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Take the help of the Worksheet on Multiplying Binomials to better understand the principles of large numbers multiplication and multiplication of binomials. Practice the Problems in Multiplying Binomials Worksheet PDF meticulously and solve binomial multiplication numbers fastly in your exams. Math Students can enhance their math skills by answering the questions from the Multiplication of Binomials Worksheet. This Math Worksheet on Binomial Multiplication can be a great thing for teaching your kids about Multiplication Facts and to know exactly what they need to learn.

Also, Refer:

• Worksheet on Multiplying Monomials
• Worksheet on Multiplying Monomial and Binomial

Multiplying Binomials Worksheet with Answers

I. Multiply a binomial by a binomial
(i) (s + 7) and (3s + 5)
(ii) (6q – 5) and (q + 2)
(iii) (n – 2) and (n + 4)
(iv)(3p + 8 ) and (p – 6)
(v) (x – 3) and (2x+ 2)
(vi) (m – n) and (m + 2n)

Solution:

(i) Given binomials are  (s + 7) and (3s + 5)
=s(3s) +5(s) + 7(3s) +5(7)
=3s²+5s+21s+35
=3s²+26s+35
Hence, By multiplying the binomials (s + 7) and (3s + 5) we get 3s²+26s+35.
(ii) Given binomials are (6q – 5) and (q + 2)
=6q(q) +2(6q) -5(q) -5(2)
=6q²+12q-5q-10
=6q²+7q-10
Hence, By multiplying the binomials (6q – 5) and (q + 2) we get 6q²+7q-10.
(iii) Given binomials are (n – 2) and (n + 4)
=n(n) + n(4n) -2(n) -2(4)
=n²+4n²-2n-8
=5n²-2n-8
Hence, By multiplying the binomials(n – 2) and (n + 4) we get 5n²-2n-8.
(iv) Given binomials are (3p + 8 ) and (p – 6)
=3p(p)-6(3p) + 8p -48
=3p²-18p+8p-48
=3p²-10p-48
Hence, By multiplying the binomials (3p + 8 ) and (p – 6) we get 3p²-10p-48.

(v) Given binomials are (x – 3) and (2x+ 2)
=x(2x) + x(2) -3(2x)-3(2)
=2x²+x²-6x-6
=3x²-6x-6
Hence, By multiplying the binomials (x – 3) and (2x+ 2) we get 3x²-6x-6.
(vi) Given binomials are (m – n) and (m + 2n)
=m(m)+m(2n)-n(m)-n(2n)
=m²+2mn-nm-2n²
=m²+mn-2n²
Hence, By multiplying the binomials (m – n) and (m + 2n) we get m²+mn-2n².

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II. Multiply the following binomials:
(i) (mx – ny) (mx + ny)
(ii) (a + 4) by (b + 7)
(iii) (20 – xy) by (xy + 4)
(iv) (xy + 5) by (2xy + 7)
(v) (ab + bc) by (ab – bc)
(vi) (10 – 4x) by (4 – 2y)

Solution:

(i) Given binomials are (mx – ny) (mx + ny)
=mx(mx)+ny(mx)-ny(mx)-ny(ny)
=m²x²+mnxy-mnxy-n²y²
=m²x²-n²y²
Therefore, By multiplying the binomials (mx – ny) (mx + ny) we get m2x2-n2y2.
(ii) Given binomials are (a + 4) by (b + 7)
=a(b)+a(7)+4(b)+4(7)
=ab+7a+4b+28
Therefore, By multiplying the binomials (a + 4) by (b + 7) we get ab+7a+4b+28.
(iii) Given binomials are (20 – xy) by (xy + 4)
=20(xy)+20(4)-xy(xy)-xy(4)
=20xy+80-x²y²-4xy
=16xy-x²y²+80
Therefore, By multiplying the binomials (20 – xy) by (xy + 4) we get 16xy-x2y2+80.
(iv) Given binomials are (xy + 5) by (2xy + 7)
=xy(2xy)+xy(7)+5(2xy)+5(7)
=2x²y²+7xy+10xy+35
=2x²y²+17xy+35
Therefore, By multiplying the binomials (xy + 5) by (2xy + 7) we get 2x2y2+17xy+35.
(v) Given binomials are (ab + bc) by (ab – bc)
=ab(ab)+ab(-bc)+bc(ab)+bc(-bc)
=a²b²-ab²c+ab²c-b²c²
=a²b²-b²c²
Therefore, By multiplying the binomials (ab + bc) by (ab – bc) we get a²b²-b²c².
(vi) Given binomials are (10 – 4x) by (4 – 2y)
=10(4)+10(-2y)-4x(4)-4x(-2y)
=40-20y-16x+8xy
Therefore, By multiplying the binomials (10 – 4x) by (4 – 2y) we get 40-20y-16x+8xy.

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III. Find the product of the binomials
(i) (m + 10) × (m + 5)
(ii) (x² – 1) × (2x + 3)
(iii) (1 – a³ ) × (a² + 2)
(iv) (m² + 4) × (m – 3)
(v) (c² + 3) × (c² – 5)
(vi) (3a + 5b) × (a + 2b)

Solution:

(i) Given binomials are (m + 10) × (m + 5)
=m(m) +m(5)+10(m)+10(5)
=m²+5m+10m+50
=m²+15m+50
Hence, By multiplying the binomials (m + 10) × (m + 5) we get m²+15m+50.
(ii) Given binomials are (x² – 1) × (2x + 3)
=x²(2x)+x²(3)-1(2x)-1(3)
=2x³+3x²-2x-3
Hence, By multiplying the binomials (x² – 1) × (x + 3) we get 2x³+3x²-2x-3.
(iii) Given binomials are (1 – a³ ) × (a² + 2)
=1(a²)+1(2)-a³(a²)-a³(2)
=a²+2-a⁵-2a³
Therefore, By multiplying the binomials (1 – a³ ) × (a² + 2) we get a²+2-a⁵-2a³.
(iv) Given binomials are (m² + 4) × (m – 3)
=m²(m)+m²(-3) +4(m) +4(-3)
=m³-3m²+4m-12
Therefore, By multiplying the binomials (m² + 4) × (m – 3) we get m³-3m²+4m-12.
(v) Given binomials are (c² + 3) × (c² – 5)
=c²(c²)+c²(-5)+3(c²)+3(-5)
=c⁴-5c²+3c²-15
=c⁴-2c²-15
Hence, By multiplying the binomials (c² + 3) × (c² – 5) we get c⁴-2c²-15.
(vi) Given (3a + 5b) × (a + 2b)
=3a(a) +3a(2b) +5b(a) +5b(2b)
=3a²+6ab+5ab+10b²
=3a²+11ab+10b²
Therefore, By multiplying the binomials (3a + 5b) × (a + 2b) we get 3a²+11ab+10b².

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In this article, students will learn about units of volume. Volume is the term that measures capacity and it mostly defines the volume of liquids. Students should know that the terms mass and volume are two different quantities. Let us learn the units of volume we used while measuring the quantities. Like Area measurement, a volume measurement also requires a unit.

Math Students will get familiar with all the units of volume we measured from this page. Also, it helps kids to focus on the concept of Volume of Cubes and Cuboids, conversion of units of volume and capacity, etc. & solve the problems on units of volume easily. Moreover, you can practice consistently by taking the help of si metric units of volume measurements examples.

What is Volume?

The volume is defined as the amount of space occupied by any three-dimensional object and volume measures capacity. The unit of volume is basically a unit for measuring any capacity of an object. To calculate the volume, we use a formula that makes students easy while calculating or finding any volume or capacity. The Formula of Volume that we use is as below:

Volume = Length × Width × Height

Volume is always written in ‘cubic’ units because it has three dimensions length, width, and height.

Units of Volume

A unit of volume is the unit measurement of measuring any volume or capacity in larger units, the extent of an object or space in three dimensions. Units of volume are used to specify or define the volume in bulk or fluids. For example, sugar, water, and flour.

In units of volume, mostly used unit terms are in cubic meters, liters, or milliliters. One cubic centimeter is written as 1 cu cm or 1 cc or 1 cm³ and each cube is called a unit cube. Standard unit systems and metric systems are the two measuring unit systems used to measure volume in units.

Relationship Between Various Units of Volume

• 1 m = 100 cm
• 1 cm = 10 mm = 1 cu cm = 10 mm × 10 mm = 1000 cu mm
• 1 cu m = 100 cm × 100 cm × 100 cm = 1000000 cu cm

List of SI Units of Volume

Volume measures in a three-dimensional space that occupies and is measured in cubic units. In the SI unit of measurement, we have different types of volume. However, the most used SI unit of volume is a liter and the metric unit of volume is used to derive large or small volume quantities. Milliliter (ml) is also one of the unit measurements of volume. Let us see some SI Units of Volume and their common measurement of capacity.

Common Units of Volume

1000 cubic millimeters (mm³) = 1 cubic centimeter (cm³) 1 cubic centimeter (cm³) = 1 milliliter (mL) 10 milliliters (mL) = 1 centiliter (cL) 10 centiliter (cL) = 1 deciliter (dL) 1000 cubic centimeter (cm³ )= 1 cubic decimeter (dm³) 1 cubic decimeter (dm³) = 1 liter (L) 10 dL = 1 liter (L) 1000 milliliter (mL) = 1 liter (L) 10 liters (L) = 1 dekaliter (daL) 10 dekaliter (daL) = 1 hectoliter (hL) 1 hectoliter (hL) = 100 liters (L) 1000 cubic decimeter (dm³) = 1 cubic meter (m³) 1000 liters (L) = 1 cubic meter (m³) 1000 liters (L) = 1 kiloliter (kL) 1 kiloliter (kL) = 10 hectoliters (hL)

Other Units of Volume List

Also, there are few other units that are utilized to represent the volume. However, it covers the British system of measurements such as drams, barrels, bushels, and pecks, gills, etc. used for larger quantities. Mostly, standard unit measurements used in the British System are the cubic inch, cubic foot, and the gallon.

Examples of Conversion of Units of Volume and Capacity

Example 1: 
Convert 140dm³ to cm³.
Solution: 
Given 140dm³ to convert into cm³
Now, we have the formula, 1000 cubic centimeter (cm³ )= 1 cubic decimeter (dm³)
Here, multiply cubic decimeter by 1000 cubic centimeters to get the answer by the known formula
140dm³ = 140 × 1000 cm³ = 140000 cubic centimeters (cm³)
Thus, 140dm³ is 1,40,000 cm³.

Example 2: 
Convert 25daL to liters (L).
Solution: 
Given 25 daL to convert into liters.
The formula we used is one dekaliter = 1 × 10 L = 10 liters (L)
Now, convert 25 daL to liters
25 daL = 25 × 10 L = 250 liters.
Hence, 25 daL is 250 L.

Example 3: 
The height, width, and length of the big barrel are 4 m, 8 m, and 600 cm respectively. There is another smaller barrel that has a volume of 100000 dm³. By how many cubic meters, is the volume of the big barrel more than the small barrel?
Solution: 
Here, we have not given the volume of the large barrel rather, we have given the measurements of three dimensions of the barrel.
The height of the barrel is 4 meters
The width of the barrel is 8 meters
The length of the barrel is 600 cm
Now, convert the length of the barrel into meters
600 cm = 6 m (100 cm = 1 m)
The volume of the big barrel = Length × Height × Width = 4 × 8 × 6 = 192 cubic meters (m³).
Now, the volume of the smaller barrel is given in cubic decimeters. So, will convert this dm³ into cubic meters m³.
The volume of the smaller barrel = 10000 dm³.
The formula, we apply here is 1 dm³ = 1 / 1000 m³.
10000 dm³ = 100000 / 1000 m³ = 100 m³.
Now, we have the volumes of both barrels. We can calculate the number of cubic meters that a bigger barrel contains and which is greater than the volume of the smaller barrel.
The volume of the bigger barrel – the volume of the smaller barrel = 192 m³ – 100 m³ = 92m³.
Thus, the volume of the bigger barrel is 92m³ greater than the volume of the smaller barrel.

Strengthen your basics in algebra by practicing the numerous questions from the Worksheet on Multiplying Monomials. Solve Algebraic Expressions fluently by attempting the Multiplication of Monomials Worksheet over here. The Printable Math Worksheet on Multiplication of Monomials will give constructive engagement and ample practice for the theories revolving around the monomials. Practice the Multiplying Monomials Worksheet with Answers in PDF Formats for free and identify the different types of algebraic expressions.

Do Refer:

• Worksheet on Dividing Monomials
• Worksheet on Multiplying Binomials
• Worksheet on Multiplying Monomial and Binomial

Multiplication of Monomials Practice Worksheet

I. Find the product of the monomials:
(i) 5a × 4a
(ii) 2x × 7x × 3
(iii) 3xy × 8by
(iv) x × 3x² × 4x³
(v) 5 × 3m²
(vi) 7 × 3p² × 4p²q²
(vii) (-3x) × 7x²y
(viii) (- 6x²y²) × (- 6xy)

Solution:

(i) Given monomials are 5a,4a.
Multiply the coefficients i.e. 4 × 5=20
Multiply the variables by adding the exponents i.e. a.a=a²
Therefore,5a.4a=20a²
(ii) Given monomials are 2x , 7x,  3
Multiply the coefficients i.e. 2 ×  7 × 3=42
Multiply the variables by adding the exponents i.e. x × x=x²
Therefore, 2x × 7x × 3=42x²
(iii) Given monomials are 3xy , 8by
Multiply the coefficients i.e. 3 × 8=24
Multiply the variables by adding the exponents i.e. xy × by=bxy²
Therefore, 3xy × 8by=24bxy²
(iv) Given monomials are x, 3x², 4x³
Multiply the coefficients i.e. 3 × 8=24
Multiply the variables by adding the exponents i.e. x × x² × x³=x⁶
=12x⁶
(v) Given monomials are 5 , 3m²
Multiply the coefficients i.e. 5 × 3=15
=15m²
(vi) Given monomials are 7, 3p²,4p²q²
Multiply the coefficients i.e. 7 × 3 × 4 =84
Multiply the variables by adding the exponents i.e. p² × p²q²=p⁴q²
Therefore, 7 × 3p² × 4p²q² = 84p4q²
(vii) Given monomials are -3x × 7x²y
Multiply the coefficients i.e. -3 × 7 =21
Multiply the variables by adding the exponents i.e. x × x²y=x³y
Therefore, (-3x) × 7x²y =-21x³y
(viii) Given monomials are (- 6x²y²) × (- 6xy)
Multiply the coefficients i.e. -6 × -6 =36
Multiply the variables by adding the exponents i.e. x²y² × xy=x³y³
=36x³y³

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II. Find the value of the following:
(i) 15x³ × 2x⁴
(ii) (-2m⁴) × 5n⁵
(iii) 4xyz × 2x²y³
(iv) abcd × a²b²c
(v) 4 × 3p² × 2p²q²
(vi) 0 × (15x⁴y⁴z²)

Solution:

(i) Given 15x³ × 2x⁴
Multiply the coefficients i.e. 15 × 2=30
Multiply the variables by adding the exponents i.e. x⁷
The value of 15x³ × 2x⁴  is 30x⁷.

(ii) Given (-2m⁴) × 5n⁵
Multiply the coefficients i.e. -2 × 5=-10
Multiply the variables by adding the exponents i.e. m⁴ × n⁵
The value of (-2m⁴) × 5n⁵ is -10m⁴ × n⁵.
(iii) Given 4xyz × 2x²y³
Multiply the coefficients i.e. 4 × 2=8
Multiply the variables by adding the exponents i.e. xyz × x²y³=x³y⁴z
The value of 4xyz × 2x²y³ is 8x³y⁴z

(iv) Given a²bc × a²b²c
Multiply the coefficients i.e. 1 × 1=1
Multiply the variables by adding the exponents i.e. a²bc × a²b²c=a⁴b³c²
The value of a²bc × a²b²c is a⁴b³c²
(v) Given 4 × 3p² × 2p²q²
Multiply the coefficients i.e. 4 × 3 × 2=24
Multiply the variables by adding the exponents i.e. p² × p²q²=p⁴q²
The value of 4 × 3p² × 2p²q² is 24p⁴q².
(vi) Given 0 × (15x⁴y⁴z²)
Multiply the coefficients i.e. 0 × 15=0
Multiply the variables by adding the exponents i.e. x⁴y⁴z² is x⁴y⁴z²
The value of 0 × (15x⁴y⁴z²) is 0.

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III. Find the product of the following two monomials:
(i) 10mn and -3mn
(ii) ab⁶ and (-a⁵b³)
(iii) 6ab and 2ac
(iv) 8mp² and 2mn²p

Solution:

(i) Given two monomials are 10mn and -3mn
Multiply the coefficients i.e. 10 × -3 =-30
Multiply the variables by adding the exponents i.e. mn × mn=m²n²
The product of the two monomials is 10mn × -3mn=-30m²n².

(ii) Given two monomials are ab⁶ and (-a⁵b³)
Multiply the coefficients i.e. 1. (-1)=-1.
Multiply the variables by adding the exponents i.e. ab⁶ × (a⁵b³)= a⁶b⁹
The product of the two monomials is -a⁶b⁹.

(iii) Given two monomials are 6ab and 2ac
Multiply the coefficients i.e. 6 × 2=12
Multiply the variables by adding the exponents i.e. ab × ac=a²bc
The product of the two monomials is 12a²bc.

(iv) Given two monomials are 8mm²p and 2mn²
Multiply the coefficients i.e. 8 × 2=16
Multiply the variables by adding the exponents i.e. mp² × mn²p=m²n²p³
The product of the two monomials is 16m²n²p³

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IV. Find the product of the three monomials:
(i) 15ab³c⁵, 3a²b³c³ and 2abc⁴
(ii) 7mn³, 2m²n² and 3mn²
(iii) 2a⁴, b³a⁵ and 12a²b²c
(iv) (-a²b³), (-5a²b) and 12a²b

Solution:

(i) Given three monomials are 15ab³c⁵, 3a²b³c³ and 2abc⁴
Multiply the coefficients i.e. 15 × 3 × 2=90
Multiply the variables by adding the exponents i.e. ab³c⁵ × a²b³c³ ×abc⁴=a⁴b⁷c¹²
The product of three monomials is 90a⁴b⁷c¹²

(ii) Given three monomials are 7mn³, 2m²n² and 3mn²
Multiply the coefficients i.e. 7 × 2 × 3=42
Multiply the variables by adding the exponents i.e. mn³ × m²n² × mn²=m⁴n⁷
The product of three monomials is 42m⁴n⁷.

(iii) Given three monomials are 2a⁴, b³a⁵ and 12a²b²c
Multiply the coefficients i.e. 2 × 1 × 12=24
Multiply the variables by adding the exponents i.e. a⁴ × b³a⁵ × a²b²c=a¹¹b⁵c
The product of three monomials is 24a¹¹b⁵c.

(iv) Given three monomials are (-a²b³), (-5a²b) and 12a²b
Multiply the coefficients i.e. -1 × -5 × 12=60
Multiply the variables by adding the exponents i.e. a²b³ × a²b × a²b=a⁶b⁵
The product of three monomials is 60a⁶b⁵.

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V. Multiply a monomial by a monomial:
(i) 9x by 6
(ii) 5a² by 9
(iii) 16mn by 2
(iv) –mn by 18

Solution:

(i) Given 9x,6
9x × 6=54x
(ii) Given 5a² by 9
5a2 × 9=45a2
(iii) Given 16mn by 2
16mn × 2=32mn
(iv) Given –mn by 18
–mn × 18=-18mn

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Are you interested in knowing how to Estimate a Nearest Tens, this article will give you clear information on estimating the nearest tens. It includes the basic knowledge for an estimate to the nearest Tens, how to estimate, and so on. An estimate is a type of Roundoff and is used in subjects like mathematics and physics. Students will learn in detail about the concept of Estimation of Numbers to Nearest Tens Place.

Estimate to nearest 10 is nothing but making the unit digits of the number to zero and getting the estimated nearest 10 for that number. Furthermore, check out the detailed steps, solved examples, and understand how do you estimate nearest tens.  

Also, Read:

• Estimate to Nearest Hundreds
• Estimate to Nearest Thousands

Estimate to Nearest Tens – Definition

Estimate to Nearest Tens is a process of making a number simpler to read and remember. It is done for the whole numbers, decimals for various places of tens, hundreds, thousands, etc. Estimate to Nearest 10 means writing the nearest 10 of the given number. By using this, you can easily estimate the answer quickly and easily.

Rules for Estimation of Numbers to Nearest Tens

For Estimate to Nearest Tens, we have two different rules. Rules are explained in the below section,

• Rule 1: When we estimate the numbers to the nearest 10, if the digit in the unit’s place is less than 5 or between 0 and 4, then the unit’s place of the number is replaced by 0.
• Rule 2: If the digit in the unit’s place is greater than 5 or equal to 5 or between 5 and 9, then the unit’s place is replaced by 0, and the ten’s place digit number is increased by 1.

How to Estimate Numbers to Nearest 10?

Follow the below procedure to learn Estimate to Nearest 10. They are along the lines

• Get a whole number for rounding.
• Identify the digit in the unit’s place.
• If the digit is in between 0 and 4, then place zero in the units place of the given number.
• If the digit is 5 or 6 or 7 or 8 or 9, then place zero in the units place and add 1 to the tens place of the number.
• Next, write the new number as a rounded or estimated number.

Examples on Estimation of Numbers to Nearest Tens Place

Problem 1:
Estimate the given numbers to Nearest Tens. The given numbers are,
(i) 31
(ii) 68
(iii) 45
Solution:
Given in the question, the values are
Now, we have to round the Nearest Tens of given values.
(i) 31
The value is 31, 1 is less than 5. So, we have to replace zero units place.
So, the value is 31.
After replacing the value is 30.
(ii) 68
The given value is 68, 8 is greater than 5. Now, we have to replace zero in the unit’s place and increase 1 in one’s place.
So, the value is 68.
After replacing 0 and 1, the value is 70.
(iii) 45
The value 45 has 5 is equal to 5. So, replace 5 with zero in the unit’s place and increase by 1 in one’s place.
So, the value is 45.
After replacing units place and increasing one’s place, the value is 50.

Problem 2:
Estimate the following digits to the nearest tens, 0.346
Solution:
Given the value is 0.346
Now, we have to find the nearest tens value.
The hundred’s digit is 3 and therefore the tenth digit will remain unchanged.
As the hundredth digit is less than 5, we have to drop off the remaining digits.
Hence, the final value is 0.3

Problem 3:
What is the Roundoff nearest tens value of 789 and 1265?

Solution: 
The values are given in the question,
Now, we have to find the nearest tens of that values.
(i) 789
In this value, the unit’s place 9 is greater than 5. So, we have to place zero in the unit’s place and increased one by ten’s place.
The value 789 after replacing the value will be 790.
Thus, the given number nearest tens value is 790.
(ii) 1265
The value in units 5 is equal to 5. So, we will replace the units place with 0 and tens place is increased by one.
Then the number after replacing it will be 1270.
Therefore, the nearest tens value of 1265 is 1270.

Problem 4:
The value 50014 is Estimated to nearest tens.
Solution:
The given value is 50014,
The value in units place is less than 5, we need to replace the units place with 0.
So, after the value 50014 is replaced it will be 50010.
Thus, the given value nearest to tens is 50010.

Problem 5:
Round the number 518 to the nearest 10.
Solution:
As given in the question, the value is 518
Now, we have to write the nearest Ten’s value.
The value in the unit’s place is greater than 8, so we need to replace the unit’s place with 0 and ten’s place digit is increased by 1.
Then, after the value 518 is replaced it will be 520.
Therefore, the 518 is rounded nearest to ten’s is 520.

FAQ’s on Estimation of Numbers to its Nearest Tens Place

1. What are the advantages of Estimating?
The following are the advantages of estimating as follows,
1. Estimating helps to improve mental math.
2. Your fluency in calculation will be improved.
3. You can understand the concept of roundoff numbers in the number system by learning the concept of estimation.

2. How do you estimate the sum?
We estimate the addition by rounding off to the nearest numbers.

3. How do you find the nearest estimate? 
The basic rule for estimating is to look at the digit to the right of the digit you want to estimate. Estimating to the nearest whole number means looking at the digit to the right of the decimal. If you see a digit greater than 5 is round up and if it’s less than 5 is round down.

In this article, you have to learn about Estimate to Nearest Hundreds, Rules, how to estimate to nearest hundreds, solved example problems, and so on. The estimation is making a number simple but keeping its value nearest to what it was. Its result is less accurate and easier to use especially while performing arithmetic operations. Estimating a number to the nearest hundred means replacing the units and tens place as zeros and either increasing or decreasing the remaining part of the number by one.

The purpose of rounding (or) estimating the numbers is to make the numbers easier to understand and remember, and the calculations become easier. The application of rounding numbers is when you want to estimate an answer, then rounding is used.

Also, Read:

• Estimate to Nearest Thousands

Estimate to Nearest Hundreds – Definition

Estimate to Nearest 100 is a process where you need to convert the given number into an easy form for various reasons. The obtained number is not the actual number but it is an approximate value of the original number. Estimating to the nearest hundred means writing the nearest 100 of the given number.

Rules for Estimation of Numbers to Nearest Hundred

The following are the 2 different rules for estimating the nearest hundred. The 2 different rules are:

• Rule 1: While Estimate to nearest 100, if the digit in the tens place is between 0 to 4 or less than 5, then the tens place and units place digits are replaced by 0.
• Rule 2: If the digit in the ten’s place is equal to 5 or greater than 5, then the tens place is replaced by 0 and the hundreds place is increased by 1.

How to Estimate a Number to Nearest 100?

The below are steps for the process of Estimating Nearest to 100. They are along the lines

• First, get the number we want to round.
• Next, we have to identify the digit in the tens place.
• If the digit in the tens place is less than 5, then replace zero’s in the tens place, units place digits of the number.
• If the digit in the tens place is more than five or equal to 5, then place zero’s in the tens place and units place and increase the hundredth place digit by 1.
• Now, write the obtained number.

Explore all other Math Concepts similar to Estimation, Rounding Off Numbers all under one roof and practice the activity sheets, practice problems, worksheets, etc. to get a good hold of them.

Examples on Estimation by Rounding Off to Nearest 100’s

Problem 1:
Estimate the following numbers to the nearest 100,
(i) 239
(ii) 3510
(iii) 75

Solution:
The values are given in the question,
Now, we have to round the nearest hundreds of the values.
(i)239
The value is 239, ten’s place digit is 3 means that is less than 5. So, we have to replace the ten’s place digit and units place digit with 0.
So, after replacing digits the value is 200.
Therefore, 239 estimate to nearest 100 is 200.
(ii) The number is 3510
We can see that the digit in tens place is 1 which is less than 5. So, we have to replace the ten’s place and units place digits with 0 and write the remaining digits as it is.
So, after replacing digits the value is 3510.
Thus, 3510 is estimating nearest to 100 is 3500.
(iii) The given number is 75
In this, the tens place digit is 7, which is greater than 5. So, we will place zeros in the tens place, units place, and increase the hundreds digit by one.
So, the obtained value is 800.
Hence, the estimating nearest to 100 is 800.

Problem 2:
To estimate the number to the nearest hundred. The number is 2,00, 579
Solution: 
As given in the question, the value is 2,00,579
Now, we have to round the nearest 100 of the given value.
First, we can identify the tens place digit, which is 7, which is greater than 5. So, we have to replace zero in tens place digit, units place digit and increase by 1 of the hundreds place digit.
So, the number is 2,00,579 after replacing and increasing the value is 2,00,600.
Thus, the rounding-off of 2,00,579 to the nearest hundred is 2,00,600.

Problem 3:
Find the value 12,06,769 estimated to nearest hundred?
Solution: 
The given number is 12,06,769
Now, we have to find the nearest hundred value of that number.
First, identify the digits in the tens place is 6, which is greater than 5. So, we have to replace the value in the tens place, units place digit with zero and increase by 1 in the hundredth place digit.
So, the value is 12,06,769.
After replacing the value is 12,06,800.
Hence, the value 12,06,769 is nearer to hundreds is 12,06,800.

FAQ’s on Estimate to Nearest Hundreds

1. What is meant by estimate to nearest hundreds?
Estimate to nearest hundred means the estimate of any decimal number to its nearest hundredth value. In decimal numbers, the hundredth means 1/100 (or) 0.01.

2. How many rules are there for estimating nearest to hundreds?
There are 2 rules for estimating nearest to a hundred.

3. How do you round decimals to the nearest 100? 
Round a number to the nearest hundred, look at the next place value to the right. If it’s 4 or less than 5, just remove all the digits to the right. If it’s 5 or greater than 5, add 1 to the digit in the hundredths place, and then replace zeros in one place and units place digit.

Parents and Teachers can provide their kids with extra practice and help them master the skill of multiplying a monomial with a binomial. You can solve complex math expressions quite easily if you know the concept of Multiplying Monomial and Binomial. In the Worksheet on Multiplying Monomial and Binomial, we have solved the questions by simply multiplying monomial with every individual term of the binomial and then further simplified.

Answer the problems in the Multiplying Monomial and Binomial Worksheet PDF and gain proficiency in the multiplication of monomial and a binomial. Multiplication of Monomial and Binomial Worksheet with Answers and improve your problem-solving ability with the interactive exercises provided.

Do Refer:

• Worksheet on Multiplying Monomial and Polynomial
• Worksheet on Multiplying Binomials
• Worksheet on Dividing Monomials

Multiplying a Monomial by a Binomial Practice Worksheet

1. Multiply a monomial by a binomial
(i) 5ab × (3ab + 4)
(ii) (-2y²) × (x + y²)

Solution:

(i) Given, 5ab × (3ab + 4)
Step 1: Multiply the monomial with the first term of the binomial.
=5ab × (3ab) =15a²b²  
Step 2: Multiply the monomial with the second term of the binomial.
=4(5ab)=20ab
Step 3: Write both the terms obtained in step 1 and step 2 together with their corresponding signs.
=15a²b² + 20ab
Hence, By multiplying 5ab, (3ab + 4) we get 15a²b²+20ab.
(ii) Given, (-2y²) × (x + y²)
Step 1: Multiply the monomial with the first term of the binomial.
=(-2y²) × x =-2xy²
Step 2: Multiply the monomial with the second term of the binomial.
=(-2y²) × y²=-2y⁴
Step 3: Write both the terms obtained in step 1 and step 2 together with their corresponding signs.
=-2xy²-2y⁴
Hence, By multiplying (-2y²), (x + y²) we get -2xy²-2y⁴.

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2. Find the Multiplication Result of a Monomial and a Binomial
(i) 5a²b³ × (a + b)
(ii) (-5x²) × (-2x²+8y)
(iii) (-4m) × (2m³ + m²n)

Solution:

(i) Given, 5a2b3 × (a + b)
Step 1: Multiply the monomial with the first term of the binomial.
=5a²b³ × (a)=5a³b³
Step 2: Multiply the monomial with the second term of the binomial.
= 5a²b³ × (b)=5a²b⁴
Step 3: Write both the terms obtained in step 1 and step 2 together with their corresponding signs.
=5a³b³+5a²b⁴
Hence, By multiplying 5a²b³, (a + b) we get 5a³b³+5a²b⁴.
(ii) Given, (-5x²) × (-2x²+8y)
Step 1: Multiply the monomial with the first term of the binomial.
=(-5x²) × (-2x²)=10x⁴
Step 2: Multiply the monomial with the second term of the binomial.
= (-5x²) × 8y=- 40x²y
Step 3: Write both the terms obtained in step 1 and step 2 together with their corresponding signs.
=10x⁴ – 40x²y
Hence, By multiplying (-5x²), (-2x²+8y) we get 10x⁴ – 40x²y.
(iii) Given, (-4m) × (2m³ + m²n)
Step 1: Multiply the monomial with the first term of the binomial.
=(-4m) × (2m³)=-8m⁴
Step 2: Multiply the monomial with the second term of the binomial.
=(-4m) × (m²n)=-4m³n
Step 3: Write both the terms obtained in step 1 and step 2 together with their corresponding signs.
=-8m⁴-4m³n
Hence, By multiplying (-4m) × (2m³ + m²n) we get -8m⁴-4m³n.

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3. Multiply a binomial by a monomial:
(i) (6a + 2b) by ab²
(ii) (4m²n – 3m²) by 2mn

Solution:

(i) Given, (6a + 2b) by ab²
Step 1: Multiply the monomial with the first term of the binomial.
=6a × (ab²)=6a²b²
Step 2: Multiply the monomial with the second term of the binomial.
= 2b × (ab²)=2ab³
Step 3: Write both the terms obtained in step 1 and step 2 together with their corresponding signs.
=6a²b²+2ab³
Hence, By multiplying 6a + 2b, ab² we get 6a²b²+2ab³.
(ii) Given, (4m²n – 3m²) by 2mn
Step 1: Multiply the monomial with the first term of the binomial.
=2mn × (4m²n)=8m³n²
Step 2: Multiply the monomial with the second term of the binomial.
-(3m²) × 2mn=-6m³n
Step 3: Write both the terms obtained in step 1 and step 2 together with their corresponding signs.
=8m³n²-6m³n
Hence, By multiplying 4m²n – 3m²,2mn we get 8m³n²-6m³n.

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4. Multiply the following monomial with a binomial

(i) (5xy + 2y) by 2x²y²
(ii) (2b² + 6c) by 5bc
(iii) (2mn + 5mp) by mnp

Solution:

(i) Given, (5xy + 2y) by 2x²y²
Step 1: Multiply the monomial with the first term of the binomial.
=5xy × (2x²y²)=10x³y³
Step 2: Multiply the monomial with the second term of the binomial.
=2y(2x²y²)=4x²y³
Step 3: Write both the terms obtained in step 1 and step 2 together with their corresponding signs.
=10x³y³+4x²y³
Hence, By multiplying 5xy + 2y,2x²y² we get 10x³y³+4x²y³.
(ii) Given, (2b² + 6c) by 5bc
Step 1: Multiply the monomial with the first term of the binomial.
=5bc(2b² )=10b³c
Step 2: Multiply the monomial with the second term of the binomial.
=6c(5bc)=30bc²
Step 3: Write both the terms obtained in step 1 and step 2 together with their corresponding signs.
=10b³c + 30bc²
Hence, By multiplying 2b² + 6c, 5bc we get 10b³c + 30bc².
(iii) Given, (2mn + 5mp) by mnp
Step 1: Multiply the monomial with the first term of the binomial.
=mnp(2mn) =2m²n²p
Step 2: Multiply the monomial with the second term of the binomial.
= mnp(5mp)=5m²np²
Step 3: Write both the terms obtained in step 1 and step 2 together with their corresponding signs.
=2m²n²p + 5m²np²
Hence, By multiplying 2mn + 5mp, mnp we get 2m²n²p + 5m²np².

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5. Find the product of the following algebraic expressions
(i) 3b(12a + b⁴)
(ii) (-a²) (5 – 4a²)
(iii) 3x²(4x + 2y)

Solution:

(i) Given, 3b(12a + b⁴)
Multiply the monomial with the first term, the second term of the binomial and write both terms obtained with their corresponding signs.
=3b(12a) + 3b(b⁴)
=36ab + 3b⁵
Hence, the product of 3b, 12a + b⁴ we get 36ab + 3b⁵.
(ii) Given, (-a²) (5 – 4a²)
Multiply the monomial with the first term, the second term of the binomial and write both terms obtained with their corresponding signs.
=(-a²)(5) – (4a²)(-a²)
=-5a² + 4a⁴
Hence, the product of (-a²), (5 – 4a²) we get -5a² + 4a^4.
(iii) Given, 3x²(4x + 2y)
Multiply the monomial with the first term, the second term of the binomial and write both terms obtained with their corresponding signs.
=3x²(4x) + 3x²(2y)
=12x³+6x²y
Hence, the product of 3x²(4x + 2y) we get 12x³+6x²y.

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6. Find the product of the following monomial and binomial
(i) x²y(x² + y²z)
(ii) (-12abc) (3ab + 4bc)

Solution:

(i) Given, x²y(x² + y²z)
Multiply the monomial with the first term, the second term of the binomial and write both terms obtained with their corresponding signs.
=x²y(x²) + x²y(y²z)
=x⁴y + x²y³
Hence, the product of x²y,(x² + y²z) we get x⁴y + x²y³.
(v) Given, (-12abc) (3ab + 4bc)
Multiply the monomial with the first term, the second term of the binomial and write both terms obtained with their corresponding signs.
=(-12abc) (3ab) + (-12abc)(4bc)
=-36a²b²c + (-48ab²c²)
= -36a²b²c -48ab²c²
Hence, the product of (-12abc) (3ab + 4bc) we get -36a²b²c -48ab²c²

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Students should have a good grip on all basic concepts of maths in primary school only as it helps them in further studies. Kids who are studying in 5th Standard should be particular about important topics like Shapes. It is the most fundamental concept in maths which should be practiced and understood by every kid. When we draw plane figures with their shape, region, and boundary, we obviously compare the objects with their size and area.

There are different geometrical shapes like squares, circles, and rectangles, etc. While solving geometrical problems, the area of a rectangle plays an important role. On this page, you will understand the concept of the area of a rectangle in detail along with steps to find Length or Breadth when the Area of a Rectangle is given.

Do Refer:

• Practice Test on Circle
• Units of Volume

What is the Area of a Rectangle?

The area occupied by a rectangle within its boundary is called the Area of a Rectangle and it can be defined by its sides. It is measured in terms of the square units. The area of a rectangle is the region sheltered by the rectangle in a two-dimensional plane as the amount of space covered by a flat surface of a particular shape.

Few examples of a rectangle are planks, laptops, LEDs, and blackboards, etc. You can use the formula to find the size of these objects. Let us represent a rectangle how it looks.

Area of Rectangle Formula

The size of any object can be measured in different ways. To solve the problems related to the area of the rectangle easily, children should be aware of the area of the rectangle formula.

The formula of the area of a rectangle is calculated in units by multiplying the length and width of the rectangle given.

Area of a Rectangle = Length × Width square units (i.e., A = l×w sq units).

Where a rectangle is of two sides. The length of a rectangle is the largest side and the width is the smallest side. The width of a rectangle is sometimes referred to as breadth(b). Square units may be square centimeters, square inches, etc.

The area of any rectangle is calculated if the length and width of the size are known. By multiplying both the length and the width, the area of a rectangle is measured and it is represented in square units.

Total surface areas can be multiplied or calculated only if the shapes are three-dimensional figures. We cannot calculate for the rectangle because they are two-dimensional figures.

Perimeter of a rectangle = 2(length + breadth)  (i.e., P = 2(l+b)).

How to Calculate the Area of a Rectangle?

To find the area of a rectangle, students should know the length and width of a rectangle and it will be easy to multiply and get the area of a rectangle in square units. Let us follow some simple steps to calculate the area of a rectangle.

Step 1: Observe the rectangle and write the dimensions of length and width from the given data.
Step 2: Now, apply the area of a rectangle formula, a = l×w.
Step 3: Write the obtained answer in square units.

Area of a Rectangle by Diagonal

There are two diagonals in a rectangle and both are of equal length. The straight line connecting is opposite vertices is the diagonal of a rectangle. The formula we use to find the diagonal of a rectangle is as follows.

(Diagonal)² = (length)² + (breadth)²

From this,

(length)² = (diagonal)² – (breadth)²
length = √(diagonal)² – (breadth)²

(breadth)² = (diagonal)² – (length)²
breadth = √(diagonal)² – (length)²

Now, the area of a rectangle = length × breadth

Area = length × (√(diagonal)² – (length)²)

Area of a Rectangle Examples

Example 1: 
Find the area of the rectangle whose length is 14 cm and the width is 6 cm.
Solution: 
Given
length 14 cm, width 6 cm
Area of a rectangle = length × width
A = 14 × 6 = 84 sq cm.
Thus, the area of a rectangle is 84 square centimeters.

Example 2: 
The length and width of a rectangular field are 100 yards and 70 yards. Find the area of the field.
Solution:
Given
Length of the field = 100 yards
Width of the field = 70 yards
Area of the field = length of the field × width of the field
A = 100 × 70 = 700 sq yards.
Hence, the area of the field is 700 square yards.

Example 3: 
Find the area of the tennis court whose length and breadth are 55 m and 25 m respectively.
Solution: 
Given
The length of the tennis court is 55m.
The Width of the tennis court is 25m.
Area of the tennis court = length × width
A = 55m × 25m = 1375 sq mts.
Therefore, the area of the tennis court is 1375 square meters.

Example 4: 
The area of a rectangle is 36 cm². If its breadth is 8 cm then find its length?
Solution: 
Given, Area of a rectangle is 36 sq cm and the Breadth is 8 cm.
Formula for area of a rectangle = length × width
Now, to find the length
Length = Area of a rectangle / Width
Length = 48 sq cm / 8 cm = 8cm
Hence, the length of a rectangle is 8 cm.

FAQs on Area of Rectangle

1. What is the area and perimeter of a rectangle?
The perimeter of a rectangle is the addition of its four sides. Thus, Perimeter of a rectangle = 2 (length + width) units. The area of a rectangle is defined as the multiplication of length and width of a rectangle. Thus, the area of a rectangle = length × width square units.

2. What is the unit area of a rectangle?
The unit of area of a rectangle is a ‘square unit’. For example, if the dimensions of a rectangle are 3 inches × 2 inches then the area of a rectangle is 6 square inches.

3. Why do we compute the area of a rectangle?
We compute the area of a rectangle to find the area occupied by a rectangle within its perimeter.

4. Is the area of a square and area of a rectangle are same?
No, the area of a rectangle and the area of a square is different because every square is a rectangle with its length and breadth but all the rectangles are not squares. The area of a rectangle is measured in square units whereas square is measured by (side)².

Worksheet on Pictograph and Bar Graph consists of different types of questions and answers along with their explanations. All the important concepts such as Interpreting the Data of a Pictograph or Bar Graph, Representing a Pictograph or Bar Graph for the given data, etc. included here. Students can master the concept of Pictograph or Bar Graph with the help of Math Worksheets on Pictograph and Bar Graph.

Therefore, without missing any concept, practice all the problems available in the Printable Pictograph and Bar Graph Worksheet and learn the skills involving reading, drawing, grouping, and many more. Pictograph and Bar Graph Problems available will Reinforce knowledge in analyzing the data under real-life situations and you need not look further anymore.

Also, Check:

• To make a Pictograph
• Pictograph to Represent the Collected Data
• Interpreting a Pictograph

Pictograph and Bar Graph Worksheets

Check out all the Problems on Pictographs and also the Problems on Bar Graphs with a clear explanation. We also included images for a better understanding of the students.

Example 1.
The students are asked to vote for their favorite ice cream flavor. Use the information in the pictograph and answer the questions.

1. How many students voted for the chocolate chips?
2. what flavor did the students like the least?
3. what flavors did the students like the maximum?
4. How many students voted for chocolate chip than strawberry?
5. How many students voted for either chocolate or strawberry?

Solution:

Given that each ice cream flavor = 2 votes.
Therefore, Chocolate = 8 votes
Vanilla = 7 votes
Chocolate chips = 13 votes
Cookie Dough = 8 votes
Strawberry = 5 votes.
1. 13 students voted for chocolate chips.
2. Strawberry flavor students liked the least.
3. Chocolate chip students like the maximum.
4. 8 students voted for chocolate chip than strawberry.
5. 13 students voted for either chocolate or strawberry.

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Example 2.
The sales of 3 Laptop company’s in the three consecutive months is represented by the table. Draw a Bar Graph.

	Lenovo	Hp	Dell
Jan	200	300	400
Feb	500	400	600
Mar	600	500	700
Apr	800	700	900

Solution:

Given that sales of 3 Laptop company are in the three consecutive months.
Take the months on the x-axis. The sales of 3 Laptop companies are on the y – axis.

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Example 3.
The following is the bar graph of the student marks obtained in three subjects.

(i) who scored the least marks in English?
(ii) who scored the highest marks in Science?
(iii) who scored the least marks in maths?

Solution:

Given that the bar graph of the student marks obtained in three subjects.
(i) Satish scored the least marks in English.
(ii) Sanjana scored the highest marks in science.
(iii) Satish scored the least marks in maths.

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Example 4.
In the examination the time is taken by the five people to solve the question paper is as follows. Represent this information on a bar graph.

Minutes	30	35	40	45	50
People	Ram	Sam	Alex	Kailyn	Ronaldo

Solution:

Given that in the examination the time is taken by the five people to solve the question paper.
Take the people on the x-axis and time on the y-axis.

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Example 5.
How to interpret given data using pictographs. From the given table, the data of 50 students has been collected, who have different colors. The information given was as follows:

No. of Students	Color
40	Yellow
30	Black
20	Blue
50	Red

Solution:

Given that data of 50 students has been collected, who have different colors.
The above table data can be represented as a pictograph as follows:

Color	No. of Students
Yellow
Black
Blue
Red

where every circle represents 10 students.  represents a circle.

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Example 6.
Verify the below pictograph and interpret the data to answer the below questions. Where every chocolate image represents 3 chocolates.

Boxes	Number of chocolates
Box 1
Box 2
Box 3
Box 4

(i) Which box had maximum chocolates as per the given data?
(ii) Which box had minimum chocolates as per the given data?
(iii) How many chocolates are more in Box 2 than Box 4?
(iv) Which Box had a maximum number of chocolates?  Box 2 or Box 1?

Solution:

Given that a pictograph consists of different boxes with a different number of chocolates in them. Also, given that every chocolate image = 3 chocolates.
Box 1 = 3 chocolate images = 3 × 3 = 9 chocolates
Box 2 = 4 chocolate images = 3 × 4 = 12 chocolates
Box 3 = 5 chocolate images = 3 × 5 = 15 chocolates
Box 4 = 1 chocolate image = 3 × 1 = 3 chocolates
(i) Box 3 had maximum chocolates as per the given data.
(ii) Box 4 had minimum chocolates as per the given data
(iii) Box 2 – Box 4 = 12 chocolates – 3 chocolates = 9 chocolates.
Box 2 has 9 chocolates more than Box 4.
(iv) Box 2 had a maximum number of chocolates compared to Box 1.

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Example 7.
If ∎ represents 3 switches and Ψ represents 5 boards, what do the following represent?
(i) ∎Ψ ∎Ψ ∎Ψ
(ii) ∎Ψ∎Ψ ∎Ψ ∎Ψ ∎Ψ ∎
(iii) ∎Ψ ∎Ψ
(iv) ∎Ψ
(v) ∎Ψ ∎Ψ ∎Ψ∎Ψ∎Ψ∎Ψ
(vi) ∎Ψ ∎Ψ ∎Ψ ∎Ψ
(vii) ∎Ψ ∎Ψ ∎Ψ ∎Ψ ∎∎Ψ
(viii) ∎Ψ Ψ Ψ Ψ Ψ Ψ Ψ Ψ Ψ Ψ

Solution:

Given that ∎ represents 3 switches and Ψ represents 5 boards.
∎ = 3 switches; Ψ = 5 boards
(i) ∎Ψ ∎Ψ ∎Ψ
∎∎∎ (three switches) = 3 × 3 switches = 9 switches.
ΨΨΨ (three boards) = 3 × 5 boards = 15 boards.

Therefore, the final answer is 9 switches, 15 boards.

(ii) ∎Ψ∎Ψ ∎Ψ ∎Ψ ∎Ψ ∎
∎∎∎∎∎∎ (six switches) = 6 × 3 switches = 18 switches.
ΨΨΨΨΨ (five boards) = 5 × 5 boards = 25 boards.

Therefore, the final answer is 18 switches, 25 boards.

(iii) ∎Ψ∎Ψ
∎∎ (two switches) = 2 × 3 switches = 6 switches.
ΨΨ (two boards) = 2 × 5 boards = 10 boards.

Therefore, the final answer is 6 switches, 10 boards.

(iv) ∎Ψ
∎ (one switch) = 1 × 3 switches = 3 switches.
Ψ (one boards) = 1 × 5 boards = 5 boards.

Therefore, the final answer is 3 switches, 5 boards.

(v) ∎Ψ ∎Ψ ∎Ψ∎Ψ∎Ψ∎Ψ
∎∎∎∎∎∎ (six switches) = 6 × 3 switches = 18 switches.
ΨΨΨΨΨΨ (six boards) = 6 × 5 boards = 30 boards.

Therefore, the final answer is 18 switches, 30 boards.

(vi) ∎Ψ ∎Ψ ∎Ψ ∎Ψ
∎∎∎∎ (four switches) = 4 × 3 switches = 12 switches.
ΨΨΨΨ (four boards) = 4 × 5 boards = 20 boards.

Therefore, the final answer is 12 switches, 20 boards.

(vii) ∎Ψ ∎Ψ ∎Ψ ∎Ψ ∎∎Ψ
∎∎∎∎∎∎ (six switches) = 6 × 3 switches = 18 switches.
ΨΨΨΨΨ (five boards) = 5 × 5 boards = 25 boards.

Therefore, the final answer is 18 switches, 25 boards.

(viii) ∎Ψ Ψ Ψ Ψ Ψ Ψ Ψ Ψ Ψ Ψ
∎ (one switch) = 1 × 3 switches = 3 switches.
ΨΨΨΨΨΨΨΨΨΨ (ten boards) = 10 × 5 boards = 50 boards.

Therefore, the final answer is 3 switches, 50 boards.

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Example 8.

Observe the bar graph representing the number of toys sold in January month on different days of a week and answer the following.

(i) On which day were the maximum toys sold and how many?
(ii) On which day were the minimum toys sold and how many?
(iii) On which day were 25 toys sold?
(iv) How many maximum toys were sold on Tuesday then Friday?

Solution:

Given that the bar graph representing the number of toys sold in January month on different days of a week.
(i) On Wednesday, the maximum number of toys sold. 50 toys are sold on Wednesday.
(ii) On Friday, the minimum number of toys sold. 10 toys are sold on Friday.
(iii) On Tuesday, 25 toys were sold.
(iv) Tuesday = 25 toys were sold
Friday = 10 toys were sold
25 – 10 = 15 toys.
Therefore, 15 more toys were sold on Tuesday than Friday.

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Example 9.

A bookseller sold 30 books on Monday, on Tuesday 40 books, on Wednesday 20 books, on Thursday 70 books, on Friday 50 books, and on Saturday 600 books. Draw a pictograph for the books sold during the week.

Solution:

Given that a bookseller sold 30 books on Monday, on Tuesday 40 books, on Wednesday 20 books, on Thursday 70 books, on Friday 50 books, and on Saturday 60 books.

Days	Books Sold
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday

Here every book is equal to 10 books. = 10 books.
Monday = 30 books = 3 books.
Tuesday = 40 books = 4 books.
Wednesday = 20 books = 2 books.
Thursday = 70 books = 7 books.
Friday = 50 books = 5 books.
Saturday = 60 books = 6 books.

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Example 10.

If ⊗ represents 2 chairs, what do the following represent?
(i) ⊗⊗
(ii) ⊗⊗⊗⊗⊗
(iii) ⊗⊗⊗
(iv) ⊗⊗⊗⊗⊗⊗⊗
(v) ⊗⊗⊗⊗
(vi) ⊗⊗⊗⊗⊗⊗⊗⊗⊗
(vii) ⊗⊗⊗⊗⊗⊗⊗⊗⊗⊗⊗⊗

Solution:

Given that ⊗ represents 2 chairs.
(i) ⊗⊗
⊗⊗ (two cross circles) = 2 × 2 chairs = 4 chairs.
Therefore, the answer is 4 chairs.

(ii) ⊗⊗⊗⊗⊗
⊗⊗⊗⊗⊗ (five cross circles) = 5 × 2 chairs = 10 chairs.
Therefore, the answer is 10 chairs.

(iii) ⊗⊗⊗
⊗⊗⊗ (three cross circles) = 3 × 2 chairs = 6 chairs.
Therefore, the answer is 6 chairs.

(iv) ⊗⊗⊗⊗⊗⊗⊗
⊗⊗⊗⊗⊗⊗⊗ (seven cross circles) = 7 × 2 chairs = 14 chairs.
Therefore, the answer is 14 chairs.

(v) ⊗⊗⊗⊗
⊗⊗⊗⊗ (four cross circles) = 4 × 2 chairs = 8 chairs.
Therefore, the answer is 8 chairs.

(vi) ⊗⊗⊗⊗⊗⊗⊗⊗⊗
⊗⊗⊗⊗⊗⊗⊗⊗⊗ (nine cross circles) = 9 × 2 chairs = 18 chairs.
Therefore, the answer is 18 chairs.

(vii) ⊗⊗⊗⊗⊗⊗⊗⊗⊗⊗⊗⊗
⊗⊗⊗⊗⊗⊗⊗⊗⊗⊗⊗⊗ (twelve cross circles) = 12 × 2 chairs = 24 chairs.
Therefore, the answer is 24 chairs.

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Reading and Writing Large Numbers worksheets will help you learn about larger numbers. Numbers are separated into different groups such as ones, tens, hundreds, thousands, millions, billions, and so on. In each group, there exists three subgroups namely ones, tens, and hundreds. Refer to the below problems provided so that you will have an idea of groups. Thus, you can easily answer the questions on reading and writing large numbers worksheets into words, numerals, or vice versa.

Also, remember that while reading or writing large numbers you need to start from the left side with the largest group and so move towards the right side. Use and download this Worksheet on Reading and Writing Large Numbers in symbols and words for free of cost and take a printout of this downloaded page for more practicing purposes.

Reading and Writing Large Numbers Worksheet

Problem 1:
How to read the number 213624537?

Solution:

Given the value,
First, we separate the numbers into groups, then separate the given numbers by comma from the right side to the left side.
Group names are ones, thousands, millions, and billions, etc. Each and every group has hundreds, tens, and ones.
Next, Start reading the digits from left to right with group names.
Read the groups as a list of words. Write ‘and’ before tens, ones of each group.
Let’s follow these rules applying to the given number 213624537
Separating 3-digits with commas – 213,624,537
Now, we have
213 in the millions group and read in words as two hundred and thirteen million.
624 in the thousands group and read in words as six hundred and twenty-four thousand.
537 in one’s group and read in words as five hundred and thirty-seven.
Thus, the Large Number read in words as Two hundred and thirteen million, six hundred and twenty-four thousand, five hundred and thirty-seven.

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Problem 2:
Write the below numbers in words. The numbers are
(i) 3,013
(ii) 6,540
(iii) 41,215
(iv) 73,626

Solution:

As given in the question,
Now, we have written the numbers in words.
So, the given numbers are,
(i) 3,013 – Three thousand and thirteen
(ii) 6,540 – Six thousand and five hundred forty
(iii) 41,215 – Forty-one thousand and two hundred fifteen
(iv) 73,626 – Seventy-three thousand and six hundred twenty-six

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Problem 3:
Write the number name in words. The number is  321,657,732,123.

Solution:

As given in the question, the value is 321, 657, 732, 123
Now, write that number into words.
So, the number name is,
321,657,732,123 – Three hundred and twenty-one billion, six hundred and fifty-seven million, seven hundred and thirty-two thousand, one hundred and twenty-three

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Problem 4:
Write the number’s name in numerals. The number names are
(i) Five hundred thirteen
(ii) Seventy-two thousand and eight hundred six
(iii) Four thousand four

Solution:

Given the data,
Now, we have to write the number names into numerals.
So, the numbers are,
(i) Five hundred thirteen – 513
(ii) Seventy-two thousand and eight hundred six – 72,806
(iii) Four thousand four – 4004

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Problem 5:
Write the given number value in numerals. The given number name is Six hundred and eight billion, seven hundred and eighty-nine million, nine hundred and ninety-two thousand, five hundred and forty.

Solution:

Given the number name,
Now, we will write that number name into numerals.
So, the number is,
Six hundred and eight billion, seven hundred and eighty-nine million, nine hundred and ninety-two thousand, five hundred and forty = 608,789,992,540
Therefore, the numeral is 608,789,992,540.

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Problem 6:
How to read the given number. The given number is 156, 478, 910, 241

Solution:

Given the value,
First, we separate the numbers into groups then separate the given numbers by comma from the right side to the left side.
Group names are ones, thousands, millions, and billions, etc. Each and every group has hundreds, tens, and ones.
Next, Start reading the digits from left to right with group names.
Read the groups as a list of words. Write ‘and’ before tens, ones of each group.
Let’s follow these rules applying to the given number 213624537
Separating 3-digits with commas – 156,478,910,241
Now, we have to read the
156 in the billions group and read in words as one hundred and fifty- six billion.
478 in the millions group and read in words as four hundred and seventy- eight million.
910 in the thousands group and read in words as nine hundred and ten thousand.
241 in the one’s group and read in words as two hundred and forty-one.
Therefore, the given Large Number read in words as one hundred and fifty-six billion, four hundred and seventy-eight million, nine hundred and ten thousand, two hundred and forty-one.

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Problem 7:
Write the given number value in numerals. The given number name is eight hundred and forty -eight million, one hundred and ten thousand, four hundred and ten

Solution:

Given the number name in words,
Now, we have to write that number name into numerals.
So, the number name is,
eight hundred and forty -eight million, one hundred and ten thousand, four hundred and ten in words. The numeral value is 848,110,410
Therefore, the given number name in numerals is 848,110,410.

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In the Worksheet on Types of Algebraic Expressions, you will find questions on identifying if it is monomial, binomial, trinomial, or polynomial, state if an algebraic expression is true or false or not. Practice the questions on the Types of Algebraic Expressions  Worksheet and improve your math skills and grip on the concept.

Know the different models of questions framed on the topic of Algebraic Expressions Types and learn the problem-solving approach too. Students can download the Algebraic Expressions Worksheet with Answers and solve the problems effectively.

Do Read: Worksheet on Degree of a Polynomial

Types of Algebraic Expressions with Examples

I. Which of the following expressions are polynomials?
(i) 7 – 15a
(ii) 13 + 2a + 5a²
(iii) 5 + 1/x + 1/x² + 1/x³ +1/x⁴
(iv) (a² + a – 6)/(a – 2)
(v) 5 + 1/x + 1/x² + 1/x³ +1/x⁴
(vi) 1/√x + 8
(vii) x + x² + x⁵ – 3
(viii) 1 + 7k

Solution:

(i) Given 7-15a
Here there are two terms.
7,15a
It is a polynomial.
(ii) Given 13 + 2a + 5a²
Here there are three terms.
It is a polynomial.
(iii) Given 5 + 1/x + 1/x² + 1/x³ +1/x⁴
Here there are 5 terms.
It is a polynomial.
(iv) Given (a² + a – 6)/(a – 2)
In this example, we have variables in the denominators of the fraction.
Hence, it is not a polynomial.
(v) Given 5 + 1/x + 1/x² + 1/x³ +1/x⁴
In this example, we have variables in the denominators of the fraction.
Hence, it is not a polynomial.
(vi) Given 1/√x + 8
Here there are two terms.
The first term has a square root of variables.
Hence, it is not a polynomial.
(vii) Given, x + x² + x⁵ – 3
Here there are four terms.
It is a polynomial.
(viii) Given, 1 + 7k
Here there are two terms.
Hence, it is a polynomial.

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II. For each expression, given below, state whether it is a monomial, binomial or trinomial:
(i) ab
(ii) xy + x
(iii) 2x ÷ y
(iv) 7xyz
(v) 7 + m + n
(vi) 2m² – m
(vii) –x
(viii) am² + bm – 7
(ix) -3ab + u
(x) 1 + a + b
(xi) 1 + k ÷ z
(xii) x + xy – z²

Solution:

(i) In the given expression, there is only one term.
Hence it is a monomial.
(ii) In the given expression, there are two terms.
Hence it is a binomial.
(iii) In the given expression, there is only one term.
Hence it is a monomial.
(iv)In the given expression, there is only one term.
Hence it is a monomial.
(v) In the given expression, there are three terms.
Hence it is a trinomial.
(vi) In the given expression, there are two terms.
Hence it is a binomial.
(vii) In the given expression, there is only one term.
Hence it is a monomial.
(viii) In the given expression, there are three terms.
Hence it is a trinomial
(ix) In the given expression, there are two terms.
Hence it is a binomial.
(x) In the given expression, there are three terms.
Hence it is a trinomial.
(xi) In the given expression, there are two terms.
Hence it is a binomial.
(xii) In the given expression, there are three terms.
Hence it is a trinomial.

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III. Identify the following expressions, state whether they are monomials, binomials, trinomials, polynomials:
(i) 23xyz
(ii) a + 2b
(iii) x + y + z
(iv) 1 – a + a² + a⁵ – a⁷ + a⁹
(v) 2mn + p
(vi) 5 + 8a + 2a³
(vii) 3ab+ 2a+ 5

Solution:

(i) In the given expression, there is only one term.
Hence it is monomial.
(ii)In the given expression, there are two terms.
Hence it is binomial.
(iii) In the given expression, there are three terms.
Hence it is trinomial.
(iv) In the given expression, there are more than three terms.
Hence it is polynomial.
(v) In the given expression, there are two terms.
Hence it is binomial.
(vi)In the given expression, there are three terms.
Hence it is trinomial.
(vii) In the given expression, there are three terms.
Hence it is trinomial.

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Iv. State whether the following statements are true or false:
(i) Expression 3x² + x is a trinomial
(ii) Expression am² + bm + c is a trinomial
(iii) 2 × xy is a binomial
(iv) 4 + ab is a binomial
(v) x³ – 2xy + 3x + 5 is a polynomial
(vi) 7a³ + 5ab + 2a + 4 is a multinomial
(vii) 2 + 3a + 4a² + a⁴ + a⁵ is a binomial
(viii) m² + n² + p² is a trinomial

Solution:

(i) false
(ii) true
(iii) false
(iv) true
(v) true
(vi)true
(vii) false
(viii) true

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