Tag: Eureka Math Grade 3 Answer Key

Engage NY Eureka Math 3rd Grade Module 1 Lesson 10 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 10 Pattern Sheet Answer Key

Multiply.

multiply by 2 (6–10)

Answer:


Explanation:
Multiplied by 2 (6–10) as shown above.

Eureka Math Grade 3 Module 1 Lesson 10 Problem Set Answer Key

Question 1.
7 × 3 = (5 × 3) + (2 × 3) = ___21_______

(5 × 3) + (2 × 3) = 15 + ___6___
15 + ___6___ = _____21________

7 × 3 = (5 × 3) + (2 × 3) =
15 + 6 = 21 or 7 × 3 = 21.

Explanation:
Given 7 × 3 wrote 7 as (5 + 2) × 3 =
(5 × 3) + (2 × 3) =
15 + 6 = 21 or 7 × 3 = 21.

Question 2.
8 × 3 = (4 × 3) + (4 × 3) = __24____

(4 × 3) + (4 × 3) = ____12_____ + _____12____ = 24,
____8_____ × 3 = ____24______

8 × 3 = (4 × 3) + (4 × 3) =
12 + 12 = 24 or 8 × 3 = 24,

Explanation:
Given 8 × 3 wrote 8 as (4 + 4) × 3 =
(4 × 3) + (4 × 3) =
12 + 12 = 24 or 8 × 3 = 24.

Question 3.
Ruby makes a photo album. One page is shown below.
Ruby puts 3 photos in each row.
a. Fill in the equations on the right.
Use them to help you draw arrays that show the photos on the top and bottom parts of the page.


Explanation:
Filled in the equations on the right as 2 × 3 = 6, 3 × 3 =9,
Used them to help to draw arrays that showed the photos on the top and bottom parts of the page as shown above in the picture.

b. Ruby calculates the total number of photos as shown below.
Use the array you drew to help explain Ruby’s calculation.

Explanation:
Ruby calculates the total 15 number of photos as shown above,
Used the array to drew to help explain Ruby’s calculation as
(2 × 3) + (3 × 3) = 6 + 9 = 15.

Eureka Math Grade 3 Module 1 Lesson 10 Exit Ticket Answer Key

Question 1.
6 × 3 = ___18___

(4 × 3) + (2 × 3) = __12__ + ___6___
6 × 3 = _12__ + _6__
_6_ × 3 = _18__

6 × 3  = (4 × 3) + (2 × 3) = 12 + 6 = 18 or 6 × 3 = 18,

Explanation:
Given 6 × 3 wrote 6 as (4 + 2) × 3 =
(4 × 3) + (2 × 3) = 12 + 6 = 18 or 6 × 3 = 18.

Question 2.
7 × 3 = _21_

(5 × 3) + (2 × 3) = __ + __
7 × 3 = __ + __
__ × 3 = ___

7 × 3  = (5 × 3) + (2 × 3) = 15 + 6 = 21 or 7 × 3 = 21,

Explanation:
Given 7 × 3 wrote 7 as (5 + 2) × 3 =
(5 × 3) + (2 × 3) =
15 + 6 = 21 or 7 × 3 = 21.

Eureka Math Grade 3 Module 1 Lesson 10 Homework Answer Key

Question 1.
6 × 3 = ____18______

6 × 3 = (4 × 3) + (2 × 3) = 12 + 6 = 18 or 6 × 3 = 18,

Explanation:
Given 6 × 3 wrote 6 as (4 + 2) × 3 =
(4 × 3) + (2 × 3) =
12 + 6 = 18 or 6 × 3 = 18.

Question 2.
8 × 2 = _16_

8 × 2 = (4 × 2) + (4 × 2) = 8 + 8 = 16 or 8 × 2 = 16,

Explanation:
Given 8 × 2 wrote 8 as (4 + 4) × 2 =
(4 × 2) + (4 × 2) =
8 + 8 = 16 or 8 × 2 = 16.

Question 3.
Adriana organizes her books on shelves. She puts 3 books in each row.
a. Fill in the equations on the right. Use them to draw arrays that show the books on Adriana’s top and bottom shelves.


Explanation:
Given Adriana organizes her books on shelves.
She puts 3 books in each row.
a. Filled in the equations on the right as (5 × 3), (1 × 3)
Used them to draw arrays that show the books on
Adriana’s top and bottom shelves as shown above.

b. Adriana calculates the total number of books as shown below.
Use the array you drew to help explain Adriana’s calculation.


Explanation:
Adriana calculates the total 18 number of books as shown above,
Used the array to drew to help explain Adriana’s calculation as
6 × 3 = (5 × 3) + (1 × 3) = 15 + 3 = 18 or 6 × 3 = 18.

Engage NY Eureka Math 3rd Grade Module 7 Lesson 16 Answer Key

Eureka Math Grade 3 Module 7 Lesson 16 Pattern Sheet Answer Key

Multiply.

multiply by 9 (6–10)
Answer:

Explanation:
9 × 5 = 45
9 × 6 = 54
9 × 7 = 63
9 × 8 = 72
9 × 9 = 81
9 × 10 = 90.

Eureka Math Grade 3 Module 7 Lesson 16 Problem Set Answer Key

Question 1.
Find the perimeter of 10 circular objects to the nearest quarter inch using string. Record the name and perimeter of each object in the chart below.

Object	Perimeter (to the nearest quarter inch)
Cap of my jam jar	13\(\frac{1}{4}\) inches
Bangle	17\(\frac{1}{4}\) inches
Plate of my Dog	34\(\frac{1}{4}\) inches
Plastic glass Mouth	21\(\frac{3}{4}\) inches
Lid of my Lunch Box	29\(\frac{3}{2}\) inches
Ear ring	2\(\frac{2}{4}\) inches
Water Bottle cap	7\(\frac{2}{4}\) inches
Pencil Mouth	1\(\frac{2}{4}\) inches
Pen Cap	3\(\frac{1}{4}\) inches
Perfume Bottom surface	12\(\frac{3}{4}\) inches

a. Explain the steps you used to find the perimeter of the circular objects in the chart above.

Answer:
Step 1: Took a string and wrapped around the object.
Step 2: Marked the string met.
Step 3: Measured the length of the string.

Explanation:
First I rolled the String around the object. Later, I marked the string met. Afterwards, I took a ruler to measure the length of the string.

b. Could the same process be used to find the perimeter of the shape below? Why or why not?

Answer: Yes, the same steps would be followed to find the perimeter of the given shape because I use the string to find the perimeter.

Explanation:
The same steps would be followed to find the perimeter of the given shape because I use the string to find the perimeter. First I rolled the String around the object. Later, I marked the string met. Afterwards, I took a ruler to measure the length of the string.

Question 2.
Can you find the perimeter of the shape below using just your ruler? Explain your answer.

Answer:
No, I cant find the perimeter of this given shape because it got curve line in it as which a ruler cant measure it.

Explanation:
A tool used to rule straight lines and measure distances is called as ruler.
No, I cant find the perimeter of this given shape because it got curve line in it as a ruler measures only straight lines.

Question 3.
Molly says the perimeter of the shape below is 6 \(\frac{1}{4}\) inches. Use your string to check her work. Do you agree with her? Why or why not?

Answer:
No, she is not correct has I have used my string and found the perimeter of the shape as 5 \(\frac{3}{4}\) inches.

Explanation:
Molly says the perimeter of the shape below is 6 \(\frac{1}{4}\) inches.
No, she is not correct has I have used my string and found the perimeter of the shape as 5 \(\frac{3}{4}\) inches.

Question 4.
Is the process you used to find the perimeter of a circular object an efficient method to find the perimeter of a rectangle? Why or why not?
Answer:
No, I don’t think this process to find the perimeter of a circular object an efficient method to find the perimeter of a rectangle, because I can just use a ruler to measure length of the straight lines.

Explanation:
No, I don’t think this process to find the perimeter of a circular object an efficient method to find the perimeter of a rectangle. A ruler is used to find the lengths of the straight lines which help in finding the perimeter of the shape.

Eureka Math Grade 3 Module 7 Lesson 16 Exit Ticket Answer Key

Use your string to the find the perimeter of the shape below to the nearest quarter inch.

Answer:
The perimeter of this circular shape is 26 3/4 inches.

Explanation:
I used my string to measure this circular shape. The perimeter of this circular shape is 26 3/4 inches.

Eureka Math Grade 3 Module 7 Lesson 16 Homework Answer Key

Question 1.
a. Find the perimeter of 5 circular objects from home to the nearest quarter inch using string. Record the name and perimeter of each object in the chart below.

Object	Perimeter (to the nearest quarter inch)
Example:  Peanut Butter Jar Cap	9\(\frac{1}{2}\) inches
Cap of my jam jar	13\(\frac{1}{4}\) inches
Plate of my Dog	34\(\frac{1}{4}\) inches
Lid of my Lunch Box	29\(\frac{3}{2}\) inches
Water Bottle cap	7\(\frac{2}{4}\) inches
Pen Cap	3\(\frac{1}{4}\) inches

b. Explain the steps you used to find the perimeter of the circular objects in the chart above.
Answer:
Well, as previously discussed I have used a string to measure the perimeter of the circular shapes. I wrapped string around the circular bodies and  noted the values where they met. Later I have used a ruler to measured the values.

Explanation:
Used string to measure the perimeter of the circular shapes. I wrapped string around the circular bodies and  noted the values where they met. Later I have used a ruler to measured the values.

Question 2.
Use your string and ruler to find the perimeter of the two shapes below to the nearest quarter inch.

a. Which shape has a greater perimeter?
b. Find the difference between the two perimeters.

Answer:
a. The perimeter of the given shape B is greater by 3inches than The perimeter of the given shape A.
b. The perimeter of the shape B given using string – The perimeter of the shape A given using string
= 17 1/4 inches – 14 1/4 inches
= 3 inches

Explanation:
a. The perimeter of the given shape A using string = 14 1/4 inches.
The perimeter of the given shape B using string = 17 1/4 inches.

b. Difference:
The perimeter of the shape  B given using string – The perimeter of the shape  A given using string
= 17 1/4 inches – 14 1/4 inches
= 3 inches.

Question 3.
Describe the steps you took to find the perimeter of the objects in Problem 2. Would you use this method to find the perimeter of a square? Explain why or why not.
Answer:
Step 1: Took a string and wrapped around the given shape A.
Step 2: Marked the string met.
Step 3: Measured the length of the string using ruler.
Step 4: Took a string and wrapped around the given shape B.
Step 5: Marked the string met.
Step 6: Measured the length of the string using ruler.

No, I cant use this process of finding perimeter for Square because Square’s perimeter can be found easily using ruler directly. For finding the perimeter of circular shapes we use string n later the ruler.

Explanation:
First I rolled the String around the given shape A. Later, I marked the string met. Afterwards, I took a ruler to measure the length of the string. same I did with the given  shape B.

For finding the perimeter of circular shapes we use string n later the ruler, to know their measurement value.  Square is a straight line shape, we can use ruler directly to find its perimeter no need of string.

Engage NY Eureka Math 3rd Grade Module 1 Lesson 9 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 9 Pattern Sheet Answer Key

multiply by 2 (1–5)

Answer:


Explanation:
Multiply by 2 (1–5) as shown above.

Eureka Math Grade 3 Module 1 Lesson 9 Problem Set Answer Key

Question 1.
The team organizes soccer balls into 2 rows of 5.
The coach adds 3 rows of 5 soccer balls.
Complete the equations to describe the total array.


Explanation:
Given the team organizes soccer balls into 2 rows of 5 as
2 × 5 = 10 and the coach adds 3 rows of 5 soccer balls as
3 × 5 = 15, Completed the equations to described the
total array as 5 × 5 = (2 + 3) × 5 = (2 × 5) + (3 × 5) =
10 + 15 = 25 or 5 × 5 = 25.

Question 2.
7 × 2 = __14___

7 × 2 = 14,

Explanation:
Given 7 × 2, we wrote 7 × 2 as (5 + 2) × 2 =
(5 × 2) + ( 2 × 2) = 10 + 4 = 14 or
7 × 2 = 14.

Question 3.
9 × 2 = __18___

9 × 2 = 18,

Explanation:
Given 9 × 2  we wrote 9 × 2 as
(10 – 1) × 2 = (10 × 2) – (1 × 2) =
20 – 2 = 18 or 9 × 2 = 18.

Question 4.
Matthew organizes his baseball cards in 4 rows of 3.
a. Draw an array that represents Matthew’s cards using an x to show each card.
b. Solve the equation to find Matthew’s total number of cards.
4 × 3 = __12__,
a.

Explanation:
Given Matthew organizes his baseball cards in 4 rows of 3,
So array shown as 4 × 3,
Drawn an array that represents Matthew’s cards using an x to show each card as shown above.

b. Total number of  Matthew’s cards are 12,

Explanation:
Solved the equation to find Matthew’s total number of cards as 4 × 3 = 12.

Question 5.
Matthew adds 2 more rows. Use circles to show his new cards on the array in Problem 4(a).
a. Write and solve a multiplication equation to represent the circles you added to the array.
___2___ × 3 = __6____

Explanation:
Given Matthew adds 2 more rows.
Used circles to show his new cards on the array in
Problem 4(a) as shown above.

b. Add the totals from the equations in Problems 4(b) and 5(a) to find Matthew’s total cards.
__12____ + ___6___ = 18

Explanation:
Added the totals from the equations in Problems 4(b) and 5(a) to find Matthew’s total cards as
(4 × 3) + (2 × 3) = 12 + 6 = 18.

c. Write the multiplication equation that shows Matthew’s total number of cards.
___6___ × ___3___ = 18,

The multiplication equation to show total number of cards is 6 × 3 = 18,

Explanation:
The multiplication equation that shows Matthew’s total number of cards is 6 X 3 = (4 X 3) + (2 X 3) =
12 + 6 = 18 or 6 × 3 = 18.

Eureka Math Grade 3 Module 1 Lesson 9 Exit Ticket Answer Key

Question 1.
Mrs. Stern roasts cloves of garlic. She places 10 rows of
two cloves on a baking sheet. Write an equation to describe
the number of cloves Mrs. Stern bakes.
___10____ × ___2____ = __20____ cloves

Answer:
The number of cloves Mrs. Stern bakes is 20 cloves,

Explanation:
Given Mrs. Stern roasts cloves of garlic. She places 10 rows of two cloves on a baking sheet. The equation to describe the number of cloves Mrs. Stern bakes is 10 × 2 = 20 cloves.

Question 2.
When the garlic is roasted, Mrs. Stern uses some for a recipe.
There are 2 rows of two garlic cloves left on the pan.
a. Complete the equation below to show how many garlic cloves Mrs. Stern uses.
____10____ twos – ____2____ twos = ____8___twos,

10 twos – 2 twos = 8 twos,

Explanation:
Completed the equation below to show how many garlic cloves Mrs. Stern uses as 10 twos – 2 twos = 8 twos.

b. 20 – ________ = 16
20 – 4 = 16,

Explanation:
Subtracted 4 from 20 we get 16 as 20 – 4 =16 cloves.

c. Write an equation to describe the number of garlic cloves
Mrs. Stern uses. ___8____ × 2 = ___16_____,

The number of garlic cloves Mrs. Stern uses are 16 cloves,

Explanation:
Given Mrs. Stern roasts cloves of garlic,
She places 10 rows of two cloves on a baking sheet and
Mrs. Stern uses some for a recipe there are 2 rows of two garlic cloves left on the pan, 10 x 2 = 20 cloves and left are
2 × 2 = 4 cloves, Means used cloves are ( 10 × 2) – (2 × 2) =
20 – 4 = 16 cloves or (10 – 2) × 2 = 8 × 2= 16 cloves.
Therefore, the equation for number of garlic cloves Mrs. Stern uses are 8 × 2 = 16 cloves.

Eureka Math Grade 3 Module 1 Lesson 9 Homework Answer Key

Question 1.
Dan organizes his stickers into 3 rows of four.
Irene adds 2 more rows of stickers. Complete the equations to describe the total number of stickers in the array.

a. (4 + 4 + 4) + (4 + 4) = __12 + 8 = ____20_____
b. 3 fours + ___2___ fours = _____5______ fours
c. ___5_____ × 4 = ____20____

a. (4 + 4 + 4) + (4 + 4) = 12 + 8 = 20,
b. 3 fours + 2 fours = 5 fours,
c. 5 X 4 = 20,

Explanation:
Given Dan organizes his stickers into 3 rows of four.
Irene adds 2 more rows of stickers.
Completed the equations to describe the total number of stickers in the array as 5 X 5 =
(3 × 5) + (2 × 5) = 15 + 10 = 25 or 5 × 5 = 25,
a. We wrote 5 × 5 as (4 + 4 + 4) + (4 + 4) = 12 + 8 = 20,
b. We wrote 5 × 5 as 3 fours + 2 fours = 5 fours,
c. We wrote 5 × 4 as 20, 5 × 4 = 20.

Question 2.
7 × 2 = __14____

7 × 2 = 14,

Explanation:
Given 7 × 2 we wrote 7 × 2 as (6 + 1) × 2 =
(6 × 2) + (1 × 2) = 12 + 2 = 14 or 7 × 2 = 14.

Question 3.
9 × 3 = __27___

9 × 3 = 27,

Explanation:
Given 9 × 3 we wrote 9 × 3 as (10 – 1) × 3 =
(10 × 3) – (1 × 3) = 30 – 3 = 27 or 9 x 2 = 27.

Question 4.
Franklin collects stickers. He organizes his stickers in 5 rows of four.
a. Draw an array to represent Franklin’s stickers.
Use an x to show each sticker.
b. Solve the equation to find Franklin’s total number of stickers. 5 × 4 = ___20___
a.

b. Franklin’s total number of stickers are 20.

Explanation:
Given Franklin collects stickers. He organizes his stickers in 5 rows of four.
a. Drawn an array to represent Franklin’s stickers,
Used an x to show each sticker as shown above.
b. Solved the equation to find Franklin’s total number of stickers as 5 × 4 = 20.

Question 5.
Franklin adds 2 more rows. Use circles to show his new stickers
on the array in Problem 4(a).
a. Write and solve an equation to represent the circles you added to the array.
___2___ × 4 = ___8___

Explanation:
Given Franklin adds 2 more rows. Used circles to show his new stickers on the array in Problem 4(a).
a. Wrote and solved an equation to represent the circles I added to the array as 2 × 4 = 8.

b. Complete the equation to show how you add the totals of 2 multiplication facts to find Franklin’s total number of stickers. __20____ + ___8___ = 28,

Equation of totals of 2 multiplication facts to find Franklin’s total number of stickers are 20 + 8 = 28,

Explanation:
Completed the equation to show how I added the totals of 2 multiplication facts to find Franklin’s total number of stickers as (5 × 4) + (2 × 4) = 20 + 8 = 28.

c. Complete the unknown to show Franklin’s total number of stickers.
__5___ × 4 = 20,

The unknown Franklin’s total number of stickers are 20,

Explanation:
To show unknown Franklin’s total number of stickers as given Franklin collects stickers. He organizes his stickers in 5 rows of four, So, Franklin’s total number of stickers are 5 × 4 = 20.

Engage NY Eureka Math 3rd Grade Module 7 Lesson 15 Answer Key

Eureka Math Grade 3 Module 7 Lesson 15 Pattern Sheet Answer Key

Multiply.

multiply by 9 (1–5)
Answer:

Explanation:
9 × 1 = 9
9 × 2 = 18
9 × 3 = 27
9 × 4 = 36
9 × 5 = 45.

Eureka Math Grade 3 Module 7 Lesson 15 Problem Set Answer Key

Question 1.
Mrs. Kozlow put a border around a 5-foot by 6-foot rectangular bulletin board. How many feet of border did Mrs. Kozlow use?
Answer:
Mrs. Kozlow used 22ft of border.

Explanation:

Length of the side of the rectangular bulletin board = 6ft
Width of the side of the rectangular bulletin board = 5ft
Perimeter of the rectangular bulletin board = 2 (Length + Width)
= 2 ( 6ft + 5ft )
= 2 × 11ft
= 22ft.

Question 2.
Jason built a model of the Pentagon for a social studies project. He made each outside wall 33 centimeters long. What is the perimeter of Jason’s model pentagon?
Answer:
Perimeter of the Jason’s model Pentagon = 165 centimeters.

Explanation:

Length of the Side of the Jason’s model Pentagon = 33 centimeters
Perimeter of the Jason’s model Pentagon = 5 × Side
= 5 × 33 centimeters
= 165 centimeters.

Question 3.
The Holmes family plants a rectangular 8-yard by 9-yard vegetable garden. How many yards of fencing do they need to put a fence around the garden?
Answer:
Perimeter of the rectangular vegetable garden = 34-yard.

Explanation:

Length of the rectangular vegetable garden=  9-yard
Width of the rectangular vegetable garden = 8-yard
Perimeter of the rectangular vegetable garden = 2 ( Length + Width )
= 2 (9-yard + 8-yard)
= 2 × 17-yard
= 34-yard.

Question 4.
Marion paints a 5-pointed star on her bedroom wall. Each side of the star is 18 inches long. What is the perimeter of the star?

Answer:
Perimeter of the Star = 180 inches.

Explanation:

Length of the side of the Star = 18 inches
Number of sides of Star = 10
Perimeter of the Star = 10 × Side
= 10 ×18 inches
=180 inches.

Question 5.
The soccer team jogs around the outside of the soccer field twice to warm up. The rectangular field measures 60 yards by 100 yards. What is the total number of yards the team jogs?
Answer:
The total number of yards the team jogs = 640 yards.

Explanation:

Length of the rectangular soccer field = 100 yards
Width of the rectangular soccer field = 60 yards
Perimeter of the rectangular soccer field = 2 ( Length + Width )
= 2 ( 100 yards + 60 yards )
= 2 × 160 yards
= 320 yards.
Number of rounds the soccer team jogs around the outside of the soccer field to warm up = twice
The total number of yards the team jogs = 320 yards × 2
= 640 yards.

Question 6.
Troop 516 makes 3 triangular flags to carry at a parade. They sew ribbon around the outside edges of the flags. The flags’ side lengths each measure 24 inches. How many inches of ribbon does the troop use?
Answer:
The troop used  216inches  of ribbon.

Explanation:

Number of triangular flags Troop 516 makes = 3 or Three
Length of the triangular flags Troop 516 makes = 24inches
Perimeter of the triangular flags Troop 516 makes =  3 × Side
= 3 × 24inches
= 72inches.
Perimeter of the 3 triangular flags Troop 516 makes = 3 × 72inches
= 216inches.

Eureka Math Grade 3 Module 7 Lesson 15 Exit Ticket Answer Key

Marlene ropes off a square section of her yard where she plants grass. One side length of the square measures 9 yards. What is the total length of rope Marlene uses?
Answer:
The total length of rope Marlene uses = 36 yards.

Explanation:

Length of the side of the square = 9 yards.
Perimeter of the square = 4 × Side
= 4 × 9 yards
= 36 yards.

Eureka Math Grade 3 Module 7 Lesson 15 Homework Answer Key

Question 1.
Miguel glues a ribbon border around the edges of a 5-inch by 8-inch picture to create a frame. What is the total length of ribbon Miguel uses?
Answer:
The total length of ribbon Miguel uses = 26-inch.

Explanation:

Length of the rectangular picture = 8-inch
Width of the rectangular picture = 5-inch
Perimeter of the rectangular picture = 2 ( Length + Width )
= 2 ( 8-inch + 5-inch )
= 2 × 13-inch
= 26-inch.

Question 2.
A building at Elmira College has a room shaped like a regular octagon. The length of each side of the room is 5 feet. What is the perimeter of this room?
Answer:
The perimeter of this room = 40 feet.

Explanation:

length of each side of the regular octagon room = 5 feet
Perimeter of the regular octagon room = 8 × Side
= 8 × 5 feet
= 40 feet.

Question 3.
Manny fences in a rectangular area for his dog to play in the backyard. The area measures 35 yards by 45 yards. What is the total length of fence that Manny uses?
Answer:
The total length of fence that Manny uses = 160 yards.

Explanation:

Length of the rectangular fence  = 45 yards
Width of the rectangular fence  = 35 yards
Perimeter of rectangular fence  = 2 ( Length + Width )
= 2 (45 yards + 35 yards )
= 2 × 80 yards
= 160 yards.

Question 4.
Tyler uses 6 craft sticks to make a hexagon. Each craft stick is 6 inches long. What is the perimeter of Tyler’s hexagon?
Answer:
Perimeter of the Tyler’s hexagon = 36 inches.

Explanation:

Length of the side of the Tyler’s hexagon = 6 inches
Perimeter of the Tyler’s hexagon = 6 × Side
= 6 × 6 inches
= 36 inches.

Question 5.
Francis made a rectangular path from her driveway to the porch. The width of the path is 2 feet. The length is 28 feet longer than the width. What is the perimeter of the path?
Answer:
Perimeter of the rectangular path = 60 feet.

Explanation:

Length of the rectangular path = 28 feet
Width of the rectangular path = 2 feet
Perimeter of the rectangular path = 2 ( Length  + Width )
= 2 ( 28 feet + 2 feet )
= 2 × 30 feet
= 60 feet.

Question 6.
The gym teacher uses tape to mark a 4-square court on the gym floor as shown. The outer square has side lengths of 16 feet. What is the total length of tape the teacher uses to mark Square A?

Answer:
Perimeter of the Square court = 64 ft.
The total length of tape the teacher uses to mark Square A = 16 ft.

Explanation:

Length of the Square court = 16 ft
Perimeter of the Square court =  4 × Side
= 4 × 16 ft
= 64 ft.
The total length of tape the teacher uses to mark Square A = Perimeter of the Square court ÷ 4
= 64 ft ÷ 4
= 16 ft.

Engage NY Eureka Math 3rd Grade Module 1 Lesson 8 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 8 Problem Set Answer Key

Question 1.
Draw an array that shows 5 rows of 3.

Explanation:
Drawn an array that shows 5 rows of 3 as 5 × 3 as shown above.

Question 2.
Draw an array that shows 3 rows of 5.

Explanation:
Drawn an array that shows 3 rows of 5 as 3 × 5 as shown above.

Question 3.
Write multiplication expressions for the arrays in Problems 1 and 2. Let the first factor in each expression represent the number of rows. Use the commutative property to make sure the equation below is true.

Explanation:
Wrote multiplication expressions for the arrays in Problems 1 and 2 as 5 × 3, 3 × 5 and let the first factor in each expression represent the number of rows.
Used the commutative property to make sure the equation below is true as 5 × 3 = 15, 3 × 5 = 15,
So,  5 × 3 = 3 × 5 is true equals to 15.

Question 4.
Write a multiplication sentence for each expression.
You might skip-count to find the totals. The first one is done for you.
a. 2 threes: 2 × 3 = 6
b. 3 twos: _________________
c. 3 fours: ________________
d. 4 threes: ________________
e. 3 sevens: ________________
f. 7 threes: ________________
g. 3 nines: _________________
h. 9 threes: ________________
i. 10 threes: _______________

b. 3 twos : 3 x 2 = 6,
Explanation:
Given 3 twos, 3 multiplied by 2 gives 6,
So, 3 × 2 = 6.

c. 3 fours: 3 × 4 = 12,
Explanation:
Given 3 fours, 3 multiplied by 4 gives 12,
So, 3 × 4 = 12.

d. 4 threes: 4 × 3 = 12,
Explanation:
Given 4 threes, 4 multiplied by 3 gives 12,
So, 4 × 3 = 12.

e. 3 sevens:  3 x 7 = 21,
Explanation:
Given 3 sevens, 3 multiplied by 7 gives 21,
So, 3 × 7 = 21.

f. 7 threes:  7 × 3 = 21,
Explanation:
Given 7 threes, 7 multiplied by 3 gives 21,
So, 7 × 3 = 21.

g. 3 nines: 3 × 9 = 27,
Explanation:
Given 3 nines, 3 multiplied by 9 gives 27,
So, 3 × 9 = 27.

h. 9 threes: 9 × 3 = 27,
Explanation:
Given 9 threes, 9 multiplied by 3 gives 27,
So, 9 × 3 = 27.

i. 10 threes: 10 x 3 =30,
Explanation:
Given 10 threes, 10 multiplied by 3 gives 30,
So, 10 × 3 = 30.

Question 5.
Find the unknowns that make the equations true.
Then, draw a line to match related facts.
a. 3 + 3 + 3 + 3 + 3 = _________
b. 3 × 9 = _________
c. 7 threes + 1 three = _________
d. 3 × 8 = _________
e. _________ = 5 × 3
f. 27 = 9 × _________

a. 3 + 3 + 3 + 3 + 3 = 15,

Explanation:
Given 3 + 3 + 3 + 3 + 3  adding 3, 5 times gives 15,
So 3 + 3 + 3 + 3 + 3 = 15.

b. 3 × 9 = 27,

Explanation:
Given 3 × 9, 3 multiplied by 9 gives 27,
So 3 × 9 = 27.

c. 7 threes + 1 three = 8 threes = 24,

Explanation:
Given 7 threes + 1 three = 7 × 3 + 1 × 3 = 21 + 3 = 24,
So, 7 threes + 1 three = 8 threes = 24.

d. 3 × 8 = 24,

Explanation:
Given 3 × 8, 3 multiplied by 8 gives 24,
So 3 × 8 = 24.

e. 15 = 5 × 3,

Explanation:
Given 5 × 3, 5 multiplied by 3 gives 15,
So 15 = 5 × 3.

f. 27 = 9 × 3

Explanation:
Given 27 = 9 × ___, Lets take missing number as x,
27 = 9 × x, So x = 27 ÷ 9 = 3, So 27 = 9 × 3.

Question 6.
Isaac picks 3 tangerines from his tree every day for 7 days.
a. Use circles to draw an array that represents the
tangerines Isaac picks.

Explanation:
Given Isaac picks 3 tangerines from his tree every day for 7 days.
a. Used circles to draw an array that represents the tangerines Isaac picks as 3 × 7 as shown above.

b. How many tangerines does Isaac pick in 7 days?
Write and solve a multiplication sentence to find the total.

Isaac picks 21 tangerines in 7 days,
Multiplication sentence to find the total is 3 x 7 = 21.

Explanation:
Wrote and solved a multiplication sentence to find the total as 3 x 7 = 21, therefore Isaac picks 21 tangerines in 7 days.

c. Isaac decides to pick 3 tangerines every day for 3 more days.
Draw x’s to show the new tangerines on the array in Part (a).

Explanation:
Given Isaac decides to pick 3 tangerines every day for 3 more days.
Drawn x’s to show the new tangerines on the array in Part (a) as shown above in the picture.

d. Write and solve a multiplication sentence to find the total number of tangerines Isaac picks.

Multiplication sentence is 3 x 10 = (3 × 7) + (3 × 3) = 21 + 9 = 30,
The total number of tangerines Isaac picks are 30,

Explanation:
Wrote and solved a multiplication sentence as
3 x 10 = (3 × 7) + (3 × 3) = 21 + 9 = 30,
Therefore the total number of tangerines Isaac picks are 30.

Question 7.
Sarah buys bottles of soap. Each bottle costs $2.
a. How much money does Sarah spend if she buys 3 bottles of soap?
_____$2_____ × ____3______ = $___6_____
b. How much money does Sarah spend if she buys 6 bottles of soap?
_____$2_____ × _____6_____ = $__12______

a. Sarah spends $6 if she buys 3 bottles of soap,

Explanation:
Given Sarah buys bottles of soap. Each bottle costs $2,
So money spent by Sarah if she buys 3 bottles of soap are
$2 × 3 = $6.

b. Sarah spends $12 if she buys 6 bottles of soap,

Explanation:
Given Sarah buys bottles of soap. Each bottle costs $2,
So money spent by Sarah if she buys 6 bottles of soap are $2 × 6 = $12.

Eureka Math Grade 3 Module 1 Lesson 8 Exit Ticket Answer Key

Mary Beth organizes stickers on a page in her sticker book.
She arranges them in 3 rows and 4 columns.
a. Draw an array to show Mary Beth’s stickers.
b. Use your array to write a multiplication sentence to find Mary Beth’s total number of stickers.
c. Label your array to show how you skip-count to solve your multiplication sentence.
d. Use what you know about the commutative property to write a different multiplication sentence for your array.

a.
Explanation:
Given Mary Beth organizes stickers on a page in her sticker book. She arranges them in 3 rows and 4 columns.
Drawn an array as 3 × 4 to show Mary Beth’s stickers.

b. Mary Beth’s total number of stickers are 12,

Explanation:
Used my array to write a multiplication sentence as 3 × 4 = 12, in finding Mary Beth’s total number of stickers.

c.

Explanation:
Labeled my array to show skip-count to 12 and solved my multiplication sentence as 3 x 4 = 12.

d. Commutative property to write a different multiplication sentence for my array is 3 × 4 = 4 x 3,

Explanation:
Used what I know about the commutative property to write a different multiplication sentence for my array as
3 × 4 = 4 x 3.

Eureka Math Grade 3 Module 1 Lesson 8 Homework Answer Key

Question 1.
Draw an array that shows 6 rows of 3.

Explanation:
Drawn an array that shows 6 rows of 3 as 6 × 3 as shown above.

Question 2.
Draw an array that shows 3 rows of 6.

Explanation:
Drawn an array that shows 3 rows of 6 as 3 × 6 as shown above.

Question 3.
Write multiplication expressions for the arrays in Problems 1 and 2. Let the first factor in each expression represent the number of rows. Use the commutative property to make sure the equation below is true.


Explanation:
Wrote multiplication expressions for the arrays in Problems 1 and 2 as 6 × 3, 3 × 6 and let the first factor in each expression represent the number of rows.
Used the commutative property to make sure the equation below is true as 6 × 3 = 18, 3 × 6 = 18,
So,  6 × 3 = 3 × 6 is true equals to 18.

Question 4.
Write a multiplication sentence for each expression.
You might skip-count to find the totals. The first one is done for you.
a. 5 threes: 5 × 3 = 15
b. 3 fives: __________________
c. 6 threes: ________________
d. 3 sixes: ___________________
e. 7 threes: __________________
f. 3 sevens: __________________
g. 8 threes: ________________
h. 3 nines: _________________
i. 10 threes: _______________

b. 3 fives: 3 × 5 = 15,

Explanation:
Given 3 fives, 3 multiplied by 5 gives 15,
So, 3 × 5 = 15.

c. 6 threes: 6 × 3 = 18,

Explanation:
Given 6 threes, 6 multiplied by 3 gives 18,
So, 6 × 3 = 18.

d. 3 sixes: = 3 × 6 = 18,

Explanation:
Given 3 sixes, 3 multiplied by 6 gives 18,
So, 3 × 6 = 18.

e. 7 threes: 7 × 3 = 21,

Explanation:
Given 7 threes, 7 multiplied by 3 gives 21,
So, 7 × 3 = 21.

f. 3 sevens: 3 × 7 = 21,

Explanation:
Given 3 sevens, 3 multiplied by 7 gives 21,
So, 3 × 7 = 21.

g. 8 threes: 8 × 3 = 24,

Explanation:
Given 8 threes, 8 multiplied by 3 gives 24,
So, 8 × 3 = 24.

h. 3 nines: 3 × 9 = 27,

Explanation:
Given 3 nines, 3 multiplied by 9 gives 27,
So, 3 × 9 = 27.

i. 10 threes: 10 × 3 = 30,

Explanation:
Given 10 threes, 10 multiplied by 3 gives 30,
So, 10 X 3 = 30.

Question 5.
Find the unknowns that make the equations true. Then, draw a line to match related facts.
a. 3 + 3 + 3 + 3 + 3 + 3 = _________
b. 3 × 5 = _________
c. 8 threes + 1 three = _________
d. 3 × 9 = _________
e. _________ = 6 × 3
f. 15 = 5 × _________

a. 3 + 3 + 3 + 3 + 3 + 3 = 18,

Explanation:
Given 3 + 3 + 3 + 3 + 3 + 3 adding 3, 6 times gives 18,
So 3 + 3 + 3 + 3 + 3 + 3 = 18.

b. 3 × 5 = 15,

Explanation:
Given 3 × 5, 3 multiplied by 5 gives 15,
So 3 × 5 = 15.

c. 8 threes + 1 three =9 threes = 27,

Explanation:
Given 8 threes + 1 three = 8 × 3 + 1 × 3 = 24 + 3 = 27,
So, 8 threes + 1 three = 9 threes = 27.

d. 3 × 9 = 27,

Explanation:
Given 3 × 9, 3 multiplied by 9 gives 27,
So 3 × 9 = 27.

e. 18 = 6 × 3,

Explanation:
Given 6 × 3, 6 multiplied by 3 gives 18,
So 18 = 6 × 3.

f. 15 = 5 × 3,

Explanation:
Given 15 = 5 × ___, Lets take missing number as x,
15 = 5 × x, So x = 15 ÷ 5 = 3, So 15 = 5 × 3.

Question 6.
Fernando puts 3 pictures on each page of his photo album.
He puts pictures on 8 pages.
a. Use circles to draw an array that represents the total number of pictures in Fernando’s photo album.
b. Use your array to write and solve a multiplication sentence to find Fernando’s total number of pictures.
c. Fernando adds 2 more pages to his book. He puts 3 pictures on each new page. Draw x’s to show the new pictures on the array in Part (a).
d. Write and solve a multiplication sentence to find the new the total number of pictures in Fernando’s album.
a.

Explanation:
Given Fernando puts 3 pictures on each page of his photo album.
He puts pictures on 8 pages.
a. Used circles to draw an array that represents the total number of pictures in Fernando’s photo album as 3 X 8.

b. Multiplication sentence for Fernando’s total number of pictures is 3 X 8 = 24,

Explanation:
Used my array to write and solve a multiplication sentence to find Fernando’s total number of pictures as 3 X 8 = 24.

c.

Explanation:
Fernando adds 2 more pages to his book. He puts 3 pictures on each new page. Drawn x’s to show the new pictures on the array in Part (a) as shown above.

d. Multiplication sentence is 3 x 10 =
(3× 8) + (3 × 2) = 24 + 6 = 30,
The new total number of pictures in Fernando’s album are 30,

Explanation:
Wrote and solved a multiplication sentence as
3 x 10 = (3 × 8) + (3 × 2) = 24 + 6 = 30,
Therefore the new total number of pictures in Fernando’s album are 30.

Question 7.
Ivania recycles. She gets 3 cents for every can she recycles.
a. How much money does Ivania make if she recycles 4 cans?
____3______ × ___4_______ = ___12_____ cents
b. How much money does Ivania make if she recycles 7 cans?
____3______ × ____7______ = ____21____ cents

a. Ivania makes 12 cents if she recycles 4 cans,

Explanation:
Given Ivania recycles and she gets 3 cents for every can she recycles,
So money Ivania makes if she recycles 4 cans is
3 cents × 4 = 12 cents.

b. Ivania makes 21 cents if she recycles 7 cans,

Explanation:
Given Ivania recycles and she gets 3 cents for every can she recycles,
So money Ivania makes if she recycles 7 cans is
3 cents × 7 = 21 cents.

Engage NY Eureka Math 3rd Grade Module 7 Lesson 14 Answer Key

Eureka Math Grade 3 Module 7 Lesson 14 Pattern Sheet Answer Key

Multiply.

multiply by 8 (6–10)
Answer:

Explanation:
8 × 5 = 40
8 × 6 = 48
8 × 7 = 56
8 × 8 = 64
8 × 9 = 72
8 × 10 = 80.

Eureka Math Grade 3 Module 7 Lesson 14 Problem Set Answer Key

Question 1.
Label the unknown side lengths of the regular shapes below. Then, find the perimeter of each shape.
a.

Perimeter = _______ in
Answer:
Length of the side of the ABCDEFG Octogen = 64in.

Explanation:

Length of the side of the ABCDEFG Octogen = 8in
Perimeter of the ABCDEFG Octogen = 8 × side
= 8 × 8in
= 64in.

b.

Perimeter = _______ ft
Answer:
Perimeter of the ABC Triangle = 21ft.

Explanation:

Length of the side of the ABC Triangle = 7ft
Perimeter of the ABC Triangle = 3 × Side
= 3 × 7ft
= 21ft.

c.

Perimeter = _______ m
Answer:
Perimeter of the ABCD Square = 36m.

Explanation:

Length of the Side of the ABCD Square = 9m
Perimeter of the ABCD Square = 4 × Side
= 4 × 9m
= 36m.

d.

Perimeter = _______ in
Answer:
Perimeter of the ABCDE Pentagon = 36in.

Explanation:

Length of the side of the ABCDE Pentagon = 6in
Perimeter of the ABCDE Pentagon = 6 × Side
= 6 × 6in
= 36in.

Question 2.
Label the unknown side lengths of the rectangle below. Then, find the perimeter of the rectangle.

Perimeter = _______ cm
Answer:
Perimeter of the side of the ABCD Rectangle = 18cm.

Explanation:

Length of the side of the ABCD Rectangle = 7cm
Width of the side of the ABCD Rectangle = 2cm
Perimeter of the side of the ABCD Rectangle = 2 (Length + Width)
= 2 ( 7cm + 2cm )
= 2 × 9cm
= 18cm.

Question 3.
David draws a regular octagon and labels a side length as shown below. Find the perimeter of David’s octagon.

Answer:
Perimeter of the ABCDEFGH Octogen = 36cm.

Explanation:

Length of the side of the ABCDEFGH Octogen = 6cm
Perimeter of the ABCDEFGH Octogen = 6 × Side
= 6 × 6cm
= 36cm.

Question 4.
Paige paints an 8-inch by 9-inch picture for her mom’s birthday. What is the total length of wood that Paige needs to make a frame for the picture?
Answer:
Perimeter of the Paige’s paints = 34inch.

Explanation:
Length of the Paige’s paints = 9inch
Width of the Paige’s paints = 8inch
Perimeter of the Paige’s paints = 2 (Length + Width )
= 2 (9inch + 8inch)
= 2 × 17inch
= 34inch.

Question 5.
Mr. Spooner draws a regular hexagon on the board. One of the sides measures 4 centimeters. Giles and Xander find the perimeter. Their work is shown below. Whose work is correct? Explain your answer.
Giles’s Work
Perimeter = 4 cm + 4 cm + 4 cm + 4 cm + 4 cm + 4 cm
Perimeter = 24 cm

Xander’s Work
Perimeter = 6 × 4 cm
Perimeter = 24 cm
Answer:
Perimeter of the Hexagon = 24 centimeters.
Both Giles and Xander’s work is correct because they got the calculation value correct even though the methodology was different of each.

Explanation:
Length of the side of the Hexagon = 4 centimeters
Perimeter of the Hexagon = 6 × Side
= 6 × 4 centimeters
= 24 centimeters.
Both are correct.

Eureka Math Grade 3 Module 7 Lesson 14 Exit Ticket Answer Key

Travis traces a regular pentagon on his paper. Each side measures 7 centimeters. He also traces a regular hexagon on his paper. Each side of the hexagon measures 5 centimeters. Which shape has a greater perimeter? Show your work.
Answer:
Perimeter of the Pentagon is greater than the Perimeter of the regular hexagon because the measurement value of the Pentagon is more than the measurement value of the regular hexagon.

Explanation:

Length of the side of the Pentagon = 7 centimeters
Perimeter of the Pentagon = 5 × Side
= 5 × 7 centimeters
= 35 centimeters.
Length of the side of the regular hexagon = 5 centimeters
Perimeter of the regular hexagon = 6 × Side
= 6 × 5 centimeters
= 30 centimeters.

Eureka Math Grade 3 Module 7 Lesson 14 Homework Answer Key

Question 1.
Label the unknown side lengths of the regular shapes below. Then, find the perimeter of each shape.
a.

Perimeter = _______ in
Answer:
Perimeter of the ABC Triangle = 16in.

Explanation:

Length of the Side of the ABC Triangle = 4in
Perimeter of the ABC Triangle = Side + Side + Side
= AB + BC + CA
= 4in + 4in + 4in
= 12in + 4in
=16in.

b.

Perimeter = _______ cm

Answer:
Perimeter of the square = 32cm.

Explanation:

Length of the Side of the square = 8cm
Perimeter of the square = 4 × Side
= 4 × 8cm
= 32cm.

c.

Perimeter = _______ m
Answer:
Perimeter of the Octagon = 72m.

Explanation:

Length of the side of the Octagon = 9m
Perimeter of the Octagon = 8 × Side
= 8 × 9m
= 72m.

d.

Perimeter = _______ in
Answer:
Perimeter of the Hexagon = 36in.

Explanation:

Length of the side of the Hexagon = 6in
Perimeter of the Hexagon = 6 × Side
= 6 × 6in
= 36in.

Question 2.
Label the unknown side lengths of the rectangle below. Then, find the perimeter of the rectangle.

Perimeter = _______ cm
Answer:
Perimeter of the rectangle =  2 ( Length + Width )
= 2 ( 9cm + 4cm )
= 2 × 13cm
= 26cm.

Explanation:
Length of the rectangle = 9cm
Width of the rectangle = 4cm
Perimeter of the rectangle =  2 ( Length + Width )
= 2 ( 9cm + 4cm )
= 2 × 13cm
= 26cm.

Question 3.
Roxanne draws a regular pentagon and labels a side length as shown below. Find the perimeter of Roxanne’s pentagon.

Answer:
Perimeter of the Roxanne’s pentagon = 5 × Side
= 5 × 7cm
= 35cm.

Explanation:
Length of the side of the Roxanne’s pentagon = 7cm
Perimeter of the Roxanne’s pentagon = 5 × Side
= 5 × 7cm
= 35cm.

Question 4.
Each side of a square field measures 24 meters. What is the perimeter of the field?
Answer:
Perimeter of the Square field = 4 × Side
= 4 × 24meters
= 96meters.

Explanation:
Length of the side of the Square field = 24 meters
Perimeter of the Square field = 4 × Side
= 4 × 24meters
= 96meters.

Question 5.
What is the perimeter of a rectangular sheet of paper that measures 8 inches by 11 inches?
Answer:
Perimeter of the rectangular sheet of paper = 2 ( Length  + Width )
= 2 (11inches + 8inches)
= 38inches.

Explanation:
Length of the rectangular sheet of paper = 11inches
Width of the rectangular sheet of paper = 8inches
Perimeter of the rectangular sheet of paper = 2 ( Length  + Width )
= 2 (11inches + 8inches)
= 2 × 19inches
= 38inches.

Engage NY Eureka Math 3rd Grade Module 1 Lesson 7 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 7 Problem Set Answer Key

Question 1.
a. Draw an array that shows 6 rows of 2.
b. Write a multiplication sentence where the first factor represents the number of rows.
________ × ________ = ________

Answer:
a.

Explanation:
Drawn an array that shows 6 rows of 2 as 6 × 2 as shown above in the picture.

b. Multiplication sentence where the first factor represents the number of rows as 6 × 2 = 12.

Explanation:
Wrote a multiplication sentence where the first factor represents the number of rows as 6 × 2 = 12.

Question 2.
a. Draw an array that shows 2 rows of 6.
b. Write a multiplication sentence where the first factor represents the number of rows.
____2____ × ____6____ = ___12_____

Answer:
a.

Explanation:
Drawn an array that shows 2 rows of 6 as 2 × 6 as shown above in the picture.

b. Multiplication sentence where the first factor represents the number of rows as 2 × 6 = 12.

Explanation:
Wrote a multiplication sentence where the first factor represents the number of rows as 2 × 6 = 12.

Question 3.
a. Turn your paper to look at the arrays in Problems 1 and 2 in different ways. What is the same and what is different about them?
b. Why are the factors in your multiplication sentences in a different order?

Answer:

a. Same is the result because both have the same value as 12, Different is number of rows and number of columns are not the same,

Explanation:
Turned my paper to look at the arrays in Problems 1 and 2 in different ways.
Is the same because both got same results as 12 and different about them is in problems 1 and 2 the number of rows and number of columns are not the same as one is 6 rows × 2 columns  and in other it is  2 rows × 6 columns.

b. The factors in my multiplication sentences are in a different order because first we are taking 6 rows × 2 columns and in other it is  2 rows × 6 columns.

Explanation:
The factors in my multiplication sentences are in a different order because first we are taking 6 rows X 2 columns  and in other it is  2 rows × 6 columns,
So, rows and columns differ in the both.

Question 4.
Write a multiplication sentence for each expression.
You might skip-count to find the totals.

Explanation:
Wrote a multiplication sentence for each expression as shown above, might have skipped-count to find the totals.

Question 5.
Write and solve multiplication sentences where the second factor represents the size of the row.

Explanation:
Wrote and solved multiplication sentences where the second factor represents the size of the row as
2 × 4 = 8 and 4 × 2= 8.

Question 6.
Ms. Nenadal writes 2 × 7 = 7 × 2 on the board.
Do you agree or disagree? Draw arrays to help explain your thinking.

Answer:
Yes, I agree Ms. Nenadal writings as 2 × 7 = 7 × 2,

Explanation:
Given Ms. Nenadal writes 2 × 7 = 7 × 2 on the board,
Yes, I agree because 2 × 7 = 14 and 7 × 2 = 14,
Drawn arrays to help and explained my thinking as shown above.

Question 7.
Find the missing factor to make each equation true.

Explanation:
Found the missing factor to make each equations true as below
5 × 2 = 2 × 5,
As each equation is true, so the missing factor in 2 X ___ is 5 as LHS is equal to RHS, So 5 X 2 = 2 X 5.

2 × 8 = 8 × 2,
As each equation is true, so the missing factor in ___ X 8 is 2 as LHS is equal to RHS, So 2 X 8 = 8 X 2.

2 × 10 = 10 × 2,
As each equation is true, so the missing factor in ___ X 2  is 10 as LHS is equal to RHS, So 2 X 10 = 10 X 2.

2 × 9 = 9 × 2,
As each equation is true, so the missing factor in 2 × ___ is 9 as LHS is equal to RHS, So 2 × 9 = 9 × 2.

Question 8.
Jada gets 2 new packs of erasers. Each pack has 6 erasers in it.
a. Draw an array to show how many erasers Jada has altogether.
b. Write and solve a multiplication sentence to describe the array.
c. Use the commutative property to write and solve a different multiplication sentence for the array.

Answer:

a.

Jada has 12 erasers altogether,

Explanation:
Given Jada gets 2 new packs of erasers. Each pack has 6 erasers in it.
So, drawn an array to show 2 × 6 = 12 number of erasers
Jada has altogether.

b. Multiplication sentence is 2 × 6 = 12, the array has 12 erasers.

Explanation:
Wrote and solved a multiplication sentence as 2 × 6 = 12 erasers, describing the array as 2 packs in rows and 6 erasers in columns, So 2 packs X 6 erasers = 12 erasers.

c.  Commutative property for multiplication sentence for the array is 2 × 6 = 6 × 2 = 12.

Explanation:
Used the commutative property to write and solve a different multiplication sentence for the array as 2 × 6 = 6 x 2 = 12 erasers.

Eureka Math Grade 3 Module 1 Lesson 7 Exit Ticket Answer Key

Do you agree or disagree with the statement in the box?
Draw arrays and use skip-counting to explain your thinking.

Answer:
Yes, I agree with the statement in the box as 2 x 5 = 5 X 2,

Explanation:
Yes, I agree or disagree with the statement in the box as 2 × 5 = 10 = 5 × 10,
Drawn arrays and used skip-counting as 2 × 5 = 10 = 5 × 10.

Eureka Math Grade 3 Module 1 Lesson 7 Homework Answer Key

Question 1.
a. Draw an array that shows 7 rows of 2.
b. Write a multiplication sentence where the first factor represents the number of rows.
____7____ × ___2____ = ____14____

Explanation:
Drawn an array that shows 7 rows of 2 as 7 × 2 as shown above in the picture.

b. Multiplication sentence where the first factor represents the number of rows as 7 × 2 = 14.

Explanation:
Wrote a multiplication sentence where the first factor represents the number of rows as 7 × 2 = 14.

Question 2.
a. Draw an array that shows 2 rows of 7.
b. Write a multiplication sentence where the first factor represents the number of rows.
____2____ × ___7_____ = ___14____

Explanation:
Drawn an array that shows 2 rows of 7 as 2 × 7 as shown above in the picture.

b. Multiplication sentence where the first factor represents the number of rows as 2 × 7 = 14.

Explanation:
Wrote a multiplication sentence where the first factor represents the number of rows as 2 × 7 = 14.

Question 3.
a. Turn your paper to look at the arrays in Problems 1 and 2 in different ways. What is the same and what is different about them?
b. Why are the factors in your multiplication sentences in a different order?

a. Same is the result because both have the same value as 14,
Different is number of rows and number of columns are not the same,

Explanation:
Turned my paper to look at the arrays in Problems 1 and 2 in different ways.
Is the same because both got same results as 14 and different about them is in problems 1 and 2 the number of rows and number of columns are not the same as one is 7 rows × 2 columns  and in other it is  2 rows × 7 columns.

b. The factors in my multiplication sentences are in a different order because first we are taking 7 rows × 2 columns and in other it is  2 rows × 7 columns.

Explanation:
The factors in my multiplication sentences are in a different order because first we are taking 7 rows X 2 columns  and in other it is  2 rows × 7 columns,
So, rows and columns differ in the both.

Question 4.
Write a multiplication sentence to match the number of groups.
Skip-count to find the totals. The first one is done for you.


Explanation:
Wrote a multiplication sentence to match the number of groups as shown above, might have skipped-count to find the totals.

Question 5.
Write and solve multiplication sentences where the second factor represents the size of the row.


Explanation:
Wrote and solved multiplication sentences where the second factor represents the size of the row as
2 × 6 = 12 and 6 × 2= 12.

Question 6.
Angel writes 2 × 8 = 8 × 2 in his notebook.
Do you agree or disagree? Draw arrays to help explain your thinking.

Explanation:
Given Angel writes 2 × 8 = 8 × 2 in his notebook,
Yes, I agree because 2 × 8 = 16 and 8 x 2 = 16,
Drawn arrays to help and explained my thinking as shown above.

Question 7.
Find the missing factor to make each equation true.


Explanation:
Found the missing factor to make each equations true as below
2 × 6 = 6 × 2,
As each equation is true, so the missing factor in 6 × ___ is 2 as LHS is equal to RHS, So 2 × 6 = 6 × 2.

7 × 2 = 2 × 7,
As each equation is true, so the missing factor in ___ × 2 is 7 as LHS is equal to RHS, So 7 × 2 = 2 × 7.

9 × 2 = 2 × 9,
As each equation is true, so the missing factor in ___ × 9 is 2 as LHS is equal to RHS, So 9 × 2 = 2 × 9.

2 × 10 = 10 × 2,
As each equation is true, so the missing factor in 2 × ___ is 10 as LHS is equal to RHS, So 2 × 10 = 10 × 2.

Question 8.
Tamia buys 2 bags of candy.
Each bag has 7 pieces of candy in it.
a. Draw an array to show how many pieces of candy Tamia has altogether.
b. Write and solve a multiplication sentence to describe the array.
c. Use the commutative property to write and solve a different multiplication sentence for the array.

a.

Tamia has 14 candy’s altogether,

Explanation:
Given Tamia buys 2 bags of candy. Each bag has 7 pieces of candy in it.
So, drawn an array to show 2 × 7 = 14 number of candy’s Tamia has altogether.

b. Multiplication sentence is 2 × 7 = 14, the array has 14 candy’s.

Explanation:
Wrote and solved a multiplication sentence as 2 × 7 = 14 candy’s, describing the array as 2 bags in rows and 7 candy’s in columns, So 2 bags X 7 candy’s = 14 candy’s.

c.  Commutative property for multiplication sentence for the array is 2 × 7 = 7 × 2 = 14.

Explanation:
Used the commutative property to write and solve a different multiplication sentence for the array as 2 × 7 = 7 x 2 = 14 candy’s.

Engage NY Eureka Math 3rd Grade Module 7 Lesson 13 Answer Key

Eureka Math Grade 3 Module 7 Lesson 13 Pattern Sheet Answer Key

Multiply.

multiply by 8 (1–5)
Answer:

Explanation:
8 × 1 = 8
8 × 2 = 16
8 × 3 = 24
8 × 4 = 32
8 × 5 = 40.

Eureka Math Grade 3 Module 7 Lesson 13 Problem Set Answer Key

Question 1.
Find the perimeter of the following shapes.
a.

P = 3 in + 8 in + 3 in + 8 in
= _________ in
Answer:
Perimeter of the ABCD Rectangle = Side + Side + Side + Side
= AB + BC +CD + DA
= 8in + 3in + 8in + 3in
= 22in.

Explanation:

Length of the AB side of the Rectangle = 8in
Length of the BC side of the Rectangle = 3in
Length of the CA side of the Rectangle = 8in
Length of the DA side of the Rectangle = 3in
Perimeter of the ABCD Rectangle = Side + Side + Side + Side
= AB + BC +CD + DA
= 8in + 3in + 8in + 3in
= 11in + 8in + 3in
= 19in + 3in
= 22in.

b.

P = ____ cm + ____ cm + ____ cm + ____ cm
= _________ cm
Answer:
Perimeter of the ABCD Square = Side + Side + Side + Side
= AB + BC +CD + DA
= 4cm + 4cm + 4cm + 4cm
= 16cm.

Explanation:

Length of the AB side of the ABCD Square = 4cm
Length of the BC side of the ABCD Square = 4cm
Length of the CD side of the ABCD Square = 4cm
Length of the DA side of the ABCD Square = 4cm
Perimeter of the ABCD Square = Side + Side + Side + Side
= AB + BC +CD + DA
= 4cm + 4cm + 4cm + 4cm
= 8cm + 4cm + 4cm
= 12cm + 4cm
= 16cm.

c.

P = ____ cm + ____ cm + ____ cm
= _________ cm
Answer:
Perimeter of the ABC Triangle = Side + Side + Side
= AB + BC + CA
= 9cm + 11cm + 6cm
= 26cm.

Explanation:

Length of the AB side of the ABC Triangle =  9cm
Length of the BC side of the ABC Triangle = 11cm
Length of the CA side of the ABC Triangle = 6cm
Perimeter of the ABC Triangle = Side + Side + Side
= AB + BC + CA
= 9cm + 11cm + 6cm
= 20cm + 6cm
= 26cm.

d.

P = ____ m + ____ m + ____ m + ____ m
= _________ m
Answer:
Perimeter of the ABCD Trapezium = Side + Side + Side + Side
= AB + BC + CD + DA
= 15m + 9m + 5m + 7m
= 36m.

Explanation:

Length of the AB side of the ABCD Trapezium = 15cm
Length of the BC side of the ABCD Trapezium = 9cm
Length of the CD side of the ABCD Trapezium = 5cm
Length of the DA side of the ABCD Trapezium = 7cm
Perimeter of the ABCD Trapezium = Side + Side + Side + Side
= AB + BC + CD + DA
= 15m + 9m + 5m + 7m
= 24m + 5m + 7m
= 29m + 7m
= 36m.

e.

P = ____ in + ____ in + ____ in + ____ in + ____ in
= _________ in
Answer:
Perimeter of the ABCDE Pentagon = Side + Side + Side + Side + Side
= AB + BC + CD + DE + EA
= 9in + 2in + 2in + 9in + 3in
= 25in.

Explanation:

Length of the AB side of the ABCDE Pentagon = 9in
Length of the BC side of the ABCDE Pentagon = 2in
Length of the CD side of the ABCDE Pentagon = 2in
Length of the DE side of the ABCDE Pentagon = 9in
Length of the EA side of the ABCDE Pentagon = 3in
Perimeter of the ABCDE Pentagon = Side + Side + Side + Side + Side
= AB + BC + CD + DE + EA
= 9in + 2in + 2in + 9in + 3in
= 11in + 2in + 9in + 3in
= 13in + 9in + 3in
= 22in + 3in
= 25in.

Question 2.
Alan’s rectangular swimming pool is 10 meters long and 16 meters wide. What is the perimeter?

Answer:
Perimeter of the ABCD Rectangle = Side + Side + Side + Side
= AB + BC + CD + DA
= 16m + 10m + 16m + 10m
= 52m.

Explanation:

Length of the AB side of the ABCD Rectangle = 16m
Length of the BC side of the ABCD Rectangle = 10m
Length of the CA side of the ABCD Rectangle = 16m
Length of the DA side of the ABCD Rectangle = 10m
Perimeter of the ABCD Rectangle = Side + Side + Side + Side
= AB + BC + CD + DA
= 16m + 10m + 16m + 10m
= 26m + 16m + 10m
= 42m + 10m
= 52m.

Question 3.
Lila measures each side of the shape below.

a. What is the perimeter of the shape?
b. Lila says the shape is a pentagon. Is she correct? Explain why or why not.
Answer:
a. The perimeter of the ABCDE shape = Side + Side + Side + Side + Side
= AB + BC + CD + DE + EA
= 9in + 6in + 3in + 2in + 4in
= 24in.

b. Lila is correct because Pentagon is a figure which has five sides in it and her figure is a five sided shape.

Explanation:

a. The perimeter of the ABCDE shape = ???
Length of the AB side of the ABCDE shape = 9in
Length of the BC side of the ABCDE shape = 6in
Length of the CD side of the ABCDE shape = 3in
Length of the DE side of the ABCDE shape = 2in
Length of the EA side of the ABCDE shape = 4in
The perimeter of the ABCDE shape = Side + Side + Side + Side + Side
= AB + BC + CD + DE + EA
= 9in + 6in + 3in + 2in + 4in
= 15in + 3in + 2in + 4in
= 18in + 2in + 4in
= 20in + 4in
= 24in.

b. Pentagon is a figure which has  five side in it. Lila is correct.

Eureka Math Grade 3 Module 7 Lesson 13 Exit Ticket Answer Key

Which shape below has the greater perimeter? Explain your answer.

Answer:
Perimeter of shape A = 14in
Perimeter of shape B = 15in
Perimeter of shape B is greater than the Perimeter of shape A because the measurement value of Perimeter of shape B is more than the the measurement value of Perimeter of shape A.

Explanation:

Perimeter of shape A = Side + Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE + EF + FG + GA
= 2in + 2in + 2in + 2in + 2in + 2in + 2in
= 4in + 2in + 2in + 2in + 2in + 2in
= 6in + 2in + 2in + 2in + 2in
= 8in + 2in + 2in + 2in
= 10in + 2in + 2in
= 12in + 2in
= 14in.
Perimeter of shape B = Side + Side + Side + Side + Side
= HI + IJ + JK + KL +LH
= 4in + 2in + 2in + 4in + 3in
= 6in + 2in + 4in + 3in
= 8in +  4in + 3in
= 12in + 3in
= 15in.

Eureka Math Grade 3 Module 7 Lesson 13 Homework Answer Key

Question 1.
Find the perimeters of the shapes below. Include the units in your equations. Match the letter inside each shape to its perimeter to solve the riddle. The first one has been done for you.

What kind of meals do math teachers eat?

Answer:
Perimeter of q Triangle shape = 21in.
Perimeter of r Pentagon shape = 36ft.
Perimeter of s Parallelogram shape = 24cm.
Perimeter of a Trapezium shape = 28yd.
Perimeter of m rhombus shape = 16in.
Perimeter of e Rectangular shape = 26cm.
Perimeter of u quadrilateral shape = 20m.
Perimeter of l Pentagon shape = 15m.

Square meals kind of meals  math teachers eats.

Explanation:

 

Perimeter of q Triangle shape = Side + Side + Side
= 7in + 7in + 7in
= 14in + 7in
= 21in.

Perimeter of r Pentagon shape = Side + Side + Side + Side + Side
= 6ft + 9ft + 6ft + 6ft + 9ft
= 15ft + 6ft + 6ft + 9ft
= 21ft + 6ft + 9ft
= 27ft + 9ft
= 36ft.

Perimeter of s Parallelogram shape = Side + Side + Side + Side
= 7cm + 5cm + 7cm + 5cm
= 12cm + 7cm + 5cm
= 19cm + 5cm
= 24cm.

Perimeter of a Trapezium shape = Side + Side + Side + Side
= 9yd + 7yd + 5yd + 7yd
= 16yd + 5yd + 7yd
= 21yd +  7yd
= 28yd.

Perimeter of m rhombus shape = Side + Side + Side + Side
= 4in + 4in + 4in + 4in
= 8in  + 4in + 4in
= 12in + 4in
= 16in.

Perimeter of e Rectangular shape = Side + Side + Side + Side
= 8cm + 5cm + 8cm + 5cm
= 13cm + 8cm + 5cm
= 21cm + 5cm
= 26cm.

Perimeter of u quadrilateral shape = Side + Side + Side + Side
= 6m + 4m + 7m + 3m
= 10m + 7m + 3m
= 17m + 3m
= 20m.

Perimeter of l Pentagon shape= Side + Side + Side + Side + Side
= 4m + 2m + 2m + 4m + 3m
= 6m + 2m + 4m + 3m
= 8m + 4m + 3m
= 12m + 3m
= 15m.

Question 2.
Alicia’s rectangular garden is 33 feet long and 47 feet wide. What is the perimeter of Alicia’s garden?

Answer:
Perimeter of Alicia’s rectangular garden = 160ft.

Explanation:

Length of the ABCD Alicia’s rectangular garden = 33ft
Width of the ABCD Alicia’s rectangular garden = 47ft
Perimeter of ABCD Alicia’s rectangular garden = 2 (Length + Width)
= 2 ( 33ft + 47ft )
= 2 × 80ft
= 160ft.

Question 3.
Jaque’s measured the side lengths of the shape below.

a. Find the perimeter of Jaques’s shape.
b. Jaques says his shape is an octagon. Is he right? Why or why not?
Answer:
a. Perimeter of Jaques’s shape =  Side + Side + Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE + EF + FG + GH +HA
= 7 in + 3 in + 4 in+ 5 in + 4 in + 3 in + 2 in +5 in
= 33in.

Explanation:
a.

Perimeter of Jaques’s shape =  Side + Side + Side + Side + Side + Side + Side + Side
= AB + BC + CD + DE + EF + FG + GH +HA
= 7 in + 3 in + 4 in+ 5 in + 4 in + 3 in + 2 in +5 in
= 10 in + 4 in + 5 in + 4 in + 3 in + 2 in +5 in
= 14 in + 5 in + 4 in + 3 in + 2 in +5 in
= 19 in + 4 in + 3 in + 2 in +5 in
= 23 in + 3 in + 2 in +5 in
= 26 in + 2 in+5 in
= 28 in + 5 in
= 33 in.

b.  Yes, he is correct. Jaques says his shape is an octagon because in geometry , an octagon is an eight-sided  polygon or 8-gon. His shape has eight sides.

Explanation:

His shape has eight sides. In geometry , an octagon is an eight-sided  polygon or 8-gon.

Engage NY Eureka Math 3rd Grade Module 1 Lesson 6 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 6 Problem Set Answer Key

Question 1.
Rick puts 15 tennis balls into cans. Each can holds 3 balls.
Circle groups of 3 to show the balls in each can.

Rick needs ____5___ cans.
___5___ × 3 = 15
15 ÷ 3 = __5____

Answer:
Rick needs 5 cans,

Explanation:
Given Rick puts 15 tennis balls into cans and each can holds 3 balls. Circled groups of 3 to show the balls in each can.
Rick needs 15 ÷ 3 = 5, 5 × 3 = 15,
Therefore Rick needs 5 cans.

Question 2.
Rick uses 15 tennis balls to make 5 equal groups.
Draw to show how many tennis balls are in each group.
There are ____3___ tennis balls in each group.
5 × __3____ = 15
15 ÷ 5 = ___3___

Answer:
There are 3 tennis balls in each group.

Explanation:
Given Rick uses 15 tennis balls to make 5 equal groups.
Drawn circles to show 3 number of tennis balls are in each group as 15 ÷ 5 = 3, 5 × 3 = 15,
Therefore, there are 3 tennis balls in each group.

Question 3.
Use an array to model Problem 1.
a. ___5___ × 3 = 15
15 ÷ 3 = ___5___
The number in the blanks represents
_____5 groups___________________.

Explanation:
Used an array to model Problem 1 as
5 × 3 = 15 or 15 ÷ 3 = 5,
The number in the blanks represents groups as
5 groups of balls.

b. 5 × __3____ = 15
15 ÷ 5 = __3____
The number in the blanks represents
____________3 balls in each group________.

Answer:
The number in the blanks represents 3 balls in each group,

Explanation:
Rick uses 15 tennis balls to make 5 equal groups,
15 ÷ 5 = 3 or 5 × 3= 15, The number in the blanks
represents 3 balls in each group.

Question 4.
Deena makes 21 jars of tomato sauce. She puts 7 jars in
each box to sell at the market. How many boxes does Deena need?
21 ÷ 7 = __3____
___3___ × 7 = 21
What is the meaning of the unknown factor and quotient?
_________3______________

Answer:
Deena needs 3 boxes,

Explanation:
Given Deena makes 21 jars of tomato sauce and she puts 7 jars in each box to sell at the market.
So number of boxes Deena needs are 21 ÷ 7 = 3, (3 × 7 = 21)
Therefore, Deena needs 3 boxes.

Question 5.
The teacher gives the equation 4 × __3__ = 12. Charlie finds the answer by writing and solving 12 ÷ 4 = __3__.
Explain why Charlie’s method works.

Answer:
Charlie method is 12 ÷ 4 = 3,

Explanation:
Given The teacher gives the equation 4 × ____ = 12.
Charlie finds the answer by writing and solving as
12 ÷ 4 = 3,
because multiplication and division are closely related,
given that division is the inverse operation of multiplication.
When we divide, we look to separate into equal groups,
while multiplication involves joining equal groups.
If we divide this product by one of the factors,
we get the other factor as a result. So Charlie uses division method to solve missing factor in the given equation as 3. So the equation is 4 X 3 = 12.

Question 6.
The blanks in Problem 5 represent the size of the groups.
Draw an array to represent the equations.

Explanation:
Drawn an array that represents the equations as  4 × 3 = 12.

Eureka Math Grade 3 Module 1 Lesson 6 Exit Ticket Answer Key

Cesar arranges 12 notecards into rows of 6 for his presentation.
Draw an array to represent the problem.
12 ÷ 6 = ___2_____
____2____ × 6 = 12
What do the unknown factor and quotient represent?
___2 notecards in each row____

Explanation:
Given Cesar arranges 12 notecards into rows of 6 for his presentation.
Drawn an array to represent the problem as 12 ÷ 6 = 2, 2 × 6 = 12, means 6 rows  and 2 columns as shown above, The unknown factor and quotient 2
represents 2 notecards in each row.

Eureka Math Grade 3 Module 1 Lesson 6 Homework Answer Key

Question 1.
Mr. Hannigan puts 12 pencils into boxes. Each box holds 4 pencils.
Circle groups of 4 to show the pencils in each box.

Mr. Hannigan needs ___3____ boxes.
___3___ × 4 = 12
12 ÷ 4 = __3____

Mr. Hannigan needs 3 boxes,

Explanation:
Given Mr. Hannigan puts 12 pencils into boxes.
Each box holds 4 pencils.
Circled groups of 4 to show the pencils in each box as shown above, So Mr. Hannigan needs 12 ÷ 4 = 3 boxes,
( 3 x 4 = 12).

Question 2.
Mr. Hannigan places 12 pencils into 3 equal groups.
Draw to show how many pencils are in each group.
There are ___4____ pencils in each group.
3 × ___4___ = 12
12 ÷ 3 = ___4___

There are 4 pencils in each group,

Explanation:
Given Mr. Hannigan places 12 pencils into 3 equal groups,
Drawn to show 4 number of pencils are in each group as 12 ÷ 3 = 4, 3 × 4 = 12, Therefore 4 pencils are there in each group.

Question 3.
Use an array to model Problem 1.
a. ___3___ × 4 = 12
12 ÷ 4 = ___3___
The number in the blanks represents
___________3 groups______________.
a.

The number in the blanks represents 3 groups,

Explanation:
Used an array to model Problem 1 as 3 x 4 = 12, or 12 ÷ 4 = 3, therefore the number in the blanks represents 3 groups.

b. 3 × ___4___ = 12
12 ÷ 3 = __4____
The number in the blanks represents
___________4 pencils in each group_____________.

Answer:

The number in the blanks represents 4 pencils in each group,

Explanation:
Given Mr. Hannigan places 12 pencils into 3 equal groups.
12 ÷ 3 = 4 or 3 × 4= 12, The number in the blanks represents 4 pencils in each group.

Question 4.
Judy washes 24 dishes. She then dries and stacks the dishes equally into 4 piles. How many dishes are in each pile?
24 ÷ 4 = ___6____
4 × ___6_____ = 24
What is the meaning of the unknown factor and quotient?
__________6 dishes are in each pile____________________

Answer:
There are 6 dishes in each pile,

Explanation:
Given Judy washes 24 dishes and she then dries and stacks the dishes equally into 4 piles.
So number of dishes in each pile are 24 ÷ 4 = 6, or 4 × 6 = 24, The meaning of the unknown factor and quotient is 6 dishes are there in each pile.

Question 5.
Nate solves the equation _____ × 5 = 15 by writing and solving 15 ÷ 5 = ____. Explain why Nate’s method works.

Answer:
Nate solves the equation as 3 × 5 = 15,

Explanation:
Given Nate solves the equation _____ × 5 = 15 by writing and solving as 15 ÷ 5 = 3, Nate’s method is correct because in the given equation ___ X 5 = 15 , we bring 5 to the other side it becomes as
15 ÷ 5 = 3 now we check as 3 × 5 it becomes 15 only.
So Nate’s method work.

Question 6.
The blanks in Problem 5 represent the number of groups.
Draw an array to represent the equations.

Explanation:
Drawn an array to represent the equations as
3 × 5 = 15 or 15 ÷ 5 = 3 as shown in the picture above.

Engage NY Eureka Math 3rd Grade Module 1 Lesson 1 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 1 Problem Set Answer Key

Question 1.
Fill in the blanks to make true statements.

a. 3 groups of five = ___15______
3 fives = ____15_____
3 × 5 = ___15______

3 groups of five = 15,
3 fives = 15,
3 × 5 = 15,

Explanation:
Given expressions as
3 groups of five means 3 × 5 = 15,
3 fives = 3 × 5 = 15 and
3 × 5 = 15 as 3 is multiplied by 5 we get 15.

b. 3 + 3 + 3 + 3 + 3 = __15___
5 groups of three = ___15______
5 × 3 = ____15_____

3 + 3 + 3 + 3 + 3 = 15,
5 groups of three = 15,
5 × 3 = 15,

Explanation:
Given expressions as
3 + 3 + 3 + 3 + 3 = 15 as 3 is added 5 times,
5 groups of three = 15 means 5  × 3 = 15,
5 × 3 = 15 as 5 is multiplied by 3 we get 15.

c. 6 + 6 + 6 + 6 = ___24________
____4___ groups of six = ___24_______
4 × __6____ = ____24______

6 + 6 + 6 + 6 = 24,
4 groups of six = 24,
4 × 6 = 24,

Explanation:
Given expressions as
6 + 6 + 6 + 6 = 24 as 6 is added 4 times,
4 groups of six = 24 means 4  × 6 = 24,
4 × 6 = 24 as 4 is multiplied by 6 we get 24.

d. 4 +__4__ + __4__ + __4__ + _4___ + __4__ = ___24______
6 groups of ____4____ = ____24_______
6 × __4____ = ____24______

4 + 4 + 4 + 4 + 4 + 4 = 24,
6 groups of 4 = 24,
6 × 4 = 24,

Explanation:
Given expressions as
4 + ___ + ___ +___ +___ +___ =  means by seeing the picture
4 is added 6 times as 4 + 4 + 4 + 4 + 4 + 4 we get 24,
6 groups of ______ =  means again by seeing the picture
6 × 4 we get 24, So 6 groups of 4 = 24,
6 × 4 = 24 as 6 is multiplied by 4 we get 24.

Question 2.
The picture below shows 2 groups of apples. Does the picture show 2 × 3? Explain why or why not.

No, the picture does not show 2 × 3,

Explanation:
Given the picture below shows 2 groups of apples but in the 2 groups the number of apples are not the same as in group -1 we have 3 apples, group 2 we have 2 apples in total there are 5 apples and 2 × 3 = 6 whose value is not same as shown in the picture.
So, the picture does not show 2 × 3.

Question 3.
Draw a picture to show 2 × 3 = 6.

Explanation:
Drawn a picture above to show 2 × 3 = 6 .

Question 4.
Caroline, Brian, and Marta share a box of chocolates. They each get the same amount. Circle the chocolates below to show 3 groups of 4. Then, write a repeated addition sentence and a multiplication sentence to represent the picture.

Addition sentence is 4 + 4 + 4 = 12,
Multiplication sentence is 3 × 4 = 12,

Explanation:
Given Caroline, Brian, and Marta share a box of chocolates and they each get the same amount as 12 ÷ 3 = 4 each.
Circled the chocolates above to show 3 groups of 4 and wrote a repeated addition sentence as  4 + 4 + 4 = 12 and a multiplication sentence to represent the picture as 3 × 4 = 12.

Eureka Math Grade 3 Module 1 Lesson 1 Exit Ticket Answer Key

Question 1.
The picture below shows 4 groups of 2 slices of watermelon.
Fill in the blanks to make true repeated addition and multiplication sentences that represent the picture.

True repeated addition is 2 + 2 + 2 + 2 = 8,
Multiplication sentences is 4 X 2 = 8.

Explanation:
Given the picture above shows 4 groups of 2 slices of watermelon.
Filled in the blanks to make true repeated addition as 2 + 2 + 2 + 2 = 8 and multiplication sentences that represent the picture as 4 X 2 = 8.

Question 2.
Draw a picture to show 3 + 3 + 3 = 9. Then, write a multiplication sentence to represent the picture.

Addition sentence is  3 + 3 + 3 = 9,
Multiplication sentences is 3 X 3 = 9.

Explanation:
Drawn a picture above as  3 groups of 3 birds  to show 3 + 3 + 3 = 9 and writing a multiplication sentence to represent the picture as 3 X 3 = 9 birds in total.

Eureka Math Grade 3 Module 1 Lesson 1 Homework Answer Key

Question 1.
Fill in the blanks to make true statements.

a. 4 groups of five = ____20_____
4 fives = ___20______
4 × 5 = ____20_____

4 groups of five = 20,
4 fives = 20,
4 × 5 = 20,

Explanation:
Given expressions as
4 groups of five means 4 × 5 = 20,
4 fives = 4 × 5 = 20 and
4 × 5 = 20 as 4 is multiplied by 5 we get 20.

b. 5 groups of four = _________
5 fours = _________
5 × 4 = _________

5 groups of four = 20,
5 fours = 20,
5 × 4 = 20,

Explanation:
Given expressions as
5 groups of four means 5 × 4 = 20,
5 fours = 5 x 4 = 20 and
5 × 4 = 20 as 5 is multiplied by 4 we get 20.

c. 6 + 6 + 6 = ___18________
____3___ groups of six = ____18______
3 × __6____ = ____18______

6 + 6 + 6  = 18,
3 groups of 6 = 18,
3 × 6 = 18,

Explanation:
Given expressions as 6 + 6 + 6 = 18 as
6 is added 3 times we get 18,
_____ groups of six =  means again by seeing the picture
3 groups of 6 =  3 × 6 we get 18, So 3 groups of 6 = 18,
3 × 6 = 18 as 3 is multiplied by 6 we get 18.

d. 3 + __3__ + __3__ + __3__ + __3__ + __3__ = ___18____
6 groups of ____3____ = _____18______
6 × __3__ = ____18______

3 + 3 + 3 + 3 + 3 + 3 = 18,
6 groups of 3 = 18,
6 × 3 = 18,

Explanation:
Given expressions as
3 + ___ + ___ +___ +___ +___ =  means by seeing the picture
3 is added 6 times as 3 + 3 + 3 + 3 + 3 + 3 we get 18,
6 groups of ______ =  means again by seeing the picture
6 × 3 we get 18, So 6 groups of 3 = 18,
6 × 3 = 18 as 6 is multiplied by 3 we get 18.

Question 2.
The picture below shows 3 groups of hot dogs. Does the picture show 3 × 3? Explain why or why not.

Yes, the picture shows 3 × 3,

Explanation:
Given the picture below shows 3 groups of hot dogs,
In the 3 groups the number of hot dogs are the same as 3, So in total we have 3 X 3 = 9 hotdogs in the picture which matches with the given equation as 3 × 3 whose value is also same 9, So, the picture shows 3 × 3.

Question 3.
Draw a picture to show 4 × 2 = 8,

Explanation:
Drawn picture of flowers as shown above to show 4 × 2 = 8.

Question 4.
Circle the pencils below to show 3 groups of 6. Write a repeated addition and a multiplication sentence to represent the picture.


Repeated addition sentence is 6 + 6 + 6 = 18,
Multiplication sentence is 3 X 6 = 18,

Explanation:
Circled the pencils above to show 3 groups of 6.
Wrote a repeated addition as 6 + 6 + 6 = 18 and a multiplication sentence as 3 X 6 = 18 to represent the picture.