 ## Integers and the Number Line – Definition, Facts, Examples | How to Represent Integers on a Number Line?

An integer is a number that doesn’t have decimal or fractional parts. The set of positive and negative numbers including zero are called integers. A number line is a horizontal straight line in which the integers are placed at equal intervals. Both ends of the number line extend indefinitely at both ends. This article is helpful for the 6th-grade math students to understand the math concept i.e number line easily. Check the steps on how to represent integers on the number lines, solved examples.

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## What is a Number Line?

A number line is a graphical representation of numbers on a straight horizontal line. The numbers are placed at equal intervals in a number line. Mainly it can be used for comparing and ordering the numbers. We can represent all real numbers on a number line easily. Using the number line, you can also perform arithmetic operations of numbers such as addition, subtraction, multiplication, and division.

## Integers – Definition

An integer is a number with no fractional or decimal part, from the set of positive, negative numbers including zero. The integers are represented by the symbol Z. The three types of integers are zero, positive integers, and negative integers. The set of integers are Z = {. . . -7, -6, 5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, . . . . . }.

### How to Represent Integers on the Number Line?

Have a look at the simple steps for representing integers on the number line.

• Draw a horizontal line and both ends of the line must extend indefinitely.
• Place vertical lines at equal intervals on that line.
• Label one interval as zero.
• Keep positive integers on the right side of the zero, negative integers on the left side of zero.
• The opposite number like 2, -2 should be at an equal distance from zero. The above gives the exact explanation for integers and the number line. In the following sections, you will get the example questions.

### Questions on Integers and the Number Line

Question 1:
Represent the set of integers {-4, 0, 2, 5} on the number line.

Solution:
Mark the given set of integers on the number line. Question 2:
Write the opposite integer of each of the following:
(i) -5
(ii) 6
(iii) 1
(iv) 23
(v) 108

Solution:
(i) 5
(ii) -6
(iii) -1
(iv) -23
(v)-108

Question 3:
Write all the integers between the following.
(i) -5 and 5
(ii) 2 and 8
(iii) -7 and 10

Solution: By observing the above-number graph, we can easily solve this question.
(i) -4, -3, -2, -1, 0, 1, 2, 3, 4
(ii) 3, 4, 5, 6, 7
(iii) -6, -5, -4,-3,-2,-1, 0,1, 2, 3, 4, 5, 6, 7, 8, 9

### FAQ’s on Number Line and the Integers

1. Why do we use the number line?

A number line is used to represent integers and compare them. It can also be used to perform simple arithmetic operations.

2. What is a number line example?

A number is nothing but the horizontal straight line that represents integers on it at regular intervals. It has zero in the middle with positive and negative integers on both sides.

3. What are some real-life examples of the number line?

Some of the examples of number lines are ruler, protractor, barometer, pressure gauge, scales, micrometer, and so on.

4. How many numbers can be represented in a number line?

A number line extends indefinitely on both sides. So, we can represent indefinite numbers on a number line. ## Worksheet on Concept of Ratio | Free Printable Ratio Problems with Solutions PDF

Worksheet on Concept of Ratio has problems on part-to-part, part-to-whole ratios, reducing ratios, dividing quantities, generating equivalent ratios, and more. Practice the questions in the Ratio Worksheet and solidify your understanding of the topic. We have included several questions in different formats to keep your learning process engaging and interesting. These Math Worksheets on Ratio Concept follow a stepwise format for explaining the questions so that you don’t feel any difficulty in understanding them.

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I. A group of friends went out for dinner. 13 of the diners ordered vegetarian food and 15 ordered non-vegetarian food. What is the ratio of the number of vegetarian meals to the number of nonvegetarian meals?

Solution:

Given,
No. of people ordered vegetarian food=13
No. of people ordered nonvegetarian food=15
The ratio of the number of vegetarian meals to the number of nonvegetarian meals=13/15
Hence, the ratio of the number of vegetarian meals to the number of nonvegetarian meals is 13/15.

II. In the group of 60 people, 40 people are educated and the remaining people are not educated. What is the ratio of the number of people who are not educated to those who are educated?

Solution:

Given,
No. of people in the group=60
No. of educated people=40
No. of noneducated people=60-40=20
The ratio of non educated people to educated people is=20/40=1/2
Hence, the ratio of non-educated people to educated people is 1/2.

III. 150 employees were working on the computer and 50 employees were playing games on their computers. What is the ratio of the number of employees playing games on the computer to the number of employees working on the computer?

Solution:

Given,
No. of employees playing working on the computer=150
No. of employees playing games on the computer=50
The ratio of number of employees playing games to the number of employees working on the computer=50/150=1/3
Therefore, the ratio of the number of employees playing games to the number of employees working on the computer is 1/3.

IV. The ratio of coins to notes in the handbag is 2: 5. If there are a total of 12 coins, find the number of notes in the handbag?

Solution:

Given,
The ratio of coins to notes in the handbag is= 2: 5
Let the number of coins be 2x.
Let the number of notes in the handbag=5x
No. of coins=12
2x=12
x=12/2=6
Number of notes in the hand bag=5x=5(6)=30
Hence, no. of notes in the handbag is 30.

V. In a minibus there are 30 seats, there are 18 occupied seats on the bus, remaining are empty. What is the ratio of the number of occupied seats to the number of empty seats?

Solution:

Given,
No. of seats=30
No. of occupied seats=18
No. of empty seats=30-18=12
The ratio of number of occupied seats to empty seats=18/12=6/4=3/2
Hence, the ratio of occupied seats to empty seats is 3/2.

Vi. In a box, there are oranges and apples. The ratio of oranges and apples is 3:5. If there are 18 oranges, find the number of apples?

Solution:

Given,
No. of oranges=18
The ratio of oranges and apples is= 3:5
Let the number of oranges be 3x.
Let the number of apples be 5x.
3x=18
x=18/3=6
No. of apples=5x=5(6)=30
Hence, there are 30 apples in the box.

Vii. Jay carries a bag of rice which weighs 50 kilograms. If he is going to reduce his bag weight in the ratio 6 : 5, find his new weight of the bag?

Solution:

Given,
Jay carries a bag of rice which weighs= 50 kilograms
Jay reduces his bag weight in the ratio=6:5
Apply the formula, If a quantity increases or decreases in the ratio a:b then-new quantity=b. original quantity/a
New weight=5.50/6=41.66
The new weight of the bag is 41.66 kg.

Viii. If the angles of a triangle are in the ratio 4:6:10, then find the angles?

Solution:

Given that,
Angles of the triangles are in the ratio 4 : 6 : 10, the three angles can be assumed to be
4x, 6x, 10x
In any triangle, sum of the angles = 180
So, we have 4x + 6x + 10x = 180°
20x = 180
x = 9
Then, we have
The first angle = 4x = 4 ⋅ 9 = 36°
The second angle = 6x = 6 ⋅ 9 = 54°
The third angle = 10x = 10 ⋅ 9 = 90°
Therefore, the three angles of the triangle are 36°, 54°, 90°.

IX. Sanjay, Sunil, and Sudheera are three friends. The ratio of average salaries of A and B is 3 : 5and that between A and C is 7: 8. Find the ratio between the average salaries of B and C?

Solution:

From A : B = 3 : 5 and A : C = 7 : 8, we find A in common.
The values corresponding to A in both ratios are different.
First, we have to make them be the same.
Value corresponding to A in the 1st ratio = 3
Value corresponding to A in the 2nd ratio = 7
LCM(3,7)=21
First ratio —-> A : B = 3 : 5 = (3 ⋅ 7) : (5 ⋅ 7) = 21 : 35
Second ratio —-> A : C = 7 : 8 = (7 ⋅ 3) : (8 ⋅ 3) = 21 : 24
Clearly,
A : B = 21 : 35 ———– (1)
A : C = 21 : 24 —————(2)
Now, the values corresponding to A in both ratios are the same.
From (1) and (2), we get
B : C = 35 : 24
Hence, the ratio between the average salary of B and C is 35:24

X. Two numbers are respectively 40% and 60% are more than a third number, Find the ratio of the two numbers?

Solution:

Let “x” be the third number.
Then, the first number is
= (100+40)% of x
= 140% of x
= 1.4x
The second number is
= (100+60)% of x
= 160% of x
= 1.6x
The ratio between the first number and second number is
= 1.4x : 1.6x
= 1.4 : 1.6
= 14 : 16
= 7 : 8
Hence, the ratio of the two numbers is 7: 8. ## Worksheet on Basic Problems on Proportion | Proportion Problems Worksheet with Answers PDF

In the Worksheet on Basic Problems on Proportion, you have questions on proportion, continued proportion, finding mean proportional between the numbers, simple proportions, proportions with decimals, etc. Practice the Questions on Proportion Problems Worksheet on a regular basis and enhance your math skills. Answering the Problems in the Proportion Worksheet with Answers students can develop practical skills that are necessary for day-day-life. Try to solve as much as you can and have a clear understanding of the topics within.

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## Solving Proportions Worksheets PDF

I. Find the value of x in each of the following proportions:
(i) x : 6 = 3 : 9
(ii) 30 : x = 6 : 2
(iii) 3 : 9 = x : 6
(iv) 3 : 2 = x : 4
(v) 5 : 2 = 15 : x
(vi) 6 : 8 = 3 : x

Solution:

(i)  Given x : 6 = 3 : 9
Convert colon based notation to fractional form
x/6=3/9
By cross multiplying we get,
9x=18
x=18/9
=2
Therefore, x=2.
(ii) Given 30 : x = 6 : 2
By converting colon based notation to fractional form we get,
30/x=6/2
Apply cross multiplication we get
60=6x
x=60/6=10
Therefore, x=10.
(iii) 3 : 9 = x : 6
By converting colon based notation to fractional form we get,
3/9=x/6
Apply cross multiplication we get
18=9x
x=18/9=2
Therefore, x=2.
(iv)3 : 2 = x : 4
First, convert colon based notation to fractional form,
3/2=x/4
By Applying cross multiplication we get
12=2x
x=12/2=6
Therefore, x=6.
(v) 5 : 2 = 15 : x
First, convert colon based notation to fractional form,
5/2=15/x
By Applying cross multiplication we get
5x=30
x=30/5=6
Therefore, x=6.
(vi) 6 : 8 = 3 : x
First, convert colon based notation to fractional form,
6/8=3/x
By Applying cross multiplication we get
6x=24
x=24/6=4
Therefore, x=4.

II. Find the mean proportional between:
(i) 0.5 and 3.8
(ii) 0.7 and 8.5
(iii)  16 and 25

Solution:

(i) Let the mean proportional between 0.5 and 3.8 be m.
By applying the formula b² = ac,
Therefore, m x m = 0.5 x 3.8 = 1.9
m2=1.9
m=$$\sqrt{ 1.9 }$$=1.378
Therefore, the mean of 0.5 and 3.8 is 1.378.
(ii) Let the mean proportional between 0.7 and 8.5 be m.
By applying the formula b² = ac,
Therefore, m x m = 0.7. 8.5=5.95
m=$$\sqrt{ 5.95 }$$=2.439
Therefore, the mean of 0.7 and 8.5 is 2.439.
(iii) Let the mean proportional between 16 and 25 be m.
By applying the formula b² = ac,
Therefore, m x m=16.25=400.
m=$$\sqrt{ 400 }$$=20
Therefore, the mean of 16 and 25 is 20.

III. Check whether the following quantities form a proportion or not:
(i) 45:25=35:15
(ii) 3:7=6:14
(iii) 6:3=8:4

Solution:

(i) 1. To check proportionality, we have to multiply means, multiply extremes.
45.15=675
25.35=875
2. Compare the results.
The results of 675,875 are not equal.
Hence, the fractions are not proportional because the product of means and extremes are not equal.
(ii) 1. To check proportionality, we have to multiply means, multiply extremes.
3.14=42
7.6=48
2. Compare the results.
The results 42,48 are not equal.
Hence, the fractions are not proportional because the product of means and extremes are not equal.
(iii) 1. To check proportionality, we have to multiply means, multiply extremes.
6.4=24
3.8=24
2. Compare the results.
The results are equal.
Hence, the fractions are proportional because the product of means and extremes are equal.

IV. Find the unknown value of the following proportion:
i. 3x+2:7=x+4:3

Solution:

i. Given
3x+2/7=x+4/3
First, convert colon based notation to fractional form,
3x+2/7=x+4/3
By cross multiplying we get,
3(3x+2)=7(x+4)
9x+6=7x+28
9x-7x=28-6
2x=22
x=11
Therefore, x=11.
ii. 2:3=x/20-x
First, convert colon based notation to fractional form,
2/3=x/20-x
2(20-x)=3x
40-2x=3x
40=5x
x=40/5=8
Therefore, x=8.

V.  If x : y = 3 : 4 and y : z = 6 : 7, find x : y : z.

Solution:

Given that,
x : y = 3 : 4 and y : z = 6 : 7
Since y is the common term between the two ratios;
Multiply each term in the first ratio by the value of y in the second ratio.
x: y = 3: 4 = 18:24
Also, multiply each term in the second ratio by the value of y in the first ratio.
y: z = 6: 7 = 24: 28
Therefore, the ratio x: y: z = 18:24:28.

VI. If m : n = 2 : 7 and n : s = 3 : 8, find m : s.

Solution:

Given that,
m : n = 2 : 7 and n : s = 3 : 8
Since n is the common term between the two ratios;
Multiply each term in the first ratio by the value of n in the second ratio.
m: n = 2: 7 = 6:21
Also, multiply each term in the second ratio by the value of n in the first ratio.
n: s = 3: 8= 21: 56
Therefore, the ratio m: s= 6:56.

VII. Verify if the ratio 2:4::4:8 is proportion?

Solution:

This is a case of continued proportion, therefore apply the formula a x c =b x b,
In this case, a: b:c =2:4:8, therefore a=2, b=4 and c=8
Multiply the first and third terms
2 × 8 = 16
Square of the middle terms:
(4) ² = 4× 4= 16
Here a x c =b x b is equal.
Therefore, the ratio of 2:4:8 is in proportion.

VIII. If the third proportion of the two numbers is 24. The first number is 6, then find the second number?

Solution:

Given that first number=6,
Third number=24
To find the second number, we can apply the formula a x c =b x b
Here a=6, c=24
b x b=6 .24
=144
b=12
Therefore, the second number is 12.

IX. One piece of pipe 10 meters long is to be cut into two pieces, with the lengths of the pieces being in a 2 : 3 ratio. What are the lengths of the pieces?

Solution:

Given,
Length of one piece of pipe=10 m
The ratio of length of pieces=2:3
Let the length of a short piece of pipe=x
Length of long pipe=10-x
short piece/long piece: 2/3=x/10-x
2(10-x)=3x
20-2x=3x
20=5x
x=20/5=4
Length of the short piece=4m.
Length of the long piece=10-4=6m.

X. The time taken by a vehicle is 2 hours at a speed of 40 miles/hour. What would be the speed taken to cover the same distance at 4 hours?

Solution:

Consider speed as m and time parameter as n.
If the time taken increases, then the speed decreases. This is an inverse proportional relation, hence m ∝ 1/n.
Using the inverse proportion formula,
m = k/ n
m × n = k
At speed of 40 miles/hour, time = 2 hours, from this we get,
k = 40 × 2 = 80
Now, we need to find speed when time, n = 4.
m × n = k
m × 4 = 80
∴ m = 80/4 = 20
Therefore, the speed at 4 hours is 20 miles/hour.

XI. In a construction company, a supervisor claims that 6 men can complete a task in 36 days. In how many days will 15 men finish the same task?

Solution:

Let the number of men is M and the number of days is D.
Given:
M1= 6 ,
D1= 36, and
M2= 15.
This is an inverse proportional relation, as if the number of workers increases, the number of days decreases.
M ∝ 1/D
Considering the first situation,
M1= k/D1
6 = k/36
k = 6 × 36 = 216
Considering the second situation,
M2= k/D2
15 = 216/D2
D2= 216/15 = 14
Therefore, 15 men can complete the same task in 14 days.

XII. Suppose x and y are in an inverse proportion such that when x = 100, y = 3. Find the value of y when x = 150 using the inverse proportion formula?

Solution:

Given: x = 100 when y = 3.
x ∝ 1/y
x = k / y, where k is a constant,
or k = xy
Putting, x = 100 and y = 3, we get;
k = 100 × 3 = 300
Now, when x = 120, then;
150 y = 300
y = 300/150 = 2
That means when x is increased to 150 then y decreases to 2. ## Elapsed Time – Definition, Facts, Formula, Examples | How to find Elapsed Time?

Are you feeling difficulty in calculating the elapsed time? Don’t worry we are here to help you to overcome the difficulties in calculating the time. Before that, you have to know what the elapsed time is. Elapsed time is the calculation of time that passes from starting off a program till its end.

It is very important to know how to calculate the elapsed time to know time. There are different techniques to calculate the elapsed time. One method is the time interval between the start of an event to the end of the event. Another method is the number line on which we break up the time intervals. We suggest the students of 5th grade go through the article and solve the given problems. Learn the concept with suitable examples and score good marks in the exams.

See More: Adding and Subtracting Time

## What is Elapsed Time in Math?

Elapsed time is the amount of time duration from the start of the event to the finish of the event. In simple words, the elapsed time is the time that goes from one time to another time.

### How to Calculate Elapsed time?

1. For solving the elapsed time first we would find the starting time and ending time.
2. Second, counts the hours and minutes between the starting point to noontime and from noontime to ending time.
3. The third is finding out the elapsed time by adding durations.

### Elapsed Time Examples

Example 1.
Find the elapsed time from 7.00 am to 8.00 pm?
Solution:
Given that,
Starting time = 7.00 am
Finishing time = 8.00 pm
The difference between 7.00 am to 12 noon = 5 hours
The difference between 12 noon to 8.00 pm = 8 hours
Duration time = 5 hours + 8 hours = 13 hours.
Duration time is 8 hours

Example 2.
Find the elapsed time from 7 hours 30 minutes to 3 hours 20 minutes?
Solution:
7 hours 30 minutes – 3 hours 20 minutes
7 hours – 3 hours = 4 hours
30 minutes – 20 minutes = 10 minutes
The elapsed time = 4 hours + 10 minutes = 4 hours 10 minutes

Example 3.
What time would it be 3 hours 30 minutes after 8 am?
Solution:
Given that,
Present time = 3 hours 30 minutes
After time = 8 am = 8 hours
Therefore 3 hours 30 minutes + 8 hours
Time would be 11 hours 30 minutes.

Example 4.
The starting time is 11.25 am and the finishing time is 3.40 pm. What is the duration of the time?
Solution:
Given that,
Starting time = 11.25 am
Finishing time = 3.40 pm
The difference between 11.25 am to 12 noon = 35 minutes
The difference between 12 noon to 3.40 pm = 3 hours 40 minutes.
Duration time = 35 minutes + 3 hours 40 minutes
35 minutes + 40 minutes = 1 hour 15 minutes
3 hours + 1 hour 15 minutes = 4 hours 15 minutes
Therefore, the Duration time is 4 hours 15 minutes.

Example 5.
If the movie starts at 4 pm and ends at 8 pm. How long is the movie?
Solution:
Given that,
Starting time of the movie = 4 pm
Finishing time of the movie = 8 pm
Duration of the movie time = 4 pm – 8 pm = 4
Thus, the duration of the movie is 4 hours.

### FAQs on Elapsed Time

1. What is elapsed time?

An elapsed time is the amount of time taken to travel or to start from one place to another place.

2. How do you calculate elapsed time?

1. First count in minutes from earlier time to the nearest hour.
2. Count on hours to the hour nearest to the later time.
3. Then count in minutes to reach the later time.

3. How do you subtract elapsed time?

In order to subtract the time, subtract the minutes and then subtract the hours. This article will assist the students in learning and understanding more about the geometry concept volume. Here, you will find different volume questions for different shapes such as cube, cone, cylinders, and sphere, etc. Indeed students can have the volume questions from easy ones to difficult ones.

Also, you can get an idea of volume in terms of square units and also in cubic units. Teachers and students of can use this article to find volume test questions for different shapes.

## Volume – Definition & Formulas

A volume is defined as the amount of space occupied by any three-dimensional object. An object can be a cube, a solid, a cuboid, and a sphere, etc. In volume different shapes have different volumes. In geometry, we have shapes and solids namely cube, cuboid, sphere, and cylinder, etc., all are defined in three dimensions.

 Name of geometrical shape Volume formula Cube V = a3, where a is the edge length of a cube Cuboid V = length x width x height Cone V = ⅓ πr²h, Where r is the radius and h is the height of a cone Cylinder V = πr²h, Where r is the radius and h is the height of a cylinder Sphere V = 4/3 πr3, Where r is the radius of a sphere Volume of Frustum πh/3 (R2+r2+Rr) Where ‘R’ and ‘r’ are the radii of base and top of frustum Volume of Prism Base Area x Height Volume of Pyramid ⅓ (Area of base) (Height) Volume of Hemisphere ⅔ (πr3), Where r is the radius

Let us practice some problems on volume in different shapes and solids to get a clear perception for students about the concept of volume.

### Practice Questions on Volume

1. Find the volume of a cube, having the sides of length 6 cm.

Solution:

Given, the length of the sides of the cube is 6 cm.
As we know, the volume of a cube = (length of the sides of a cube)³ = a³.
Now, volume = (6 cm)³ = 6 cm× 6 cm× 6 cm = 216 cm³.
Thus, the volume of a cube, V = 261 cm³.

2. Find the volume of a cube of a side 22 cm.

Solution:

Given, the volume of a cube of a side is 22 cm.
The volume of a cube, V = a³.
Now, V = (22 cm)³ = 22 cm× 22 cm× 22 cm = 10648 cm³.
Hence, the volume of a cube is 10648 cm³.

3. Find the volume of a cuboid of dimensions 6 cm, 4.2 cm, 12 cm.

Solution:

Given, Volume of cuboid dimensions is 6 cm, 4.2 cm, 12 cm.
The formula used for the volume of cuboid is V = length x width x height.
V = 6 cm× 4.2 cm× 12 cm = 302.4 cubic centimeters.
Therefore, the volume of a cuboid is 302.4 cu cm.

4. A cylinder has a diameter of 4.2 cm and is 6.4 cm in height. Calculate the volume of a cylinder and have to answer with one decimal place.

Solution:

Given, dimensions are the diameter of 4.2 cm and height of 6.4 cm.
The volume of a cylinder is V = πr²h.
Now, V = 3.14× 2.1× 2.1× 6.4 = 88.62336 cm³.
Thus, the volume of a cylinder is 88.6 cm³.

5. How many cubes of side 5 cm are needed to build a cube of side 15 cm? Solution:

Given
The side of a small cube is 5 cm.
The side of a large cube is 15 cm.
We required n no. of small cubes.
The volume of the cube, V = a³
n no. of cubes = Volume of a large cube / Volume of a small cube
Now, we find the volume of both the cubes
The volume of a small cube, V = (5 cm)³ = 125 cm³.
The volume of a large cube, V = (15 cm)³ = 3375 cm³.
n no. of cubes = 3375/125 = 125.
Hence, 125 cubes of side 5 cm are essential to build a cube of 15 cm.

6. Find the length of the edges of a cube, if the volume is equal to 216 m³.

Solution:

Given, a cube of a volume is 216 m³.
Let, the length of the edges is a³.
By the formula, Volume of a cube, V = (length of the edges of a cube)³ = a³.
Now, substitute the given values
216 = a³
a = ³√216 = 6 m
Hence, the length of a cube is 6 m.

7. What is the volume of a gift box, if the dimensions of a cuboidal gift box are 20 inches, 45 inches, 15 inches?

Solution:

Given that, a gift box is in cuboidal shape.
The dimensions of a gift box are
The length of the gift box is 20 in
The width of the gift box is 45 in
The height of the gift box is 15 in
The volume of the gift box is
V =  length × width × height
V = 20 in× 45 in× 15 in = 13,500 in³.
Therefore, the volume of a cuboidal gift box is 13,500 cubic inches.

8. If the cuboid volume is 212 cm³, its length is 6 cm and height is 8 cm. Find the cuboid breadth?

Solution:

Given that,
Cuboid volume is 212 cm³
Cuboid length is 6 cm
Cuboid height is 8 cm
The formula for the volume of cuboid is, V = length × breadth × height
Substitute the given values in the formula,
212 = 6 × breadth × 8
Cuboid breadth = Volume / (length × height)
Cuboid breadth = 212 / (6 × 8) = 212 / 48 = 4.416 cm.
Thus, the breadth of a cuboid is 4.416 cm. ## Worksheet on Subtraction of Literals | Subtracting Literals Practice Worksheet PDF with Answer Key

Subtracting Literals Worksheet deals with the subtraction of literals using various ways and solve the questions on the concept. Different models of questions used in the Worksheet on Subtraction of Literals make these activity sheets engaging and interesting. Subtraction of Literals is a crucial topic that you will find in algebra concepts. Subtracting Literals Worksheet PDF approaches questions in a stepwise manner so that you won’t find any difficulty in understanding the questions. Download the Subtraction of Literals Worksheet in PDF Formats for free and score well in your exams.

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## Subtracting Literals Practice Worksheet

I. Represent the following statements algebraically:

(i) 12 less than z
(ii)  x less than 54
(iii) 380 taken away from m
(iv) 67 less than x
(v) Decrease the sum of m and n by p
(vi) y less than 220
(vii) y less than the sum of z and 7

Solution:

(i) 12 less than z is represented as z-12
(ii)  x less than 54 is represented as 54-x
(iii) 380 taken away from m is represented as m-380
(iv) 67 less than x is represented as x-67
(v) Decrease the sum of m and n by p is represented as (m+n)-p
(vi) y less than 220 is represented as 220-y
(vii) y less than the sum of z and 7 is y-(z+7)

II. Write each of the following algebraically using signs and symbols:
(i) 357 less than literal y
(ii) Decrease m by o
(iii) Subtract h from 1000
(iv) 185 is diminished by w
(v) n less than 88
(vi) 43 less than p
(vii) 689 less than literal n
(viii) 42 less than v
(ix) 323 is diminished by k

Solution:

(i) 357 less than literal y is represented as y-357.
(ii) Decrease m by o is represented as m – o.
(iii) Subtract h from 1000 is represented as 1000 – h.
(iv) 185 is diminished by w is represented as 185 – w.
(v) n less than 88 is represented as 88 – n.
(vi) 43 less than p is represented as p-43.
(vii) 689 less than literal n is represented as n-689.
(viii) 42 less than v is represented as v-42.
(ix) 323 is diminished by k is represented as 323 – k.

III. Write each of the following statements using numbers, literals, and symbols:
(i) Subtract x from 770
(ii) r less than a sum of q and 20
(iii) 13 less than a sum of b and c
(iv) Decrease the sum of p and q by y
(v) Decrease the sum of k and 10 by l
(vi) x is diminished by 890
(vii) Decrease b by 7

Solution:

(i) Subtract x from 770 is represented as 770-x
(ii) r less than a sum of q and 20 is represented as (q+20)-r
(iii) 13 less than a sum of b and c is represented as (b+c)-13
(iv) Decrease the sum of p and q by y is represented as (p+q)-y
(v) Decrease the sum of k and 10 by l is represented as (K+10)-l
(vi) x is diminished by 890 is represented as x-890
(vii) Decrease b by 7 is represented as b-7

IV. Write the following statements algebraically:
(i) 19 less than literal k
(ii) Decrease m by 9
(iii) Subtract 18 from y
(iv) n took away from d
(v) 48 is diminished by v
(vi) 109 less than a sum of m and n
(vii) 500 less than literal z
(viii) Number 660 less than w

Solution:

(i) 19 less than literal k is represented as k-19
(ii) Decrease m by 9 is represented as m-9
(iii) Subtract 18 from y is represented as y-18
(iv) n took away from d is represented as d-n
(v) 48 is diminished by v is represented as 48-v
(vi) 109 less than a sum of m and n is represented as (m+n)-109
(vii) 500 less than literal z is represented as z-500
(viii) Number 660 less than w is represented as w-660 ## Worksheet on Powers of Literal Numbers | Math Powers of Literal Numbers Worksheet with Answers

Powers of Literal Numbers is nothing but the repeated product of a number with itself written in exponential form. Practice using the Worksheet on Powers of Literal Numbers and know the different models of questions framed on the topic. Use the Powers of Literal Numbers Worksheet PDF as a cheat sheet to self-examine your preparation on the concept.

Math Students are advised to solve the questions from the Powers of Literal Numbers Worksheet with Answers to master the concept as well as to enhance their general math skills. You can also check the Solutions for the Problems on Literal Numbers Powers in case of any doubts and learn how to frame quality answers in your exams and thereby score well.

## Free Printable Worksheet on Powers of Literal Numbers

I. Write each of the following in the exponential form:
(i) c × c × c × c × c
(ii) 5 × c × b × z × z × z
(iii) m × 12 × n × b × z
(iv) m× n × p × q × n × 55
(v) 230 × b × c × c × b
(vi) m × m × n × n × n × 280

Solution:

(i) Given c × c × c × c × c
Here c has written as 5 times.
It can be written as an exponent of 5.
c × c × c × c × c=c5
(ii) Given 5 × c × b × z × z × z
Here 5 is written ‘1’ time.
c has written ‘1’ time.
b has written ‘1’ time.
z has written 3 times.
5 × c × b × z × z × z=5cbz3
(iii) m × 12 × n × b × z
Here m is written ‘1’ time.
12 has written ‘1’ time.
n,b, and Z are written 1 time.
m × 12 × n × b × z=12mnbz

(iv) m× n × p × q × n × 55
Here m is written ‘1’ time.
n has written 2 times.
p,q,55 are written 1 time.
m× n × p × q × n × 55 =55pqmn2
(v) 230 × b × c × c × b
Here 230 is written ‘1’time.
b has written 2 times.
c has written 2 times.
230 × b × c × c × b= 230b2c2

(vi) m × m × n × n × n × 280
Here m is written 2 times.
n has written 3 times.
280 has written ‘1’ time.
m × m × n × n × n × 280=280m2n3

II. Convert each of the following exponential form to (expanded) product form:
(i) y3z2
(ii) p2q3r5
(iii) 5khb2
(iv) 18c3d4h5
(v) 25ab2cd3

Solution:

(i) y3z2
y3 is written in expanded form as y × y × y.
z2 is written in expanded form as z × z.
y3z2= y × y × y × z × z.
(ii) p2q3r5
p2 is written in expanded form as p × p.
q3 is written in expanded form as q × q × q.
r5 is written in expanded form as r × r × r × r × r.
p2q3r5= p × p × q × q × q × r × r × r × r × r.
(iii) 5khb2
b2 is written in expanded form as b × b.
5,h,k is written only once.
5khb2= 5 × k ×h × b × b.
(iv) 18c3d4h5
c3 is written in expanded form as c × c × c.
d4 is written in expanded form as d × d × d × d.
h5 is written in expanded form as h × h × h × h × h.
18c3d4h5=18 × c × c × c × d × d × d × d × h × h × h × h × h.
(v) 25ab2cd3
25, a, c are written once.
b2 is written in expanded form as b × b.
d3 is written in expanded form as d × d × d.
25ab2cd3=25 × a × b × b × c × d × d × d.

III. Write each of the following in product form:
(i) 3p2q4r
(ii) 73b4c2z3
(iii) a4b3c2
(iv) 7p2q3r4
(v) 17ac2dy3

Solution:

(i) Given 3p2q4r
p2=p × p
q4=q × q × q × q
3p2q4r is written in product form as 3 × p × p × q × q × q × q × r.

(ii) Given 73b4c2z3
b4= b × b × b × b
c2= c × c
z3= z × z × z
73b4c2z3 is written in product form as 73 × b × b × b × b × c × c × z × z × z.
(iii) Given a4b3c2
a4= a × a × a × a
b3= b × b × b
c2 = c × c
a4b3c2 is written in product form as a × a × a × a × b × b × b × c × c
(iv)Given 7p2q3r4
p2= p × p
q3= q × q × q
r4= r × r × r × r
7p2q3r4 is written in product form as 7 × p × p × q × q × q × r × r × r × r.
(v) Given 17ac2dy3
c2= c × c
y3= y × y × y
17ac2dy3=17 × a × c × c × d × y × y × y.

IV. Write each of the following products in index (exponential) form:
(i) p × p × p × p x a × a × b × b × b
(ii) 10 × a × a × b × b × b × c
(iii) c × c × c × c × ..… 8 times d × d × d × d × ..… 8 times.
(iv) a × a × a × ..… 15 times b × b × b × ..… 7 times c × c × c× ..… 20 times.
(v) 5 × a × a × a ×…6 times  b × b × b

Solution:

(i) Given p × p × p × p x a × a × b × b × b
Here P is written 4 times. It can be written as an exponent of 4.
p × p × p × p=p4
a was written 2 times. It can be written as an exponent of 2.
a × a=a2
b was written 3 times. It can be written as an exponent of 3.
b × b × b=b3
p × p × p × p x a × a × b × b × b in exponential form is p4a2b3.
(ii) Given 10 × a × a × b × b × b × c
Here a is written 2 times. It can be written as an exponent of 2.
a × a=a2
b was written 3 times. It can be written as an exponent of 3.
b × b × b=b3
10 × a × a × b × b × b × c=10a2b3c.
(iii) Given c × c × c × c × ..… 8 times d × d × d × d × ..… 8 times.
c was written 8 times. It can be written as an exponent of 8.
c × c × c × c × ..… 8 times=c8
d was written 8 times. It can be written as an exponent of 8.
d × d × d × d × ..… 8 times=d8
c × c × c × c × ..… 8 times d × d × d × d × ..… 8 times=c8d8.
(iv) Given a × a × a × ..… 10 times b × b × b × ..… 7 times c × c × c × ..… 14 times.
a was written 10 times. It can be written as an exponent of 10.
a × a × a × ..… 10 times=a10.
b was written 7 times. It can be written as an exponent of 7.
b × b × b × ..… 7 times=b7.
c was written 14 times. It can be written as an exponent of 14.
a × a × a × ..… 10 times b × b × b × ..… 7 times c × c × c × ..… 14 times=a10b7c14.
(v) Given 5 × a × a × a ×…6 times b × b × b
a was written 6 times. It can be written as an exponent of 6.
a × a × a ×…6 times=a6
b was written 3 times. It can be written as an exponent of 3.
b × b × b=b3
5 × a × a × a ×…6 times b × b × b=5a6b3

V. State whether true or false:
(i) m × m × m × m × m × m = m6
(ii) 5 × 2 × a × a × b × b = 52a2b2
(iii) 3 × 7 × b × b × c × c × c = 33b2c3
(iv) k × l × k × l × k × l = k3l3
(v) m × n × 10 × p × q × q× p = 10mnp2q2
(vi) a × a × a × b × b × b × c × c=a3b3c2
(vii) 6 × a × a × b × b × b × c × c=6a2b3c2

Solution:

(i) true
(ii) false
(iii) false
(iv) true
(v) true
(vi) true
(vii) true ## Worksheet on Circle | Free Printable Circle Worksheets with Answers PDF

Worksheet on Circle is a great resource for students to learn complete circle concepts in Math. Practice all the questions provided in the Circle Worksheet to score good marks in the exams. All questions including radius, diameter, Circumference, Area are given in this article.

Check out different ways to solve circle math problems in this article. The Circle Worksheet with Answers has questions on finding the circumference, area, identifying radius, chord, and diameter of a circle, etc. Download the Printable Worksheet on Circle in PDF Formats and practice the questions on a frequent basis and improve your proficiency in the concept.

Do Refer: Practice Test on Circle

1. Explain the following terms of a circle?
(ii) Centre
(iii) Chord
(iv) Diameter
(v) Interior of a Circle

Solution:

Radius – The radius is the line segment from the center of the circle to the circumference or surface of the circle.
Center – The Center of the circle is the middle point that is equidistant from all the points on the edge of the circle.
Chord – The chord on the circle that joins two points on the circumference.
Diameter – The diameter of the circle is defined as double the length of the radius of a circle.

2. The following figure shows a circle with center O and some line segments drawn in it. Classify the line segments as chord, radius, and diameter, etc. (i) OM = ………………..
(ii) OL = ………………..
(iii) ON = ………………..
(iv) AB = ………………..
(v) CD = ………………..
(vi) EF = ………………..

Solution:

(iv) AB = Diameter
(v) CD = Chord
(vi) EF = Chord

3. Find the area and the circumference of a circle whose radius is 30 cm. (Take the value of π = 3.14)

Solution:

Given that the radius is 30 cm.
Area =πr2
Area =  3.14  × (30)2
Area =  2826 cm2
Circumference, C = 2πr
Circumference = 2 × 3.14 × 30
Circumference = 188.4 cm

4. Find the area of a circle whose circumference is 15.7 cm?

Solution:

Given that the circumference is 15.7 cm.
To find the area of a circle, we must find the radius.
From the circumference, the radius can be calculated
2 π r = 15.7
(2)(3.14)r = 15.7
r = $$\frac { 15.7 }{ (2)(3.14) }$$
r = $$\frac { 15.7 }{ 6.28 }$$
r = 2.5
Therefore, the radius of the circle is 2.5 cm.
The area of a circle is πr2 square units
Now, substitute the radius value in the area of a circle formula, we get
A = π(2.5)2
A = 3.14 x 6.25
A =  19.625 cm2

Therefore, the area of a circle is 19.625 cm2.

5. Observe the circles given below and identify them. (b) Chord = …………………………
(c) Diameter = …………………………

Solution:

(a) Radius = OC, OD, OM
(b) Chord = EF, CN
(c) Diameter = CD

6. Take a point and draw a circle of radii 5 cm, 2 cm, 7 cm, each having the same center M.

Solution:

Given that three circles having a radius of 5 cm, 2 cm, 7 cm.
All 3 circles must have the same center named as M.
So, the figure with three different radii is given below. 7. Take two points Q and R on a circle. Draw a circle with Q in the center and passes through R.

Solution:

Given that two points Q and R are present on a circle.
Q is the center of the circle and it passes through R.
So, the figure with three different radii is given below. 8. To cover a distance of 5 km a wheel rotates 2500 times. Find the radius of the wheel?

Solution:

Given that a distance of 5 km a wheel rotates 2500 times.
No. of rotations = 2500.
Total distance covered = 5 km, and we have to find out the radius of the circle.
Let ‘r’ be the radius of the wheel.
Circumference of the wheel = Distance covered in 1 rotation = 2πr.
In 2500 rotations, the distance covered = 5 km = 500000 cm.
Hence, in 1 rotation, the distance covered = 500000 cm/2500 = 200cm
But this is equal to the circumference. Hence, 2πr = 200 cm
r = $$\frac { 200 }{ 2π }$$
r = $$\frac { 100 }{ π }$$
Taking the approximate value of π as 22/7, we get
r = 100 x 7/22
r = 31.82 cm approx.

9. The difference between the circumference and the diameter of a circular bangle is 10 cm. Find the radius of the bangle. (Take π = 22/7)

Solution:

Let the radius of the bangle be ’r’.
According to the question:
Circumference – Diameter = 10 cm
We know that the Circumference of a circle = 2πr
Diameter of a circle = 2r
Therefore, 2πr – 2r =10 cm
2r(π-1) = 10 cm
2r((22/7) – 1) = 10 cm
r($$\frac { 15 }{ 7 }$$) = 5 cm
r = 5 ($$\frac { 7 }{ 15 }$$)
r = 2.333 cm

The radius of the bangle is 2.333 cm.

10. Observe the circles given below and identify Label the centre as O.
Draw 2 radius OA and OB.
Draw the chord CB.

Solution:

Given that the circle and point the centre as O. OA = OB = 2.
The CB is a chord.
The below figure shows the exact details of the given information. 11. Draw a diameter, radius, chord in the given circle using the points. Also, find the length of diameter and radius.
(i) Radius = ……………… = ……………… cm
(ii) Diameter = ……………… = ……………… cm
(iii) Highlight the circumference by using green color.
(iii) Highlight the chord by using blue color. Solution:

Given that the circle and point the centre as O.
Let us consider the radius 2 cm.
OL and OM are the radii of the circle. LM is the diameter of the circle.
The MN and NL are the is a chord of the circle.
(i) OL = OM = 2 cm
(ii) LM = 4 cm = 2 (OL)
The below figure shows the exact details of the given information. (i) 6 cm
(ii) 4 cm
having the same circle.

Solution:

Given that two circles with radii of 4 cm and 6 cm.
Both circles must have the same center.
Let the centre be X. Then, the first circle will have a radius of 4 cm. The second circle will have a radius of 6 cm. 13. Draw a circle of radius 5 cm.

Solution:

Take a Circle with centre O. Draw the radius of the circle 5 cm.
The radius of the circle is the line segment from the center to any point on the circumference.
Let the point be A.
OA = 5 cm. 14. Draw a circle of diameter 7.5 cm.

Solution:

Take a Circle with centre O. Draw the diameter of the circle 7.5 cm.
The diameter of the circle is defined as double the length of the radius of a circle.
Let the points of the diameter are A and B.
AB = 7.5 cm. 15. Draw a circle with centre M and radius 2.4 cm. Mark points A, B, C such that A lies in the interior of the circle, B lies on the circle and C lies in the exterior of the circle.

Solution:

Take a Circle with centre M. Draw the given points on the circle.
A lies inside the circle. B lies on the circle and C lies outside the circle. 16. Draw a circle whose diameter is 20 cm. Find its radius?

Solution:

Given that the diameter of the circle is 20 cm.
The diameter is double the length of the radius.
d = 2r where d is the diameter of the circle and r is the radius of the circle.
20 = 2r
r = 10 cm.
Therefore, the radius of the circle is 10 cm.

17. Find the radius if the diameter of the circle is:
(i) 6 cm
(ii) 16 cm
(iii) 14 m
(iv) 18 cm

Solution:

(i) Given that the diameter of the circle is 6 cm.
The diameter is double the length of the radius.
d = 2r where d is the diameter of the circle and r is the radius of the circle.
6 = 2r
r = 3 cm.
Therefore, the radius of the circle is 3 cm.

(ii) Given that the diameter of the circle is 16 cm.
The diameter is double the length of the radius.
d = 2r where d is the diameter of the circle and r is the radius of the circle.
16 = 2r
r = 8 cm.
Therefore, the radius of the circle is 8 cm.

(iii) Given that the diameter of the circle is 14 cm.
The diameter is double the length of the radius.
d = 2r where d is the diameter of the circle and r is the radius of the circle.
14 = 2r
r = 7 cm.
Therefore, the radius of the circle is 7 cm.

(iv) Given that the diameter of the circle is 18 cm.
The diameter is double the length of the radius.
d = 2r where d is the diameter of the circle and r is the radius of the circle.
18 = 2r
r = 9 cm.
Therefore, the radius of the circle is 9 cm.

18. Find the diameter if the radius of the circle is:
(i) 22 cm
(ii) 5 cm
(iii) 20 cm
(iv) 19 cm

Solution:

(i) Given that the radius of the circle is 22 cm.
The radius is half of the diameter of the circle.
d = 2r where d is the diameter of the circle and r is the radius of the circle.
d = 2r
d = 2 (22) cm.
d = 44 cm.
Therefore, the diameter of the circle is 44 cm.

(ii) Given that the radius of the circle is 5 cm.
The radius is half of the diameter of the circle.
d = 2r where d is the diameter of the circle and r is the radius of the circle.
d = 2r
d = 2 (5) cm.
d = 10 cm.
Therefore, the diameter of the circle is 10 cm.

(iii) Given that the radius of the circle is 20 cm.
The radius is half of the diameter of the circle.
d = 2r where d is the diameter of the circle and r is the radius of the circle.
d = 2r
d = 2 (20) cm.
d = 40 cm.
Therefore, the diameter of the circle is 40 cm.

(iv) Given that the radius of the circle is 19 cm.
The radius is half of the diameter of the circle.
d = 2r where d is the diameter of the circle and r is the radius of the circle.
d = 2r
d = 2 (19) cm.
d = 38 cm.
Therefore, the diameter of the circle is 38 cm.

19. With the same circle, draw three circles first with a radius of 3 cm. second with a radius of 5 cm, and third with a radius of 7 cm.

Solution:

Given that three circles having a radius of 3 cm, 5 cm, 7 cm.
All 3 circles must have the same center named as M.
So, the figure with three different radii is given below. 20. A circle has a radius 8 cm. Find the length of the longest chord of this circle?

Solution:

Given that a circle has a radius of 8 cm.
The longest chord of this circle is a diameter.
d = 2r where d is the diameter of the circle and r is the radius of the circle.
d = 2r
d = 2 (8) cm.
d = 16 cm.
Therefore, the diameter or the length of the longest chord of the given circle is 16 cm. ## Worksheet on Addition and Subtraction of Polynomials | Adding and Subtracting Polynomials Worksheet

Worksheet on Addition and Subtraction of Polynomials will assist students to learn the concept of addition, subtraction of polynomials. Polynomials play a key role in laying the foundation for several topics related to algebra and students need to know about how to perform arithmetic operations on them.

Access the numeric questions available in the Adding and Subtracting Polynomials Worksheet that are well structured and practice using them on a regular basis. The Math Printable Worksheet for Adding and Subtracting Polynomials will develop logical and reasoning skills among you.

I. Simplify the following polynomial expressions:

(i) (10x2-5x+5)+(8x+10)
(ii) Take 3a + 9b – c from 7a + 12b – 4c

Solution:

(i) Given (10x2-5x+5)+(8x+10)
Arrange the like terms together and then add,
=10x2-5x+8x+5+10
=10x2+3x+15
Hence, By simplifying (10x2-5x+5)+(8x+10) we get 10x2+3x+15.
(ii) Take 3a + 9b – c from 7a + 12b – 4c
i.e. we have to Subtract 3a + 9b – c from 7a + 12b – 4c.
=7a+12b-4c-(3a+9b-c)
=7a+12b-4c-3a-9b+c
=7a-3a+12b-9b-4c+c
=4a+3b-3c
Hence, By simplifying 3a + 9b – c from 7a + 12b – 4c we get 4a+3b-3c.

II. The sum of two expressions is 5x2 – 4xy + 2y2. If one of them is 3x2 + y2, find the other?

Solution:

Given,
The sum of the two expressions is= 5x2 – 4xy + 2y2
one of the expression is=3x2 +y2Let the other expression be p(x).
3x2 + y2+P(x)=5x2 – 4xy + 2y2
p(x)=5x2 – 4xy + 2y2-(3x2 + y2)
p(x)=5x2 – 4xy + 2y2-3x2 – y2
=2x2 -4xy+y2
Therefore, the other expression is 2x2 -4xy+y2.

III. Evaluate the following expressions
i) ( 15a2 + 10a − 6a ) + (25a + 10)
ii) On adding 14a2 +8𝑎4 with 3a2+ 2a4+4

Solution:

i) Given ( 15a2 + 10a − 6a ) + (25a + 10)
=15a2 + 10a − 6a+25a+10
=15a2+29a+10
Hence, By adding ( 15a2 + 10a − 6a ) + (25a + 10) we get 15a2+29a+10.
ii) add 14a2 +8𝑎4 ,3a2+ 2a4+4
=14a2 +8𝑎4+3a2+ 2a4+4
=17a2+10𝑎4+4
Hence, By adding 14a2 +8𝑎4 ,3a2+ 2a4+4 we get 17a2+10𝑎4+4.

IV. Subtract the following polynomial expression
(-150a2-54a+5)-(34a2+66a)

Solution:

Given, (-150a2-54a+5)-(34a2+66a)
=-150a2-54a+5-34a2-66a
=-184a2-120a+5
Hence, By subtracting 34a2+66a from -150a2-54a+5 we get -184a2-120a+5.

V. Solve the following

i) How much is 4x + 2y greater than x – 2y?
ii) How much should p + 2q – r be increased to get 5p+q?

Solution:

i) Subtract x-2y from 4x+2y.
=4x+2y-(x-2y)
=4x+2y-x+2y
=4x-x+4y
=3x+4y
Hence, 4x + 2y greater than x – 2y by 3x+4y.
ii) Let p + 2q – r be increased by x.
p+2q-r+x=5p+q
x=5p+q-(p+2q-r)
=5p+q-p-2q+r
=4p-q+r
Hence, p + 2q – r be increased to 4p-q+r to get 5p+q.

VI. From the sum of 5x + y – 8z and 2y – 4z, subtract 2x – y – 3z.

Solution:

Sum of 5x + y – 8z and 2y – 4z is
=5x + y – 8z+2y-4z
=5x+3y-12z
Subtract 2x – y – 3z from 5x+3y-12z
=5x+3y-12z-(2x – y – 3z)
=5x+3y-12z-2x+y+3z
=3x+4y-9z
Hence, By subtracting 2x – y – 3z from the sum of 5x + y – 8z and 2y – 4z we get 3x+4y-9z.

VII. Subtract 2x – 3y – z from the sum of 6x – 7y + 2z and 2x + y – 4z.

Solution:

Sum of 6x – 7y + 2z and 2x + y – 4z is=6x-7y+2z+2x+y-4z
=6x+2x-7y+y+2z-4z
=8x-6y-2z
Now, Subtract 2x – 3y – z from 8x-6y-2z.
=8x-6y-2z-(2x-3y-z)
=8x-2x-6y+3y-2z+z
=6x-3y-z
Hence, By subtracting 2x – 3y – z from the sum of 6x – 7y + 2z and 2x + y – 4z is 6x-3y-z.

VIII. By how much 8a2 – 2b2 be diminished to give 3a2 + b2?

Solution:

Let the value be x.
x-(8a2 – 2b2)=3a2 + b2
x=3a2 + b2 +(8a2 – 2b2)
x=11a2-b2
Therefore, 11a2-b2 be diminished to give 3a2 + b2.

IX. Subtract m2 – 5n + 7 from 4m2 + 2n and add your result to m2+ n – 1.

Solution:

Subtract m2 – 5n + 7 from 4m2 + 2n
=4m2 + 2n-(m2 – 5n + 7)
=4m2 + 2n-m2 + 5n -7
=4m2-m2+2n+5n-7
=3m2+7n-7
Add 3m2+7n-7 to m2+ n – 1.
=3m2+7n-7+m2+ n – 1
=4m2+8n-8
Therefore, the sum of 3m2+7n-7 and m2+ n – 1 is 4m2+8n-8.

x. Subtract 2m2n – 3n2 from 6m2n– 2mn2+ 8n2 and subtract your result from the sum of the two expressions 3mn2 + 5n2 – 2m2n and m2n + 5m2 – 2mn2.

Solution:

Subtract 2m2n – 3n2 from 6m2n– 2mn2+ 8n2
=6m2n– 2mn2+ 8n2-(2m2n-3n2)
=6m2n-2m2n-2mn2+8n2+3n2
=4m2n-2mn2+11n2
The sum of the two expressions 3mn2 + 5n2 – 2m2n and m2n + 5m2 – 2mn2 is
= 3mn2 + 5n2 – 2m2n + m2n + 5m2 – 2mn2
=3mn2-2mn2+5n2-2m2n + m2n + 5m2
=mn2– m2n +5m2 + 5n2
Subtract 4m2n-2mn2+11n from mn2– m2n +5m2 + 5n2
=mn2– m2n +5m2 + 5n2– (4m2n-2mn2+11n2)
=mn2+2mn2-m2n-4m2n+5m2+5n2-11n2
=3mn2-5m2n+5m2-6n2
Therefore, the result is 3mn2-5m2n+5m2-6n2. ## Worksheet on Multiplying Monomial and Polynomial | Multiplying a Polynomial by a Monomial Worksheet with Answers

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## Multiplication of a Monomial and Polynomial Worksheet

I. Multiply monomial by polynomial:
(i) 4x and (2x – 3y + 5z)
(ii) (-5m) and (3m – 4n + 2p)
(iii) 6xyz and (-7xy – 2yz – zx)
(iv) 7a3b2c2 and (4a2b – 2a3c2 – b3c)
(v) (-2x3y2z4) and (4x4y3 – 3x3y2z3 – 6xy2z2y)

Solution:

(i) Given 4x and (2x – 3y + 5z)
Multiply the monomial with every term of the polynomial
=4x(2x)-3y(4x)+5z(4x)
=8x2-12xy+20xz
Hence, By multiplying 4x and (2x – 3y + 5z) is 8x2-12xy+20xz.
(ii) Given (-5m) and (3m – 4n + 2p)
Multiply the monomial with every term of the polynomial
=(-5m)(3m) -4n(-5m)+ 2p(-5m)
=-15m2+20mn-10pm
Hence, By multiplying (-5m) and (3m – 4n + 2p) we get -15m2+20mn-10pm.
(iii) Given 6xyz and (-7xy – 2yz – zx)
Multiply the monomial with every term of the polynomial
=6xyz(-7xy)-2yz(6xyz)-zx(6xyz)
=-42x2y2z-12xy2z2-6x2yz2
Hence, By multiplying 6xyz and (-7xy – 2yz -zx) we get -42x2y2z-12xy2z2-6x2yz2.
(iv) Given 7a3b2c2 and (4a2b – 2a3c2 – b3c)
Multiply the monomial with every term of the polynomial
=4a2b(7a3b2c2)-2a3c2(7a3b2c2 )-b3c(7a3b2c2)
=28a5b3c2-14a6b2c4-7a3b5c3
Hence, By multiplying 7a3b2c2 and (4a2b – 2a3c2 – b3c)  we get 28a5b3c2-14a6b2c4-7a3b5c3  .
(v) Given (-2x3y2z4) and (4x4y3 – 3x3y2z3 – 6xy2z2)
Multiply the monomial with every term of the polynomial
=4x4y3((-2x3y2z4) -3x3y2z3(-2x3y2z4) -6xy2z2(-2x3y2z4)
=-8x7y5z4+6x6y4z7+12x4y4z6
Hence, By multiplying (-2x3y2z4)  and (4x4y3 – 3x3y2z3 – 6xy2z2)  we get -8x7y5z4+6x6y4z7+12x4y4z6.

II. Multiply polynomial by monomial:
(i) (m+ m4 + 1) and 5m
(ii) (ax2 + bx3 + 5x) and x2
(iii) (2x + xz + z3) and 7z
(iv) (m – 2mn + 8n) and (–m7)
(v) (a + 2bc + ca) and (-a)

Solution:

(i) Given, (m+ m4 + 1) and 5m
Multiply the monomial with every term of the polynomial
=m(5m) + m4 (5m) + 1(5m)
=5m2+5m5+5m
Therefore, By multiplying (m+ m4 + 1) and 5m we get 5m2+5m5+5m.
(ii) Given, (ax2 + bx3 + 5x) and x2
Multiply the monomial with every term of the polynomial
=ax2(x2) + bx3(x2) + 5x(x2)
=ax4+bx5+5x3
Therefore, By multiplying (ax2 + bx3 + 5x) and x2 we get ax4+bx5+5x3.
(iii) Given, (2x + xz + z3) and 7z
Multiply the monomial with every term of the polynomial
=2x(7z) + xz(7z) + z3 (7z)
=14xz + 7xz2 + 7z4
Therefore, By multiplying (2x + xz + z3) and 7z is 14xz + 7xz2 + 7z4.
(iv) Given, (m – 2mn + 8n) and (–m7)
Multiply the monomial with every term of the polynomial
=m(–m7) – 2mn(–m7) + 8n(–m7)
=-m8+2m8n-8m7n
Therefore, By multiplying (m – 2mn + 8n) and (–m7) we get -m8+2m8n-8m7n.
(v) Given,(a + 2bc + ca) and (-a)
Multiply the monomial with every term of the polynomial
=a(-a)+2bc(-a)+ca(-a)
=-a2-2abc-a2c
Therefore, By multiplying (a + 2bc + ca) and (-a) we get -a2-2abc-a2c.

III. Find the product of the following:
(i) 3ab(2ab + b2c + 5ca)
(ii) (-13m2)(5 + mx + ny)
(iii) 2m2n(mn + n – n2)
(iv) mn(m2+n2)
(v) -6a2bc(3ab + bc – 7ca)
(vi) (x+3y)(2x+6y)

Solution:

(i) Given, 3ab(2ab + b2c + 5ca)
Multiply the monomial with every term of the polynomial
=3ab(2ab)+b2c(3ab) + 3ab(5ca)
=6a2b2 + 3ab3c + 15a2bc
Hence, By multiplying 3ab(2ab + b2c + 5ca) we get 6a2b2 + 3ab3c + 15a2bc.
(ii) Given, (-13m2)(5 + mx + ny)
Multiply the monomial with every term of the polynomial
=(-13m2)5 + mx((-13m2) + ny(-13m2)
=-65m2-13m3x-13m2ny
Hence, By multiplying (-13m2)(5 + mx + ny) we get -65m2-13m3x-13m2ny.
(iii) Given, 2m2n(mn + n – n2)
Multiply the monomial with every term of the polynomial
=2m2n(mn) + n(2m2n)-n2(2m2n)
=2m3n2+2m2n2-2m2n3
Hence, By multiplying 2m2n(mn + n – n2) we get 2m3n2+2m2n2-2m2n3.
(iv) Given, mn(3m2+2n2)
Multiply the monomial with every term of the polynomial
=mn(3m2) + mn(2n2)
=3m3n + 2mn3
Hence, By multiplying mn(m2+n2) we get 3m3n + 2mn3.
(v) Given, -6a2bc(3ab + bc – 7ca)
Multiply the monomial with every term of the polynomial
=-6a2bc(3ab) + bc(-6a2bc) -7ca(-6a2bc)
=-18a3b2c-6a2b2c2+42a3bc2
Hence, By multiplying -6a2bc(3ab + bc – 7ca) we get -18a3b2c-6a2b2c2+42a3bc2.
(vi) Given, (x+3y)(2x+6y)
By using the distributive property, multiply the polynomials,
=x(2x+6y) + 3y(2x+6y)
=2×2+6xy+6yx+18y2
=2×2+12xy+18y2
Hence, By multiplying (x+3y)(2x+6y) we get 2×2+12xy+18y2.

IV. The product of two numbers is 3m4 if one of them is 1/4m2. Find the other?

Solution:

Given,
The product of two numbers is =3m4
one of the number=1/4m2
Let the other number be x.
1/4m2 × x=3m4
x=3m4.4m2
x=12m6
Therefore, the other number is 12m6.

V. If P=5x2+2x, Q=2x, R=24. Find the value of (P × R)/Q.

Solution:

Given P=5x2+2x, Q=2x+2, R=24
P × R=24(5x2+2x)
=120x2 + 48x
(P × R)/Q= (120x2 + 48x)/2x
=60x+24
Hence, the value of (P × R)/Q is 60x+24.

VI. If the length and width of the rectangle are (-4a2+7) and (2a + 5) respectively. Find the area of the rectangle?

Solution:

Given,
The length of the rectangle=(-4a2+7)
The breadth of the rectangle=(2a + 5)
Area of the rectangle=length * breadth
=(-4a2+7) * (2a + 5)
=-4a2(2a+5) + 7(2a+5)
=-8a3-20a2+14a+35