 90 + 8
375 = 300 + 70 + 5
873 = 700 + 160 + 13

Question 12.
209 + 158 = ___

Answer:
209 = 200 + 00 + 9
158 = 100 + 50 + 8
367 = 300 + 50 + 17

Think and Grow: Modeling Real Life

A giant panda weighs 696 pounds less than a polar bear. How much does the polar bear weigh?
Addition equation:
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 55$

Answer:
Given,
A giant panda weighs 696 pounds less than a polar bear.
696 – 263 = 433
Thus the polar bear weighs 433 pounds.

Show and Grow

$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 56$
Question 15.
A herd of wildebeests has 258 fewer members than a herd of zebras has. There are 335 wildebeests in the herd. How many zebras are in their herd?

Answer:
Given,
A herd of wildebeests has 258 fewer members than a herd of zebras has. There are 335 wildebeests in the herd.
335 – 258 = 77
Therefore there are 77 zebras in their herd.

Question 16.
There are three candidates in an election. Candidate A receives 184 fewer votes than Candidate B. Who wins the election?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 57$

Answer:
Given,
There are three candidates in an election. Candidate A receives 184 fewer votes than Candidate B.
Number of votes candidate A receives = 347
347 + 184 = 631
Thus Candidate B wins the election.

Question 17.
DIG DEEPER!
You, your friend, and your cousin play a video game. Your friend scores 161 fewer points than you. Your cousin scores 213 more points than your friend. What is each player’s score? Who wins?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 58$

Answer:
Your friend score is 579
Your friend scores 161 fewer points than you.
579 + 161 = 740
Your score is 740
Your cousin scores 213 more points than your friend.
213 + 579 = 592
Your cousin’s score is the highest among all three.
So, your cousin wins the game.

Use Partial Sums to Add Homework & Practice 8.4

Use partials sums to add

Question 1.
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 59$

Answer:
We can find the sum by using the partial sum model.
479 = 400 + 70 + 9
356 = 300 + 50 + 6
835 = 700 + 120 + 15
So, 479 + 356 = 835

Question 2.
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 60$

Answer:
We can find the sum by using the partial sum model.
674 = 600 + 70 + 4
321 = 300 + 20 + 1
995 = 900 + 90 + 5

Question 3.
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 61$

Answer:
We can find the sum by using the partial sum model.
396 = 300 + 90 + 6
278 = 200 + 70 + 8
674 = 500 + 160 + 14

Question 4.
564 + 218 = ___

Answer:
We can find the sum by using the partial sum model.
564 = 500 + 60 + 4
218 = 200 + 10 + 8
782 = 700 + 70 + 12

Question 5.
190 + 123 = ___

Answer:
We can find the sum by using the partial sum model.
190 = 100 + 90 + 0
123 = 100 + 20 + 3
313 = 200 + 110 + 3

Question 6.
YOU BE THE TEACHER
Your friend uses partial sums to find 205 + 124. Is your friend correct? Explain.
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 62$

Answer: Your friend is incorrect.
205 = 200 + 00 + 5
124 = 100 + 20 + 4
329 = 300 + 20 + 9
The sum of 205 + 124 = 329

Question 7.
Patterns
Write and solve the next problem in the pattern.
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 63$

Answer:
The addend is increased by 100.
So, the next pattern is
516
+178
694

Question 8.
Modeling Real Life
There are worker bees and drone bees in a beehive. A hive has 268 fewer drones than workers. There are 351 drone bees. How many worker bees are there?
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 64$

Answer:
Given,
There are worker bees and drone bees in a beehive. A hive has 268 fewer drones than workers. There are 351 drone bees.
351
-268
83
Therefore there are 83 worker bees.

Question 9.
Modeling Real Life
Three athletes compete in Olympic weight lifting. Weight lifter A lifts 104 fewer pounds than Weight lifter B. Who lifts the most weight?
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 65$

Answer:
Three athletes compete in Olympic weight lifting. Weight lifter A lifts 104 fewer pounds than Weight lifter B.
Weight lifter A lifts 368 pounds
368 + 104 = 472
Therefore, Weight Lifter B lifts the most weight.

Review & Refresh

Circle the value of the underlined digit.

Question 10.
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 66$

Answer:
$Big-Ideas-Math-Answers-3rd-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-66$
In the given value 2 is in the ones place so the answer is 2.

Question 11.
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 67$

Answer:
$Big-Ideas-Math-Answers-3rd-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-67$
In the given value 4 is in the hundreds place so the answer is 400.

Question 12.
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 68$

Answer:
$Big-Ideas-Math-Answers-3rd-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-68$
In the given value 0 is in tens place so the answer is 0.

Lesson 8.5 Add Three-Digit Numbers

Explore and Grow

Model the equation. Draw your model. Then find the sum.
195 + 308 = ___

Answer:
195 = 100 + 90 + 5
308 = 300 + 00 + 8
503 = 400 + 90 + 13

Reasoning
How can you use an estimate to check whether your answer reasonable?

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
195 = 200
308 = 300
200
+300
500
Ths sum is about 500.
Step 2: Find the sum. Add the ones, tens, then the hundreds.
195
+308
503
503 is close to 500. So, the answer is reasonable.

Think and Grow: Add Three-Digit Numbers

Example
Find 236 + 378. Check whether your answer is reasonable.
Step 1: Estimate. Round each addend to the nearest hundred.
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 69$

Show and Grow

Find the sum. Check whether your answer is reasonable.

Question 11.
Estimate: ___
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 70$

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
457 – 500
133 – 100
500
+100
600
The sum is about 600.
Step 2: Find the sum. Add the ones, tens, then the hundreds.
457
+133
590
590 is close to 600. So, the answer is reasonable.

Question 2.
Estimate: ___
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 71$

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
269 – 300
354 – 300
300
+300
300
Step 2: Find the sum. Add the ones, tens, then the hundreds.
269
+354
623
623 is close to 600. So, the answer is reasonable.

Question 3.
Estimate: ___
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 72$

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
560 – 600
274 – 300
600
+300
900
Step 2: Find the sum. Add the ones, tens, then the hundreds.
560
+274
834
834 is close to 900. So, the answer is reasonable.

Question 4.
Estimate: ___
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 73$

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
345 – 300
286 – 300
300
+300
600
Step 2: Find the sum. Add the ones, tens, then the hundreds.
345
+286
631
631 is close to 600. So, the answer is reasonable.

Question 5.
Estimate: ___
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 74$

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
129 – 100
668 – 700
100
+700
800
Step 2: Find the sum. Add the ones, tens, then the hundreds.
129
+668
797
797 is close to 800. So, the answer is reasonable.

Question 6.
Estimate: ___
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 75$

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
383 – 400
539 – 500
400
+500
900
Step 2: Find the sum. Add the ones, tens, then the hundreds.
383
+539
922
922 is close to 900. So, the answer is reasonable.

Apply and Grow: practice

Find the sum. Check whether your answer is reasonable.

Question 7.
Estimate: ___
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 76$

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
803 – 800
179 – 200
800
+200
1000
Step 2: Find the sum. Add the ones, tens, then the hundreds.
803
+179
982
982 is close to 1000. So, the answer is reasonable.

Question 8.
Estimate: ___
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 77$

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
608 – 600
239 – 200
600
+200
800
Step 2: Find the sum. Add the ones, tens, then the hundreds.
608
+239
847
847 is close to 800. So, the answer is reasonable.

Question 9.
Estimate: ___
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 78$

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
427 – 400
385 – 400
400
+400
800
Step 2: Find the sum. Add the ones, tens, then the hundreds.
427
+385
812
812 is close to 800. So, the answer is reasonable.

Question 10.
Estimate: ___
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 79$

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
401 – 400
109 – 100
400
+100
500
Step 2: Find the sum. Add the ones, tens, then the hundreds.
401
+109
510
510 is close to 500. So, the answer is reasonable.

Question 11.
Estimate: ___
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 80$

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
265 – 300
157 – 100
300
+100
400
Step 2: Find the sum. Add the ones, tens, then the hundreds.
265
+157
422
422 is close to 400. So, the answer is reasonable.

Question 12.
Estimate: ___
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 81$

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
375 – 400
64 – 100
400
+100
500
Step 2: Find the sum. Add the ones, tens, then the hundreds.
375
+64
439
439 is close to 500. So, the answer is reasonable.

Question 13.
Estimate: ___
469 + 284 = ___

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
469 – 500
284 – 300
500
+300
800
Step 2: Find the sum. Add the ones, tens, then the hundreds.
469
+284
753
753 is close to 800. So, the answer is reasonable.

Question 14.
Estimate: ___
580 + 246 = __

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
580 – 600
246 – 200
600
+200
800
Step 2: Find the sum. Add the ones, tens, then the hundreds.
580
+246
826
826 is close to 800. So, the answer is reasonable.

Question 15.
Estimate: ___
796 + 135 = ___

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
796 – 800
135 – 100
800
+100
900
Step 2: Find the sum. Add the ones, tens, then the hundreds.
796
+135
931
931 is close to 900. So, the answer is reasonable.

Question 16.
A truck driver travels 428 miles on Monday. He travels 473 miles on Tuesday. How many miles does he travel in all on Monday and Tuesday?

Answer:
Given that,
A truck driver travels 428 miles on Monday. He travels 473 miles on Tuesday.
428
+473
901
Thus he travels 901 miles on Monday and Tuesday.

Question 17.
Reasoning
Your friend finds a sum. Is her answer reasonable? If not, describe her mistake.
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 85$

Answer:
119
+ 187
306
Your answer is not reasonable.

Think and Grow: Modeling Real Life

A construction team builds an 825-meter-long boardwalk on a beach. The team builds 408 meters one week and 377 meters the next week. Is the boardwalk complete?
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 86$
Addition equation:
The boardwalk ___ complete.

Answer:
Given,
A construction team builds an 825-meter-long boardwalk on a beach. The team builds 408 meters one week and 377 meters the next week.
408
+377
785
The boardwalk did not complete.

Show and Grow

Question 18.
A road crew repaves the road on a 547-meter-long bridge. The crew repaves 318 meters the ﬁrst day and 229 meters the second day. Is the road on the bridge completely repaved?

Answer:
Given,
A road crew repaves the road on a 547-meter-long bridge.
The crew repaves 318 meters on the ﬁrst day and 229 meters on the second day.
318
+229
547
Yes, the road on the bridge completely repaved.

Question 19.
A family drives from St. Louis to Orlando for a vacation. The family drives 363 miles the first day and 386 miles the second day. How many miles does the family have left to drive?

$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 87$
Answer:
Given,
A family drives from St. Louis to Orlando for a vacation. The family drives 363 miles the first day and 386 miles the second day.
363 + 386 = 749 miles
922 miles – 749 miles = 173 miles
173 miles the family have left to drive.

Question 20.
Which booth had more visitors in all?
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 88$

Answer:
Day 1 – 468 + 527 = 995
Day 2 – 416 + 374 = 790
995 – 790 = 205
Thus day 1 has more visitors in all.

Add Three-Digit Numbers Homework & Practice 8.5

Find the sum. Check whether your answer is answerable.

Question 1.
Estimate: ___
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 89$

Answer: 630

Explanation:
The estimated number of 493 is 490
The estimated number of 142 is 140.
490
+ 140
630

Question 2.
Estimate: ___
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 90$

Answer: 820

Explanation:
The estimated number of 763 is 760.
The estimated number of 58 is 60.
760
+ 60
820

Question 3.
Estimate: ___
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 91$

Answer: 580

Explanation:
The estimated number of 308 is 310
The estimated number of 273 is 270
310
+270
580

Question 4.
Estimate: ___
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 92$

Answer: 420

Explanation:
The estimated number of 276 is 280
The estimated number of 138 is 140
280
+ 140
420

Question 5.
Estimate: ___
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 93$

Answer: 700

Explanation:
The estimated number of 532 is 530
The estimated number of 167 is 170.
530
+170
700

Question 6.
Estimate: ___
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 94$

Answer: 980

Explanation:
The estimated number of 680
The estimated number of 296 is 300.
680
+300
980

Question 7.
Estimate: ___
595 + 280 = ___

Answer: 880

Explanation:
The estimated number of 595 is 600
The estimated number of 280
600
+ 280
880

Question 8.
Estimate: ___
419 + 295

Answer: 720

Explanation:
The estimated number of 419 is 420
The estimated number of 295 is 300
420
+ 300
720

Question 9.
Estimate: ___
498 + 305 = ___

Answer: 800

Explanation:
The estimated number of 498 is 500
The estimated number of 305 is 300
500
+300
800

Question 10.
Open-Ended
Complete the addends so you need to regroup to add. Then find the sums.
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 95$

Answer:
Let the addend be 3
479
+283
762
If you take 3 as addend then you need to regroup to find the sum.
Let the addend be 9
697
+135
832
If you take 9 as addend then you need to regroup to find the sum.

Question 11.
DIG DEEPER!
Find the missing digits.
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 96$

Answer:
$Big-Ideas-Math-Answers-3rd-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-96$
You will get 466 if you add 107 and 359.
You will get 982 if you add 748 and 234
You will get 962 if you add 670 and 292.
You will get 982 if you add 809 and 173.

Question 12.
Modeling Real Life
Newton wants to complete a 770-mile hike in 2 months. He hikes 423 miles the first month and 347 miles the second month. Does he complete the hike?

Answer:
Given,
Newton wants to complete a 770-mile hike in 2 months.
He hikes 423 miles the first month and 347 miles the second month
423
+347
770
Thus he Newton completed the hike.

Question 13.
Modeling Real Life
You ship a package 750 miles from San Diego to Salt Lake City. The package is now in Las Vegas. How many miles are left until your package is delivered?
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 97$

Answer:
Given,
You ship a package 750 miles from San Diego to Salt Lake City. The package is now in Las Vegas.
121 + 270 = 391 miles until Las Vegas.
750 – 391 = 359 miles
Therefore 359 miles are left until your package is delivered.

Review & Refresh

Find the quotient.

Question 14.
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 98$

Answer: 10

Explanation:
Divide the two numbers 100 and 10.
100/10 = 10
Thus the quotient is 10.

Question 15.
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 99$

Answer: 9

Explanation:
Divide the two numbers 45 and 5.
45/5 = 9
Thus the quotient is 9.

Question 16.
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 100$

Answer: 7

Explanation:
Divide the two numbers 14 and 7.
14/2 = 7
Thus the quotient is 7.

Question 17.
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 101$

Answer: 8

Explanation:
Divide the two numbers 10 and 80.
80/10 = 8
Thus the quotient is 8.

Question 18.
Divide 25 by 5.

Answer: 5

Explanation:
Divide the two numbers 5 and 25.
25/5 = 5
Thus the quotient is 5.

Question 19.
Divide 30 by 10.

Answer: 3

Explanation:
Divide the two numbers 10 and 30.
30/10 = 3
Thus the quotient is 3.

Question 20.
Divide 8 by 2.

Answer: 4

Explanation:
Divide the two numbers 8 and 2.
8/2 = 4
Thus the quotient is 4.

Lesson 8.6 Add Three or More Numbers

Find the sum of the numbers. Which two numbers should you add first?
$Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 102$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
348 = 350
478 = 480
152 = 150
350 + 480 + 150 = 980
The sum is about 980.
517 = 520
117 = 120
283 = 280
520 + 120 + 280 = 920
The sum is about 920
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
348
+478
+152
978
978 is close to 980. So, the answer is reasonable.
517
+117
+283
917
917 is close to 920, So, the answer is reasonable.

Reasoning
Why did you choose those numbers? Compare your strategy to your partner’s strategy.

Answer:
348 = 350
478 = 480
152 = 150
I choose these numbers because they are nearest to ten. This strategy will help to improve the mental math. After solving the problem you can verify the answer with the actual sum.

Think and Grow: Add Three or More Numbers

Example
Find 138 + 221 + 176 + 92. Check whether your answer is reasonable.
Step 1: Estimate. Round each addend to the nearest ten.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 103$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
140 + 220 + 180 + 90 = 630
The sum is about 630.
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
138
+221
+176
+92
627
627 is close to 630. So, the answer is reasonable.

Show and Grow

Find the sum. Check whether your answer is reasonable.

Question 1.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 104$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
342 = 340
73 = 70
267 = 270
340 + 70 + 270 = 680
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
342
+73
+267
682
682 is close to 680. So, the answer is reasonable.

Question 2.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 105$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
191 = 190
452 = 450
206 = 210
190 + 450 + 210 = 850
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 105$
849
849 is close to 850. So, the answer is reasonable.

Question 3.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 106$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
65 = 60
98 = 100
637 = 640
640
+100
+60
800
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
637
+98
+65
800
800 is close to 800. So, the answer is reasonable.

Question 4.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 107$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
241 = 240
394 = 390
85 = 80
193 = 190
390
240
190
+80
900
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 107$
913
913 is close to 900. So, the answer is reasonable.

Question 5.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 108$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
136 = 140
51 = 50
64 = 60
410 = 410
140
410
+60
+50
660
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 108$
661
661 is close to 660. So, the answer is reasonable.

Question 6.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 109$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
105 = 100
113 = 110
222 = 220
307 = 310
100 + 110 + 220 + 310 = 740
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 109$
747
747 is close to 740. So, the answer is reasonable.

Apply and Grow: Practice

Find the sum. Check whether your answer is answerable.

Question 7.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 110$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
557 = 560
79 = 80
283 = 280
560 + 80 + 280 = 920
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 110$
919
919 is close to 920. So, the answer is reasonable.

Question 8.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 111$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
382 = 380
357 = 360
160 + 380 + 360 = 900
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
160
382
357
899
899 is close to 900. So, the answer is reasonable.

Question 9.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 112$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
35 = 30
68 = 70
827 = 830
30 + 70 + 830 = 930
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
35 + 68 + 827 = 930
930 is close to 930. So, the answer is reasonable.

Question 10.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 113$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
153 = 150
235 = 230
458 = 460
67 = 70
150 + 230 + 460 + 70 = 910
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
153 + 235 + 458 + 67 = 913
913 is close to 910. So, the answer is reasonable.

Question 11.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 114$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
108 = 110
172 = 170
200 = 200
263 = 260
110 + 170 + 200 + 260 = 740
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
108 + 172 + 200 + 263 = 743
743 is close to 740. So, the answer is reasonable.

Question 12.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 115$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
181 = 180
629 = 630
140 = 140
23 = 20
180 + 630 + 140 + 20 = 970
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
181 + 629 + 140 + 23 = 973
973 is close to 970. So, the answer is reasonable.

Question 13.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 116$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
213 = 210
208 = 210
462 = 460
111 = 110
210 + 210 + 460 + 110 = 990
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 116$
994
994 is close to 990. So, the answer is reasonable.

Question 14.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 117$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
108 = 110
172 = 170
200 = 200
263 = 260
110 + 170 + 200 + 260 = 740
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 117$
743
743 is close to 740. So, the answer is reasonable.

Question 15.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 118$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
259 = 260
233 = 230
223 = 220
147 = 150
260 + 230 + 220 + 150 = 860
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 118$
862
862 is close to 860. So, the answer is reasonable.

Question 16.
Number Sense
Use the Associative Property of Addition to find (345 + 234) + 206.

Answer:
We can find the (345 + 234) + 206 by using the Associative Property of Addition.
(a + b) + c = a + (b + c)
(345 + 234) + 206 = 345 + (234 + 206)
345 + 440 = 785

Question 17.
YOU BE THE TEACHER
Your friend finds 364 + 109 + 27. Is your friend correct? Explain.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 119$

Answer:
Your friend is incorrect. The order of the sum is wrong. 27 is placed in the wrong pattern. 2 and 7 must be places on tens place and ones place.
364
109
+27
503

Think and Grow: Modeling Real Life

An elevator has a weight limit of 1,000 pounds. A 186-pound man has three 265-pound boxes to deliver. Can he bring all 3 boxes on the elevator at once?
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 120$

Understand the problem:
Make a plan:
Solve:
He ___ bring all 3 boxes on the elevator at once.

Answer:
Given that,
An elevator has a weight limit of 1,000 pounds.
A 186-pound man has three 265-pound boxes to deliver.
265 + 265 + 265 + 186 = 981
1000 – 981 = 19 pounds
Thus he can bring all 3 boxes on the elevator at once.

Show and Grow

Question 18.
An auditorium has 650 seats. 175 students from each of 3 schools compete in a math competition. 68 teachers assist. Are there enough seats for all of the students and teachers?

Answer:
Given,
An auditorium has 650 seats. 175 students from each of 3 schools compete in a math competition. 68 teachers assist.
175 + 175 + 175 + 68 = 593
Thus the seats are enough for all of the students and teachers.

Question 19.
DIG DEEPER!
Four students at a school organize a petition for more lunch food options. They need 500 signatures. How many more signatures do they need?
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 121$

Answer:
Add the number of signatures of all the students
A + B + C + D = 77 + 108 + 112 + 96 = 393
500 – 393 = 107
Thus they need 107 signatures more.

Add Three or More Numbers Homework & Practice 8.6

Find the sum. Check whether your answer is reasonable

Question 1.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 122$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
862 is close to 860. So, the answer is reasonable.

Question 2.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 123$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
65 = 60
41 = 40
786 = 790
60 + 40 + 790 = 890
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 123$
892
892 is close to 890. So, the answer is reasonable.

Question 3.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 124$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
409 = 410
87 = 90
463 = 460
410 + 90 + 460 = 960
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 124$
959
959 is close to 960. So, the answer is reasonable.

Question 4.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 125$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
302 = 300
253 = 250
169 = 170
18 = 20
300 + 250 + 170 + 20 = 740
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 125$
742
742 is close to 740. So, the answer is reasonable.

Question 5.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 126$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
353 = 350
121 = 120
154 = 150
116 = 120
350 + 120 + 150 + 120 = 740
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 126$
744
744 is close to 740. So, the answer is reasonable.

Question 6.
Estimate: ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 127$

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
213 = 210
251 = 250
139 = 140
210 + 270 + 250 + 140 = 870
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 127$
873
873 is close to 870. So, the answer is reasonable.

Question 7.
Structure
Which problem can you solve without regrouping?
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 128$

Answer:
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-128$

Question 8.
Reasoning
You add 602 + 125 + 231. Your friend adds 231 + 602 + 125. Do you both get the same answer? Use an additional property to explain.

Answer:
There are four mathematical properties that involve addition. The properties are the commutative, associative, additive identity and distributive properties. Additive Identity Property: The sum of any number and zero is the original number.
602 + 125 + 231 = 958
231 + 602 + 125 = 958
The sum will be the same irrespective of the change of order.

Question 9.
Modeling Real Life
A firefighter’s ladder has a weight limit of 750 pounds. One firefighter weighs 196 pounds. Another firefighter weighs 243 pounds. They each have67 pounds of gear. If both firefighters wear their gear, can they climb the ladder at the same time?
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 129$

Answer:
A firefighter’s ladder has a weight limit of 750 pounds.
One firefighter weighs 196 pounds. Another firefighter weighs 243 pounds. They each have 67 pounds of gear.
196 + 243 + 67 + 67 = 573
750 – 573 = 177 pounds
If both firefighters wear their gear, they can climb the ladder at the same time.

Question 10.
DIG DEEPER!
Your principal agrees to make a lip-sync video if the school’s social media page reaches 1,000 likes in 5 days. How many more likes does the school’s page need?
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 130$

Answer:
Given,
Your principal agrees to make a lip-sync video if the school’s social media page reaches 1,000 likes in 5 days.
Add all the number of likes
573 + 168 + 201 + 47 = 989
1000 – 989 = 19 likes
Thus the school’s page needs 19 likes more.

Review & Refresh

Question 11.
Find the area of the shape
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 131$

Answer:
From the figure, we can observe that each block = 1 sq. cm
We have to find the area of the shaded part.
The shaded region is in the form of a rectangle.
So, we have to find the area of the rectangle.
A = l × b
A = 7 × 3
A = 21 sq.cm
Thus the area of the shape is 21 sq. cm

Lesson 8.7 Use Number Lines to Subtract

Explore and Grow

Color to find 79 – 47. Then model your jumps on the number line.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 132$

Answer:
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-132 (1)$

Reasoning
How can finding 79 – 47 help you find 379 – 47?

Answer:
You can subtract 47 from 79 by using the number line.
79 – 47 = 32
Just add 300 to 79 or add 3 to the left and subtract 47 from 379.
379 – 47 = 332

Think and Grow: Subtracting on a Number Line

Example
Find 358 – 82.
One Way:
Use the count back strategy. Start at 358. Count back by tens, then by ones.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 133$

$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-133$

Another Key:
Use the count on strategy. Start at 82. Count on until you reach 358.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 134$
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-134$

Show and Grow

Question 1.
Use the count back strategy to find 273 – 36.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 135$

Answer:
Use the count back strategy. Start at 273. Count back by tens, then by ones.
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-135$

Question 2.
Use the strategy to find 124 – 45.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 136$

Answer:
Use the count on strategy. Start at 45. Count on until you reach 124.
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-136$

Apply and Grow: Practice

Question 3.
Use the count back strategy to find 961 – 38.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 137$

Answer:
Use the count back strategy. Start at 961. Count back by tens, then by ones.
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-137$

Question 4.
Use the count back strategy to find 853 – 77.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 138$

Answer:
Use the count back strategy. Start at 853. Count back by tens, then by ones.
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-138$

Find the difference.

Question 5.
316 – 24 = ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 139$

Answer:
Use the count back strategy. Start at 316. Count back by tens, then by ones.
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-139$

Question 6.
548 – 113 = ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 140$

Answer:
Use the count back strategy. Start at 548. Count back by tens, then by ones.
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-140$

Question 7.
Your friend knows 154 words in Italian. You want to know just as many words as your friend. So far, you have learned 73 words. How many words do you have left to learn?

Answer:
Given,
Your friend knows 154 words in Italian. You want to know just as many words as your friend. So far, you have learned 73 words.
154 – 73 = 81
Thus 81 words are left to learn.

Question 8.
Structure
Write the equation shown by the number line.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 141$

Answer:
By seeing the above number line we can find the subtraction equation.
Use the count on strategy. Start at 36. Count on until you reach 407.
407 – 36 = 371

Think and Grow: Modeling Real Life

Each member of a marching band and a football team is awarded a ribbon. The marching band has 123 members. The football team has 66 members. How many more ribbons are needed for the marching band than for the football team?
Subtraction equation:
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 142$
___ more ribbons are needed for the marching band.

Answer:
Given,
Each member of a marching band and a football team is awarded a ribbon. The marching band has 123 members. The football team has 66 members.
123 – 66 = 57
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-142$
Thus 57 ribbons are needed for the marching band than for the football team.

Show and Grow

Question 9.
A marine biologist feeds 435 pounds of fish to an orca and 50 pounds of fish to a sea lion. How many more pounds did the orca eat than the sea lion?
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 143$

Answer:
Given,
A marine biologist feeds 435 pounds of fish to an orca and 50 pounds of fish to a sea lion.
435 – 50 = 385
Thus 385 more pounds the orca eat than the sea lion.

Question 10.
There are 620 paper lanterns for a festival. Some are let go. There are 42 left. How many paper lanterns were let go?

Answer:
Given,
There are 620 paper lanterns for a festival. Some are let go. There are 42 left.
620 – 42 = 578
Therefore 578 paper lanterns were let go.

Question 11.
DIG DEEPER!
There are some guests at an amusement park. 387 of them leave when it rains. 474 of them stay. How many guests were there before it rained?

Answer:
Given that,
There are some guests at an amusement park. 387 of them leave when it rains. 474 of them stay.
387 + 474 = 861
Thus 861 guests were there before it rained.

Use Number Lines to Subtract Homework & Practice 8.7

Question 1.
Use the count back strategy to find 232 – 53.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 144$

Answer:
Use the count back strategy. Start at 232. Count back by tens, then by ones.
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-144$

Question 2.
Use the count back strategy to find 796 – 81.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 145$

Answer:
Use the count back strategy. Start at 796. Count back by tens, then by ones.
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-145$

Find the difference

Question 3.
474 – 19 = ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 146$

Answer:
Use the count back strategy. Start at 474. Count back by tens, then by ones.
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-146$

Question 4.
615 – 204 = ___
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 147$

Answer:
Use the count back strategy. Start at 615. Count back by tens, then by ones.
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-147$

Question 5.
Writing
Write and solve a subtraction word problem using 995 and 238.

Answer:
There are 995 guests at an amusement park. 238 of them leave when it rains. How many guests were there before it rained?
995 – 238
We can solve the problem by using the number line.
$Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-146$

Question 6.
DIG DEEPER!
Which number lines can you use to find 734 – 308?
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 148$

Answer:
Among all the number lines option ii is used to find 734 – 308.

Question 7.
Modeling Real Life
You take 107 pictures on a field trip to a zoo. Your friend takes 73 pictures. How many more pictures do you take than your friend?
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 149$

Answer:
Given that,
You take 107 pictures on a field trip to a zoo. Your friend takes 73 pictures.
107 – 73 = 34
You take 34 pictures than your friend.

Question 8.
Modeling Real Life
An author has 350 copies, not of her book. Some are signed. 115 copies are signed. How many copies are signed?

Answer:
Given,
An author has 350 copies, not of her book. Some are signed. 115 copies are signed.
350 – 115 = 235 copies
Therefore 235 copies are to be signed.

Review & Refresh

Find the quotient.

Question 9.
Divide 25 by 5.

Answer: 5

Explanation:
Divide the two numbers 25 and 5.
25/5 = 5
Thus the quotient is 5.

Question 10.
Divide 40 by 4.

Answer: 10

Explanation:
Divide the two numbers 40 and 4.
40/4 = 10
Thus the quotient is 10.

Question 11.
Divide 72 by 8.

Answer: 9

Explanation:
Divide the two numbers 72 and 8.
72/8 = 9
Thus the quotient is 9.

Lesson 8.8 Use Mental Math to Subtract

Explore and Grow

750 – 300 = ___
650 – 200 = ____
750 – 300 = ___
700 – 250 = ____
750 – 300 = ___
740 – 290 = ____

What patterns do you notice? Explain.

Answer:
750 – 300 = 450
650 – 200 = 450
750 – 300 = 450
700 – 250 = 450
750 – 300 = 450
740 – 290 = 450
Here you notice that all the answers are 450 for different patterns.

Think and Grow: Mental Math Strategies for Subtraction

Example
Find 433 – 198.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 150$
One Way: Use compensation to change both numbers.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 151$
433 + 2 = 435
198 + 2 = 200
435
-200
235
So, 433 – 198 = 235

Another Way: Use compensation to change one number.
$Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 152$
198 is close to 200 and it is easier to subtract 200.
433
-198 + 2 = 200
233
233 + 2 = 235
So, 433 – 198 = 235

Show and Grow

Use compensation to find the difference

Question 1.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 153$

Answer:
Use compensation to change both numbers.
Add 4 to both numbers and use mental math strategy.
$Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-153$

Question 2.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 154$

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
$Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-154$

Question 3.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 155$

Answer:
Use compensation to change one number.
219 is close to 220 and it is easier to subtract 220.
$Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-155$

Apply and Grow: Practice

Use compensation to find the difference

Question 4.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 156$

Answer:
Use compensation to change both numbers.
Add 1 to both numbers and use mental math strategy.
$Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-156$

Question 5.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 157$

Answer:
117 is close to 120. So it is easier to subtract 120.
$Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-157$

Use mental math to find the difference

Question 6.
643 – 115 = ___

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
643 + 5 = 648
115 + 5 = 120
648
-120
528
So, 643 – 115 = 528

Question 7.
863 – 257 = ___

Answer:
Use compensation to change both numbers.
Add 3 to both numbers and use mental math strategy.
863 + 3 = 866
257 + 3 = 260
866
-260
606

Question 8.
768 – 543 = ___

Answer:
Use compensation to change both numbers.
Add 2 to both numbers and use mental math strategy.
768 + 2 = 770
543 + 2 = 545
770
-545
225

Question 9.
688 – 414 = ___

Answer:
Use compensation to change both numbers.
Add 1 to both numbers and use mental math strategy.
688 + 1 = 689
414 + 1 = 415
689
-415
274

Question 10.
499 – 106 = ___

Answer:
Use compensation to change both numbers.
Add 4 to both numbers and use mental math strategy.
499 + 4 = 503
106 + 4 = 110
503
-110
393

Question 11.
495 – 162 = ___

Answer:
Use compensation to change both numbers.
Add 3 to both numbers and use mental math strategy.
495 + 3 = 498
162 + 3 = 165
498
-165
333

Question 12.
874 – 515 = ___

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
874 + 5 = 879
515 + 5 = 520
879
-520
359

Question 13.
637 – 228 = ___

Answer:
Use compensation to change both numbers.
Add 2 to both numbers and use mental math strategy.
637 + 2 = 639
228 + 2 = 230
639
-230
409

Question 14.
986 – 432 = ___

Answer:
Use compensation to change both numbers.
Add 3 to both numbers and use mental math strategy.
986 + 3 = 989
432 + 3 = 435
989
-435
554

Question 15.
A movie theater has 225 seats. 108 seats are not taken. How many seats are not taken?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 158$

Answer:
Given that,
A movie theater has 225 seats. 108 seats are not taken.
225 – 108 = 117
Therefore 117 seats are not taken.

Question 16.
Reasoning
Your friend starts to ﬁnd 741 – 295. What is the next step? Explain.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 159$

Answer:
The next step is to subtract the original numbers.
741
-295
446

Think and Grow: Modeling Real Life

A softball coach has $325 for new equipment. She buys the catching gear. Does she have enough money left to buy the bat?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 402$

Subtraction equation:
Compare:
The coach ___ has enough money to buy the bat.

Answer:
Given A softball coach has $325 for new equipment. She buys the catching gear.
325 + 1 = 326
219 + 1 = 220
326
-220
106
The cost of the bat is $109.
Thus she does not have enough money.

Show and Grow

Question 17.
A store owner has 550 T-shirts. He sells 333 of them. Then he receives an order for 168 T-shirts. Does he have enough T-shirts to complete the order?

Answer:
Given that,
A store owner has 550 T-shirts. He sells 333 of them.
550 – 333 = 217
Then he receives an order for 168 T-shirts.
217 – 168 = 49
Yes, he has enough T-shirts to complete the order.

Question 18.
The manager of a gaming center has $700 for new electronics. She buys the game system. Does she have enough money left for either of the other two items? If so, which one?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 160$

Answer:
Given that,
The manager of a gaming center has $700 for new electronics. She buys the game system.
$399 + $169 = $568
700 – 568 = $132
Thus she has enough money left for either of the other two items.
She can buy the game system and a bundle of games.

DIG DEEPER!
How much more money does the manager need to buy both the television and the bundle of games?

Answer:
The cost of television is $379
The cost of the bundle of games = $169
379 + 169 = 548
Thus the manager need $548 to buy both the television and the bundle of games

Use Mental Math to Subtract Homework & Practice 8.8

Use compensation to find the difference.

Question 1.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 161$

Answer:
Use compensation to change both numbers.
Add 3 to both numbers and use mental math strategy.
596 + 3 = 599
317 + 3 = 320
599
-320
279
So, 596 – 317 = 279

Question 2.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 162$

Answer:
Add 6 to 214 to make the subtraction easier.
685
-214 + 6 = 220
685
-220
465
So, 685 – 214 = 465

Use mental math to find the difference.

Question 3.
782 – 489 = ___

Answer:
Use compensation to change both numbers.
Add 1 to both numbers and use mental math strategy.
782 + 1 = 783
489 + 1 = 490
783
-490
293
So, 782 – 489 = 293

Question 4.
672 – 266 = ___

Answer:
Use compensation to change both numbers.
Add 4 to both numbers and use mental math strategy.
672 + 4 = 676
266 + 4 = 270
676
-270
406
So, 672 – 266 = 406

Question 5.
983 – 155 = ___

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
983 + 5 = 988
155 + 5 = 160
988
-160
820
So, 983 – 155 = 820

Question 6.
744 – 125 = ___

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
744 + 5 = 749
125 + 5 = 130
749
–130
619
So, 744 – 125 = 619

Question 7.
967 – 619 = ___

Answer:
Use compensation to change both numbers.
Add 1 to both numbers and use mental math strategy.
967 + 1 = 968
619 + 1 = 620
968
-620
348
So, 967 – 619 = 348

Question 8.
854 – 517 = ___

Answer:
Use compensation to change both numbers.
Add 3 to both numbers and use mental math strategy.
854 + 3 = 857
517 + 3 = 520
857
-520
337
So, 854 – 517 = 337

Question 9.
472 – 215 = ___

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
472 + 5 = 479
215 + 5 = 220
479
–220
259
So, 472 – 215 = 259

Question 10.
883 – 335 = ___

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
883 + 5 = 888
335 + 5 = 340
888
-340
548
So, 883 – 335 = 548

Question 11.
575 – 198 = ___

Answer:
Use compensation to change both numbers.
Add 2 to both numbers and use mental math strategy.
575 + 2 = 577
198 + 2 = 200
577
-200
377
So, 575 – 198 = 377

Question 12.
Reasoning
To find 765 – 246, Newton adds 5 to each number and then subtracts. To find the difference, Descartes adds 4 to each number, and then subtracts. Will they both get the correct answer? Explain.

Answer:
Given,
To find 765 – 246, Newton adds 5 to each number and then subtracts.
765 + 5 = 770
246 + 5 = 251
770 – 251 = 519
To find the difference, Descartes adds 4 to each number, and then subtracts.
765 + 4 = 769
246 + 4 = 250
769 – 250 = 519
Yes they both get the correct answer.

Question 13.
Modeling Real Life
A custodian has 350 desks to clean. She cleans 124 desks on the first floor and 147 desks on the second floor. Does she clean all of the desks?

Answer:
Given that,
A custodian has 350 desks to clean.
She cleans 124 desks on the first floor and 147 desks on the second floor.
124 + 147 = 271
350
-271
79
No, she did not clean all of the desks.

Question 14.
Modeling Real Life
A fashion designer has $ 725 to spend on new supplies. She buys the sewing machine. Does she have enough money left for either of the other two items? If so, which one?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 163$

Answer:
Given,
A fashion designer has $ 725 to spend on new supplies. She buys the sewing machine.
725 – 495 = 230
Yes she has enough money left for either of the other two items
She can buy a Mannequin.

DIG DEEPER!
How much more money does the fashion designer need to buy both the mannequin and the fashion design software?

Answer:
Given,
The cost of the Mannequin is $129
The cost of the fashion design software is $329
129 + 329 = 458
458 – 230 = 228
Thus she needs 228 to buy both the mannequin and the fashion design software.

Review & Refresh

Draw equal groups. Then complete the equations.

Question 15.
3 groups of 6
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 164$

Answer:
3 groups of 6 means 3 times 6.
6 + 6 + 6 = 18
3 × 6 = 18

Question 16.
4 groups of 9
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 165$

Answer:
4 groups of 9 means 4 times 9.
9 + 9 + 9 + 9 = 36
4 × 9 = 36

Lesson 8.9 Subtract Three-Digit Numbers

Explore and Grow

Model the equation. Draw to show your model. Then find the difference.
694 – 418 = ___

Reasoning
How can you use an estimate to check whether your answer is reasonable?

Answer:
Find 694 – 418. Check whether your answer is reasonable.
694 = 700
418 = 400
700
-400
300
The difference is about 300.
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup
$Big-ideas-math-answers-grade-3-chapter-8-img-1$
Step 3: Check 276 is close to 300, so the answer is reasonable.

Think and Grow: Subtract Three-Digit Numbers

Example
Find 604 – 215. Check whether your answer is reasonable.
Step 1: Estimate. Round each number to the nearest hundred.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 166$
600
-200
400
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 167$
$Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-167$
Step 3: Check 389 is close to 400, so the answer is reasonable.

Show and Grow

Find the difference. Check whether your answer is reasonable.

Question 1.
Estimate: ___
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 168$

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
300
200
100
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM-Grade-3-Chapter-8-Add-and-subtract-digit-numbers-img-2
Step 3: Check 136 is close to 100, so the answer is reasonable.

Question 2.
Estimate: ___
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 169$

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
538 = 500
371 = 400
500 – 400 = 100
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers img-3
Step 3: Check 167 is close to 100, so the answer is reasonable.

Question 3.
Estimate: ___
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 170$

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
500 – 300 = 200
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
$Bigideas-math-answer-key-3rd-grade-chapter-8-add-and-subtract-multi-digit-numbers-img-4$
Step 3: Check 238 is close to 200, so the answer is reasonable.

Question 4.
Estimate: ___
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 171$

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
963 = 1000
429 = 400
1000 – 400 = 600
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
$Big-ideas-math-answers-3rd-grade-chapter-8-add-and-subtract-multi-digit-numbers-img-5$
Step 3: Check 534 is close to 600, so the answer is reasonable.

Question 5.
Estimate: ___
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 172$

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
641 = 600
287 = 300
600 – 300 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM-Book-Grade-3-chapter-8-add-and-subtract-multi-digit-numbers-img-6
Step 3: Check 354 is close to 300, so the answer is reasonable.

Question 6.
Estimate: ___
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 173$

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
832 = 800
359 = 300
800 – 300 = 500
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
$Big-ideas-math-book-answers-3rd-grade-chapter-8-img-7$
Step 3: Check 473 is close to 500, so the answer is reasonable.

Apply and Grow: Practice

Find the difference. Check whether your answer is reasonable.

Question 7.
Estimate: ___
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 174$

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
518 = 500
232 = 200
500 – 200 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
$Big-ideas-math-book-grade-3-amswers-chapter-8-add-and-subtract-multi-digit-numbers-img-14$
Step 3: Check 286 is close to 300, so the answer is reasonable.

Question 8.
Estimate: ___
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 175$

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
971 = 1000
320 = 300
1000 – 300 = 700
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
$Big-ideas-math-book-grade-3-amswers-chapter-8-add-and-subtract-multi-digit-numbers-img-15$
Step 3: Check 651 is close to 700, so the answer is reasonable.

Question 9.
Estimate: ___
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 176$

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
565 = 600
289 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
$Big-ideas-math-book-grade-3-amswers-chapter-8-add-and-subtract-multi-digit-numbers-img-16$
Step 3: Check 276 is close to 300, so the answer is reasonable.

Question 10.
Estimate: ___
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 177$

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
546 = 500
341 = 300
500 – 300 = 200
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
$Big-ideas-math-book-grade-3-amswers-chapter-8-add-and-subtract-multi-digit-numbers-img-17$
Step 3: Check 205 is close to 200, so the answer is reasonable.

Question 11.
Estimate: ___
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 178$

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
707 = 700
453 = 400
700 – 400 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
$Big-ideas-math-book-grade-3-amswers-chapter-8-add-and-subtract-multi-digit-numbers-img-18$
Step 3: Check 254 is close to 300, so the answer is reasonable.

Question 12.
Estimate: ___
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 179$

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
406 = 400
77 = 100
400 – 100 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
$Big-ideas-math-book-grade-3-answers-chapter-8-add-and-subtract-multi-digit-numbers-img-19$
Step 3: Check 329 is close to 300, so the answer is reasonable.

Question 13.
Estimate: ____
552 – 381 = ___

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
552 = 600
381 = 400
600 – 400 = 200
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
$Big-ideas-math-book-grade-3-answers-chapter-8-add-and-subtract-multi-digit-numbers-img-20$
Step 3: Check 171 is close to 200, so the answer is reasonable.

Question 14.
Estimate: ___
725 – 146 = ____

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
725 = 700
146 = 100
700 – 100 = 600
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
$Big-ideas-math-book-grade-3-answers-chapter-8-add-and-subtract-multi-digit-numbers-img-21$
Step 3: Check 579 is close to 600, so the answer is reasonable.

Question 15.
Estimate: ___
800 – 486 = ___

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
486 = 500
800 – 500 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
$Big-ideas-math-book-grade-3-answers-chapter-8-add-and-subtract-multi-digit-numbers-img-22$
Step 3: Check 314 is close to 300, so the answer is reasonable.

Question 16.
The number of rings on a tree is equal to its age. A redwood tree has 473 rings. A bristlecone pine tree has 806 rings. How much older is the bristlecone pine tree than the redwood tree?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 180$

Answer:
Given that,
The number of rings on a tree is equal to its age. A redwood tree has 473 rings. A bristlecone pine tree has 806 rings
806
-473
333
Thus 333 older is the bristlecone pine tree than the redwood tree.

Question 17.
Writing
Explain how to regroup 408 to subtract 259.

Answer:
Working each column from right to left
9 is greater than 8 so you must regroup:
Take 1 from 4, so 4 becomes 3.
Add 10 to 0, so 0 becomes 10.
Take 1 from 10, so 10 becomes 9.
Add 10 to 8, so 8 becomes 18.
18 minus 9 is 9.
9 minus 5 is 4.
3 minus 2 is 1.
408
-259
149

Think and Grow: Modeling Real Life

How many more pennies does the class need to reach the goal?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 500$
Subtraction equation:
The class needs to collect ___ more pennies.

Answer:
800
-444
356
The class needs to collect 356 more pennies.

Show and Grow

Question 18.
How many more campers can attend the summer camp?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 501$

Answer:
Total summer camp openings = 405
Filled = 316
400 – 316 = 84
84 more campers can attend the summer camp.

Question 19.
A musician wants to buy a set of speakers that costs $672. She saves $224 each month for 2 months. How much money does she still need to save?

Answer:
Given that,
A musician wants to buy a set of speakers that costs $672.
She saves $224 each month for 2 months.
224 + 224 = 448
672 – 448 = 224
Thus she still need to save $224.

Question 20.
DIG DEEPER!
Newton has 442 packages to deliver. Descartes has 464. Newton delivers 174 packages, and Descartes delivers 188. Who is closer to finishing his deliveries?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 502$

Answer:
Given,
Newton has 442 packages to deliver. Descartes has 464. Newton delivers 174 packages, and Descartes delivers 188.
442 – 174 = 268
464 – 188 = 276
Thus Newton is closer to finishing his deliveries.

Subtract Three-Digit Numbers Homework & Practice 8.9

Find the difference. Check whether your answer is reasonable.

Question 1.
Estimate: ___
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 503$

Answer:
The estimated number for 571 is 600.
The estimated number for 220 is 200.
600
-200
400
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM-Grade-3-Answer-Key-Chapter-8-Add-&-Subtract-Multi-Digit-Numbers-img-8
352 is close to 400. So the answer is reasonable.

Question 2.
Estimate: ___
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 504$

Answer:
The estimated number for 421 is 400.
The estimated number for 277 is 300.
400
-300
100
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM-3rd-Grade-Solutions-Chapter-8-Add-and-subtract-multi-digit-numbers-img-9
144 is close to 100. So the answer is reasonable.

Question 3.
Estimate: ___
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 505$

Answer:
The estimated number for 534 is 500.
The estimated number for 186 is 200
500
-200
300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
$Bigideas-math-answers-3rd-grade-chapter-8-img-10$
348 is close to 300. So the answer is reasonable.

Question 4.
Estimate: ___
690 – 298 = ___

Answer:
The estimated number for 690 is 700
The estimated number for 298 is 300.
700
-300
400
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM-grade-3-solution-key-chapter-8-add-and-subtract-multi-digit-numbers-img-11
392 is close to 400. So the answer is reasonable.

Question 5.
Estimate: ___
613 – 472 = ___

Answer:
The estimated number for 613 is 600.
The estimated number for 472 is 500
600
-500
100
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
$Big-ideas-math-answer-key-grade-3-8th-chapter-add-&-subtract-multi-digit-numbers-img-12$
141 is close to 100. So the answer is reasonable.

Question 6.
Estimate: ___
835 – 189 = ___

Answer:
The estimated number for 835 is 800.
The estimated number for 189 is 200.
800
-200
600
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM-Solution-Key-Grade-3-Chapter-8-img-13
646 is close to 600. So the answer is reasonable.

Question 7.
YOU BE THE TEACHER
Your friend says you have to regroup every time you subtract from a number that has a zero. Is your friend correct? Explain.

Answer:
Your friend says you have to regroup every time you subtract from a number that has a zero.
Yes, your friend is correct. Because whenever you subtract a number from 0 you have to regroup.

Question 8.
Modeling Real Life
How many more soup can labels does the school need to reach the goal?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 506$

Answer:
To find how many more soup can label does the school need to reach the goal you have to subtract label collected from school soup.
1000 – 638 = 362

Question 9.
Modeling Real Life
Newton wants to buy a couch that costs $594. He saves $198 each month for 2 months. How much money does he still need to save?

Answer:
Given,
Newton wants to buy a couch that costs $594. He saves $198 each month for 2 months.
198 × 2 = $396
594 – 396 = -198
Therefore Newton need to save $198 to buy the couch.

Question 10.
DIG DEEPER!
Find the missing digits.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 507$

Answer:
$Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-507$

Review & Refresh

Round the number to the nearest ten and to the nearest hundred.

Question 11.
64
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 508$

Answer:
The number 64 nearest ten is 60.
The number nearest hundred to 64 is 60.

Question 12.
411
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 509$

Answer:
The number 411 nearest ten is 410
The number nearest hundred to 411 is 400.

Lesson 8.10 Relate Addition and Subtraction

Explore and Grow

How can you find the missing number? How do you know you are correct?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 510$

Reasoning
How are addition and subtraction related?

Answer:
497 – 358 = 136
Check the answer by using the addition model.
136
+358
497
So, the answer is reasonable.

Think and Grow: Relate Addition and Subtraction

Inverse operations are operations that “undo” each other. Addition and subtraction are inverse operations.
Example
Find 846 – 283. Use the inverse operation to check.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 511$

Answer:
First, subtract 283 from 846 and then use the inverse operation to check the solution.
$Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-511$

Example
Find 355 + 437. Use the inverse operation to check.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 512$

Answer:
First, add 355 from 437 and then use the inverse operation to check the solution.
$Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-512$

Show and Grow

Find the sum or difference. Use the inverse operation to check.

Question 1.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 183$

Answer:
First, subtract 682 from 419 and then use the inverse operation to check the solution.
$Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-183$
So, the answer is reasonable.

Question 2.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 184$

Answer:
First, add 169 from 745 and then use the inverse operation to check the solution.
$Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-184$
So, the answer is reasonable.

Question 3.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 513$

Answer:
First, add 376 from 238 and then use the inverse operation to check the solution.
$Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-513$
So, the answer is reasonable.

Question 4.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 514$

Answer:
First, subtract 547 from 285 and then use the inverse operation to check the solution.
$Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-514$
So, the answer is reasonable.

Question 5.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 515$

Answer:
First, add 463 from 349 and then use the inverse operation to check the solution.
463
+349
812
Use the inverse operation to check the solution
812
-349
463
So, the answer is reasonable.

Question 6.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 516$

Answer:
First, subtract 790 from 317 and then use the inverse operation to check the solution.
790
-317
473
Use the inverse operation to check the solution
473
+317
790
So, the answer is reasonable.

Apply and Grow: Practice

Find the sum or difference. Use the inverse operation to check.

Question 7.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 517$

Answer:
First, subtract 857 from 567 and then use the inverse operation to check the solution.
857
-567
290
Use the inverse operation to check the solution
290
+567
857
So, the answer is reasonable.

Question 8.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 518$

Answer:
First, add 762 from 143 and then use the inverse operation to check the solution.
762
+143
905
Use the inverse operation to check the solution
905
-143
762
So, the answer is reasonable.

Question 9.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 519$

Answer:
First, add 653 from 217 and then use the inverse operation to check the solution.
653
+217
870
Use the inverse operation to check the solution
870
-217
653
So, the answer is reasonable.

Question 10.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 520$

Answer:
First, subtract 294 from 156 and then use the inverse operation to check the solution.
294
-156
138
Use the inverse operation to check the solution
138
+156
294
So, the answer is reasonable.

Question 11.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 521$

Answer:
First, add 475 from 438 and then use the inverse operation to check the solution.
475
+438
913
Use the inverse operation to check the solution
913
-438
475
So, the answer is reasonable.

Question 12.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 522$

Answer:
First, subtract 514 from 386 and then use the inverse operation to check the solution.
514
-386
128
Use the inverse operation to check the solution
128
+386
514
So, the answer is reasonable.

Question 13.
Which one Doesn’t Belong? Which equation does not belong with the other three?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 523$

Answer:
The fourth equation does not belong with the other three. Because it is not using the inverse operation of addition and subtraction.

Question 14.
Open-Ended
Write a subtraction equation that has a difference of 381.

Answer: 901 – 520 = 381
Take the number on your own and write the subtraction equation with the difference of 381.

Think and Grow: Modeling Real Life

A kayak costs $321. A customer pays $196 for the kayak after using a gift card. How much money is the gift card worth?
The gift card is worth $___.
Check:
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 524$

Answer:
Given that,
A kayak costs $321. A customer pays $196 for the kayak after using a gift card.
321 – 196 = 125
Thus the gift card worth $125.

Show and Grow

Question 15.
You print 600 flyers for an event. You hand out some of them. There are 237 left. How many flyers did you hand out?

Answer:
Given that,
You print 600 flyers for an event. You hand out some of them. There are 237 left.
600 – 237 = 363
Thus you hand out 363 flyers.
363 + 237 = 600

Question 16.
A building has 163 floors. You start on the 28th floor. You go up in the elevator 126 floors. Then you go down 145 floors. On which floor do you end?

Answer:
Given that,
A building has 163 floors. You start on the 28th floor. You go up in the elevator 126 floors. Then you go down 145 floors.
126 – 28 = 98 floors
145 – 98 = 47
Thus you end at 47th floor.

Question 17.
DIG DEEPER!
A bus travels from Boston to Washington, D.C. On the way back, the bus stops in New York City. How many miles has the bus traveled in all? How many miles does the bus have left to travel?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 525$

Answer:
Given that,
A bus travels from Boston to Washington, D.C. On the way back, the bus stops in New York City.
We have to find How many miles has the bus traveled in all
441 + 225 = 666 miles
Thus a bus travels 666 miles.
666 – 225 = 441 miles

Relate Addition and Subtraction Homework & Practice 8.10

Find the sum or difference. Use the inverse operation to check.

Question 1.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 526$

Answer:
931
-544
387
Now you have to do the inverse operation.
387
+544
931

Question 2.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 527$

Answer:
623
+285
908
Now you have to do the inverse operation.
908
-285
623

Question 3.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 528$

Answer:
523
+237
760
Now you have to do the inverse operation.
760
-237
523

Question 4.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 529$

Answer:
403
-252
151
Now you have to do the inverse operation.
151
+252
403

Question 5.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 530$

Answer:
612
+387
999
Now you have to do the inverse operation.
999
-387
612

Question 6.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 531$

Answer:
511
-371
140
Now you have to do the inverse operation.
140
+371
511

Question 7.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 532$

Answer:
437
+156
593
Now you have to do the inverse operation.
593
-156
437

Question 8.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 533$

Answer:
726
-362
364
Now you have to do the inverse operation.
364
+362
726

Question 9.
YOU BE THE TEACHER
Your friend uses an inverse operation to check her answer. Is your friend correct? Explain.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 510$

Answer:
No, your friend is incorrect.
380
-159
221
Now you have to do the inverse operation.
221
+159
380

Question 10.
Which One Doesn’t Belong? Which does not belong with the other three?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 535$

Answer:
208 + 475 = 683
Now you have to do the inverse operation.
683 – 475 = 208
The second figure does not belong to the other three expressions.

Question 11.
Modeling Real Life
A telescope costs $169. A customer pays $119 for the telescope after using a gift card. How much money is the gift card worth?

Answer:
Given that,
A telescope costs $169. A customer pays $119 for the telescope after using a gift card.
169
-119
50
The cost of the gift card is $50.

Question 12.
DIG DEEPER!
A train travels from Dallas to San Antonio. On the way back, the train stops in Austin. How many miles has the train travel? How many miles does the train have left to travel?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 536$

Answer:
Given,
A train travels from Dallas to San Antonio. On the way back, the train stops in Austin.
274 – 79 = 195 miles
The train has traveled 195 miles.
79 miles left to travel from Austin to San Antonio.

Review & Refresh

Question 13.
Use the Distributive Propertytoﬁnd the area of the rectangle.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 537$

Answer:
4 × 8 = 4 × (4 + 4)
4 × 8 = (4 × 4) + (4 × 4)
4 × 8 = 16 +16
4 × 8 = 32
Thus the area of the rectangle = 32 square foot.

Lesson 8.11 Problem Solving: Addition and Subtraction

Explore and Grow

You read 150 pages in three weeks.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 538$

what does p represent?
p = ___

Answer:
Let 9 be the number of pages read
56 + 47 + p = 150
103 + p = 150
p = 150 – 103
p = 47

Construct Arguments
Explain to your partner how to find what n represents.
250 + n = 580

Answer:
Given the expression 250 + n = 580
n = 580 – 250
n = 330

Think and Grow: Using the Problem-Solving Plan
Example
Newton has 368 baseball cards. He gives away 139 of them. He buys 26 more. How many cards does he have now?

Understand the Problem

Newton has ___ cards.
He gives away of them.
He buys ___ more.
You need to find how many ___ he has now.

Answer:

Newton has 368 cards.
He gives away of them.
He buys 26 more.
You need to find how many cards he has now.

Make a Plan

How will you solve?

Subtract ___ from ___ to find how many ___ he has left after he gives some away.
Then add ___ to the difference to find how many he has now.

Answer:

Subtract 139 from 368 to find how many cards he has left after he gives some away.
Then add 26 to the difference to find how many he has now.

Solve
Draw a part-part-whole model and write an equation.
Use a letter to represent the unknown number.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 539$

Newton has __ cards now.

Answer:
Step 1:
c is the unknown difference.
368 – 139 = c
c = 229
Step 2:
c = 229
229 + 26 = n
n = 225
Thus the unknown sum is 225.

Show and Grow

Question 1.
Explain how you can check whether your answer above is reasonable.

Answer:
You can check the answer by using addition and subtraction.
368
-139
229
Now check whether the answer is correct or not.
229
+139
368
So, the answer is reasonable.

Apply and Grow: Practice

Write equations to solve. Use letters to represent the unknown numbers. Check whether your answer is reasonable.

Question 2.
A baker makes 476 muffins. He sells 218 of them. Then he makes 390 more. How many muffins does the baker have now?

Answer:
Given,
A baker makes 476 muffins. He sells 218 of them.
476
-218
258
Then he makes 390 more.
390
+258
648
Thus the baker has 348 muffins now.

Question 3.
Newton knocks down 146 pins in his first bowling game. He knocks down 19 more pins in his second game than in his first game. How many pins does he knock down in all?

Answer:
Given that,
Newton knocks down 146 pins in his first bowling game. He knocks down 19 more pins in his second game than in his first game.
146
+19
165
He knocks down 165 pins in his second game.
To find the total number of pins we have to the points in the first game and second game.
146
+165
311
Thus he knockdowns 311 pins in all.

Question 4.
You are traveling to a campground that is 243 miles away. You travel 155 miles in the morning and 59 miles in the afternoon. How many more miles do you need to travel before you get to the campground?

Answer:
Given that,
You are traveling to a campground that is 243 miles away.
You travel 155 miles in the morning and 59 miles in the afternoon.
155
+59
214
243
-214
029
Thus you need to travel 29 miles to get to the campground.

Question 5.
There are 205 lawn tickets and 585 bleacher tickets sold for a concert. There are 680 fewer VIP tickets sold than lawn and bleacher tickets combined. How many VIP tickets are sold?

Answer:
Given,
There are 205 lawn tickets and 585 bleacher tickets sold for a concert.
205 + 585 = 790
There are 680 fewer VIP tickets sold than lawn and bleacher tickets combined.
790
-680
110
Thus 110 VIP tickets are sold.

Think and Grow: Modeling Real Life

How many more people went to see the movie on Friday than on Thursday and Saturday combined?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 542$
Understand the problem:
Make a plan:
Solve:
__ more people went to see the movie on Friday than on Thursday and Saturday combined.

Answer:
Thursday – 346 people
Saturday – 512 people
346 + 512 = 858 people
897
-858
39
39 more people went to see the movie on Friday than on Thursday and Saturday combined.

Show and Grow

Question 6.
How many more people used the ferry on Friday than on Saturday and Sunday combined?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 543$
Explain how you can check whether your answer is reasonable.

Answer:
Friday – 903 passengers
Saturday – 624 passengers
Sunday – 255 passengers
624
+255
879
Now subtract 879 from 903.
903
-879
024

Problem Solving: Addition and Subtraction Homework & Practice 8.11

Write equations to solve. Use letters to represent the unknown numbers. Check whether your answer is reasonable.

Question 1.
Newton has 387 tokens, and Descartes has 295. They use a total of 461 tokens. How many tokens do they have now?

Answer: 222 tokens

Explanation:
Given that,
Newton has 387 tokens, and Descartes has 295.
387 + 295 = 682
They use a total of 461 tokens.
682 – 461 = 222 tokens
Thus they have 222 tokens now.

Question 2.
There are 125-second graders and 118 third graders at a museum. There are 249 more adults than students at the museum. How many adults are at the museum?

Answer:
Given that,
There are 125-second graders and 118 third graders at a museum.
125 + 118 = 243 graders
There are 249 more adults than students at the museum.
243 + 249 = 492
Therefore 492 adults are at the museum.

Question 3.
You received 171 votes in a coloring contest. Your friend received 24 fewer votes than you. How many people voted for you and your friend in all?

Answer:
Given,
You received 171 votes in a coloring contest. Your friend received 24 fewer votes than you.
171 + 24 = 195 votes
The number of votes for your friend is 195.
171+ 195 = 366 votes
Thus 366 people voted for you and your friend.

Question 4.
Writing
Write and solve a two-step problem that can be solved using addition or subtraction.

Answer:
You bought 10 packs of sketches and your friend bought 4 packs less than you. Each pack contains 10 sketches. Find how many sketches you and your friend bought in all.
Sol: You bought 10 packs of sketches and your friend bought 4 packs less than you.
10 – 4 = 6 packs
Your friend bought 6 packs.
Each pack contains 10 sketches.
10 + 6 = 16
16 × 10 = 160
Thus you and your friend bought 160 sketches.

Question 5.
Modeling Real Life
How many more fish were caught on Sunday than on Friday and Saturday combined?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 544$
Explain how you can check whether your answer is reasonable.

Answer:
Number of fish caught on Friday and Saturday = 127 + 244 = 371
Number of fish caught on Sunday = 564
564 – 371 = 193
193 more fish were caught on Sunday than on Friday and Saturday combined.

Review & Refresh

Question 6.
Use the multiplication table.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 545$
Describe the pattern in the shaded row and column.
What property explains this pattern?

Answer: The pattern shows that it is the multiple of 5.
5 × 1 = 5
5 × 2 = 10
5 × 3 = 15
5 × 4 = 20
5 × 5 = 25

Add and Subtract Multi-Digit Numbers Performance Task

Your school holds a talent show.

Question 1.
You and your friend hand out programs to guests before the show. You each start with 250 programs. There are 114 programs left. How many programs did you and your friend hand out?

Answer:
250 – 114 = 136 programs
Thus you and your friend hand out 136 programs.
136 + 114 = 250 programs

Question 2.
75 students wait backstage to perform in the show. There are 336 children, 125 adults, and 14 teachers in the audience.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 216$
a. How many people are at the talent show in all? Explain how to use addition properties to find the sum.

Answer:
Given,
75 students wait backstage to perform in the show. There are 336 children, 125 adults, and 14 teachers in the audience.
75 + 336 + 125 + 14 = 550
Thus there are 550 people are in the talent show.

b. Four students perform in each of the ﬁrst 4 acts. How many students still need to perform?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 217$

Answer: 16 students

Explanation:
Given,
Four students perform in each of the ﬁrst 4 acts.
4 × 4 = 16
Thus 16 students need to perform.

c. Each performer is given a juice box backstage. Juice boxes come in packages of 10. How many packages did the teachers buy? How many juice boxes are left?

Answer:
Given,
Each performer is given a juice box backstage. Juice boxes come in packages of 10.
1 box – 10 packages
16 × 1 = 16 boxes
16 × 10 = 160 packages

Add and Subtract Multi-Digit Numbers Activity

Three in a Row: Addition and Subtraction

Directions:
1. Players take turns.
2. On your turn, spin both spinners. Add or subtract the two numbers. Cover the sum or difference.
3. If the sum or difference is already covered, then you lose your turn.
4. The first player to get three counters in a row, horizontally, vertically, or diagonally, wins!
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 218$

Answer:
Sums:
547 + 107 = 654
547 + 338 = 885
547 + 262 = 809
I got the three counters vertically.

Add and Subtract Multi-Digit Numbers Chapter Practice

8.1 Identify Addition Properties

Identify the property.

Question 1.
59 + 0 = 59

Answer:
It shows the Addition Property of Zero. The Addition Property of Zero defines the sum of any number and 0 is that number.

Question 2.
(14 + 32) + 6 = 14 + (32 + 6)

Answer:
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 3.
27 + 51 = 51 + 27

Answer:
Commutative Property of addition Changing the grouping of addends does not change the sum.

Question 4.
Structure
Which equations show the Commutative Property of Addition?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 219$

Answer:
64 + 12 = 12 + 64 – Commutative Property of addition changing the grouping of addends does not change the sum.
71 + 0 = 71 – It shows the Addition Property of Zero. The Addition Property of Zero defines the sum of any number and 0 is that number.
(56 + 21) + 34 = 56 + (21 + 34) – It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.
26 + (41 + 4) = 4 + (26 + 41) – It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

8.2 Use Number Lines to Add

Question 5.
Find 648 + 37.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 220$

Answer:
Use the count on strategy. Start at 648. Count on by tens, then by ones.
$Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-220$

8.3 Use Mental Math to Add

Use mental math to find the sum.

Question 6.
192 + 107 = ___

Answer:
You can find the sum of 192 and 107 by using mental math strategies.
192 – 2 = 190
107 + 2 = 109
190
+109
299

Question 7.
676 + 114 = ___

Answer:
You can find the sum of 676 and 114 by using mental math strategies.
676 + 4 = 680
114 – 4 = 110
680
+110
790

Question 8.
716 + 279 = ___

Answer:
You can find the sum of 716 and 279 by using mental math strategies.
716 – 1 = 715
279 + 1 = 280
715
+280
995

Question 9.
501 + 468 = ___

Answer:
You can find the sum of 501 and 468 by using mental math strategies.
501 – 1 = 500
468 + 1 = 469
500
+469
969

Question 10.
527 + 343 = ___

Answer:
You can find the sum of 527 and 343 by using mental math strategies.
527 + 3 = 530
343 – 3 = 340
530
+340
870

Question 11.
441 + 189 = ___

Answer:
You can find the sum of 441 and 189 by using mental math strategies.
441 – 1 = 440
189 + 1 = 190
440
+190
630

8.4 Use Partial Sums to Add

Use partial sums to add.

Question 12.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 221$

Answer:
586 = 500 + 80 + 6
107 = 100 + 00 + 7
693 = 600 + 80 + 13

Question 13.
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 222$

Answer:
647 = 600 + 40 + 7
293 = 200 + 90 + 3
940 = 800 + 130 + 10

Question 14.
Modeling Real Life
On Earth, your cousin weighs 207 pounds less than he would on Jupiter. Your cousin weighs 135 pounds on Earth. How much would he weigh on Jupiter?
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 223$

Answer:
Given,
On Earth, your cousin weighs 207 pounds less than he would on Jupiter. Your cousin weighs 135 pounds on Earth.
207 + 135 = 342
Thus your cousin would weigh 342 pounds on Jupiter.

8.5 Add Three-Digit Numbers

Find the sum. Check whether your answer is reasonable.

Question 15.
Estimate: ___
326 + 490 = ___

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
326 = 300
490 = 500
300 + 500 = 800
The sum is about 800.
Step 2: Find the sum. Add the ones, tens, then the hundreds.
326
+490
816
816 is close to 800. So, the answer is reasonable.

Question 16.
Estimate: ___
657 + 189 = ___

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
657 = 700
189 = 200
600 + 200 = 800
The sum is about 800.
Step 2: Find the sum. Add the ones, tens, then the hundreds.
657 + 189 = 846
846 is close to 800. So, the answer is reasonable.

Question 17.
Estimate: ___
543 + 261 = ___

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
543 = 500
261 = 300
500 + 300 = 800
The sum is about 800.
Step 2: Find the sum. Add the ones, tens, then the hundreds.
543 + 261 = 804
804 is close to 800. So, the answer is reasonable.

8.6 Add Three or More Numbers

Question 18.
Estimate: ___
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 224$

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
78 = 100
433 = 400
367 = 400
100 + 400 + 400 = 900
The sum is about 900.

Question 19.
Estimate: ___
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 225$

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
194 = 200
151 = 200
244 = 200
231 = 200
200 + 200 + 200 + 200 = 800
The sum is about 800.

Question 20.
Estimate: ___
$Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 226$

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
373 = 400
329 = 300
118 = 100
61 = 100
400 + 300 + 100 + 100 = 900
The sum is about 900.

8.7 Use Number Lines to Subtract

Question 21.
Find 856 – 29.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 511$

Answer:
Use the count back strategy. Start at 856. Count back by tens, then by ones.
$Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-511$

Question 22.
Structure
Write the equation shown by the number line.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 512$

Answer:
By seeing the above number we can find the subtraction equation.
764 – 50 = 714
714 – 4 = 710
710 – 3 = 707
The subtraction equation is 764 – 57 = 707

8.8 Use Mental Math to Subtract

Use mental math to find the difference.

Question 23.
957 – 619 = ___

Answer:
957 – 7 = 950
619 + 7 = 626
950
-626
324
The difference is 324.

Question 24.
831 – 415 = ___

Answer:
831 – 1 = 830
415 + 1 = 416
830
-416
414
The difference is 414.

Question 25.
876 – 366 = ___

Answer:
876 – 6 = 870
366 + 6 = 372
870 – 372 = 498

Question 26.
636 – 317 = ___

Answer:
636 + 6 = 642
317 – 6 = 311
642 – 311 = 331
The difference is 331.

Question 27.
965 – 528 = ___

Answer:
528 – 8 = 520
965 + 8 = 973
973 – 520 = 453
The difference is 453.

Question 28.
384 – 118 = ____

Answer:
684 + 4 = 688
118 – 4 = 114
688 – 114 = 574
The difference is 574.

8.9 Subtract Three-Digit Numbers

Find the difference. Check whether your answer is reasonable.

Question 29.
Estimate: ___
963 – 51 = ___

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
963 = 1000
51 = 100
1000 – 100 = 900
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
963 -51 = 912
912 is close to 900. So, the answer is reasonable.

Question 30.
Estimate: ___
878 – 594 = ___

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
878 = 900
594 = 600
900 – 600 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
878 – 594 = 284
284 is close to 300. So, the answer is reasonable.

Question 31.
Estimate: ___
766 – 297 = ___

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
766 = 800
297 = 300
800 – 300 = 500
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
766 – 297 = 469
469 is close to 500. So, the answer is reasonable.

Question 32.
YOU BE THE TEACHER
Your friend ﬁnds 760 – 482. Is your friend correct? Explain.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 513$

Answer: No your friend is not correct.
BIM Answer Key for Grade 3 Chapter 8 img-13

8.10 Relate Addition and Subtraction

Find the sum or difference. Use the inverse operation to check.

Question 33.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 514$

Answer:
649
+227
876
Now you have to do the inverse operation of the sum.
876
-227
649

Question 34.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 515$

Answer:
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 515$
288
Now you have to do the inverse operation of the difference.
288
+517
805

8.11 Problem Solving: Addition and Subtraction

Question 35.
There are 532 dogs enrolled in police academies. 246 dogs graduate in July, and 187 dogs graduate in August. How many dogs still need to graduate?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 516$

Answer:
Given that,
There are 532 dogs enrolled in police academies.
246 dogs graduate in July, and 187 dogs graduate in August.
246 + 187 = 433
532
-433
99
Thus 99 dogs are still needed to graduate.

Add and Subtract Multi-Digit Numbers Cumulative Practice 1 – 8

Question 1.
Which numbers round to 300 when rounded to the nearest hundred?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 517$

Answer: The numbers round to 300 when rounded to the nearest hundred is
298, 309, 347
Thus the correct answer is the option a, b, c.

Question 2.
You buy 18 cups of yogurt. The yogurt is sold in packs of 6 cups. How many packs of yogurt do you buy?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 519$

Answer:
Given,
You buy 18 cups of yogurt. The yogurt is sold in packs of 6 cups.
18/6 = 3 packs
Thus the correct answer is option c.

Question 3.
A bedroom ﬂoor is 9 feet long and 8 feet wide. What is the area of the bedroom ﬂoor?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 520$

Answer:
Given,
A bedroom ﬂoor is 9 feet long and 8 feet wide.
The area of the rectangle = l × w
A = 9ft × 8ft
A = 72 sq. ft
Thus the area of the bedroom ﬂoor is 72 sq. ft.

Question 4.
Your friend says 458 – 298 = 160. How can you use inverse operations to check your friend’s answer? Is your friend correct?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 544$

Answer:
Your friend says 458 – 298 = 160.
Your friend is correct.
The correct answer is option a.

Question 5.
Your friend says she needs (9 × 3) + (3 × 9) = 27 × 27 = 54 tiles to make the design. Why is her thinking incorrect?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 545$

Answer:
Your friend says she needs (9 × 3) + (3 × 9) = 27 × 27 = 54 tiles to make the design.
Her thinking is incorrect because she needs to add 27 and 27 but she multiplied.
(9 × 3) + (3 × 9) = 27 + 27 = 54

Question 6.
Part A
What is the least number that can be made with the digits 7, 9, and 8 using each digit only once?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 546$
Your friend says the greatest number he can make with the digits 7, 9, and 8 using each digit only once is 879. Is he correct? If not, correct his answer. Explain.

Answer:
Given,
Your friend says the greatest number he can make with the digits 7, 9, and 8 using each digit only once is 879
Your friend is incorrect because the greatest number with the digits 7, 9, and 8 is 987.

Question 7.
Which equation is shown by the number line?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 547$

Answer:
The count starts from 0.
The count jumps from 0 and skips for every 3s.
3 × 7 = 21
Thus the correct answer is option c.

Question 8.
Find the sum.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 548$

Answer:
548
+372
920

Question 9.
Which equations show the Associative Property of Addition?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 549$

Answer:
According to the associative property of addition, the sum of three or more numbers remains the same regardless of how the numbers are grouped.
Options A and D show the equation for the Associative Property of Addition.

Question 10.
A teacher takes 7 students on a field trip. Each student pays $5. How much money does the teacher collect in all?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 550$

Answer:
Given,
A teacher takes 7 students on a field trip. Each student pays$5.
7 × $5 = $35
Thus the teacher collects $35 in all.
The correct answer is option c.

Question 11.
What is the area of the shape?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 551$

Answer:
You can divide the figure into two parts.
Figure 1:
l = 4m
b = 3m
A = l × b
A = 4 × 3 = 12 sq. m
Figure 2:
l = 8m
b = 3m
A = l × b
A = 8 × 3 = 24 sq. m
Add the area of both the figures 12 + 24 = 36 sq.m
Thus the correct answer is option D.

Question 12.
There are 459 girls and 552 boys in a school. How many more boys are there than girls?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 552$

Answer:
Given that,
There are 459 girls and 552 boys in school.
Subtract the number of girls from the number of boys.
552
-459
93
Thus the correct answer is option b.

Question 13.
Look at the pattern. What rule was used to make the pattern?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 553$

Answer:
1, 3, 9, 27 are the multiples of 3.
Thus the correct answer is option c.

Question 14.
A smoothie shop sells 368 smoothies in July and 205 smoothies in August. About how many more smoothies did the shop sell in July than in August?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 554$

Answer:
Given that,
A smoothie shop sells 368 smoothies in July and 205 smoothies in August.
368
-205
163
Thus the shop sell 163 smoothies in July than in August.

Question 15.
Which shape does not have an area of 16 square units?
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 555$

Answer:
There are 16 counters in the first figure
2 × 8 = 16
There are 16 counters in the second figure
4 × 4 = 16
There are 15 counters in the third figure
5 × 3 = 15
Thus the correct answer is option c.

Complete the table

Question 16.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 556$

Answer:
$Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-556$
You can complete the table by multiplying the rows and columns.

Question 17.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 557$

Answer:
You can complete the table by multiplying the rows and columns.
$Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-557$

Add and Subtract Multi-Digit Numbers Steam Performance Task 1-8

Question 1.
The carpeting in the third-grade classrooms of an elementary school is being replaced. One roll of carpet covers 100 square yards. The map shows the classrooms that will receive the new carpet.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 560$
a. Explain two different ways to ﬁnd the area of Classroom A.

Answer:
You can find the area of classroom A by using the composite figure.
The shape of classroom A is square.
a = 10 yd
Area of the classroom is a × a
A = 10 × 10 = 100 sq.yd
Thus the area of the classroom A is 100 sq. yd
Another way:
l = 10 yd
w = 3 yd
A = 10 × 3 = 30 sq. yd
l = 10 yd
A = 8 × 4 = 32 sq. yd
l = 10 yd
w = 3 yd
A = 10 × 3 = 30 sq. yd

b. Find the total area of all of the classrooms in square yards.

Answer:
Area of the classroom is a × a
A = 10 × 10 = 100 sq.yd
Thus the area of the classroom A is 100 sq. yd
Area of Classroom D = 10 × 7 = 70 sq.yd
Area of Classroom B = 11 × 7 = 77 sq.yd
Area of Classroom C = 12 × 7 = 84 sq. yd, 8 × 3 = 24 sq.yd
Area of Classroom C = 84 + 24 = 108 sq. yd
Total area of the classrooms = 100 + 70 + 77 + 108 = 355 sq. yd

c. Estimate the number of rolls of carpet needed for the classrooms. Explain.

Answer:
There are 4 classrooms. So, the estimated number of rolls of carpet is 4.

d. Find the area of the hallway in square yards. Is there enough carpet for the hallway?

$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 561$
Answer:
First, divide the hallway into 3 parts.
i. It is in the form of a square.
a = 3 yd
A = 3 × 3 = 9 sq.yd
ii. It is in the form of rectangle
A = l × b
A = 5 × 4 = 20 sq.yd
iii. It is in the form of rectangle
A = l × b
A = 5 × 3 = 15 sq.yd
The area of the hallway in square yards = 9 + 20 + 15 = 44 sq.yd

Question 2.
Each school keeps a record of the total number of students in each class and grade.
a. Use the number of students in your class to estimate the total number of students in your grade. Explain.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 562$

Answer:
Let the number of students in your class is 47.
By this, we can estimate the total number of students in your grade i.e., 50.
The total number of students in your grade is 50.

b. Use the table to write the number of students in each grade of your school.
$Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 563$

Answer:
$Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-563$
You can prepare the table by estimating the number of students in your school.

c. How does your estimate compare to the actual number of students in your grade? Explain.

Answer: You can compare the number of students in the above table with the actual number of students in your grade.

d. What is the total number of students in your school?

Answer:
Add the number of students of all the grades
40+ 50 + 52 + 47 + 47 + 50 = 286
Thus there are 286 students in your school.

e. Write and answer a question using the information from the table above.

Answer:
Compare the number of students in grade 3 with the actual number of students in your school in grade 3?
The estimated number of students in grade 3 in the above table is 50.
The actual number of students in your school is 48.
50 – 48 = 2

f. What is one reason your principal may want to know the total number of students in your class, grade, or school?

Answer: Shaping a vision of academic success for all students.

Conclusion:

I wish the information provided in this BIM Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers pdf is helpful for you. This pdf helps you to score the highest marks in the exams. Get the answers of Bigideas Math Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers from here. Follow our page to get the solutions with a brief explanation for all the 3rd-grade chapters from 1 to 15.