Class 8 Maths

• Chapter 1 Rational Numbers
• Chapter 2 Linear One Equations In Variable
• Chapter 3 Understanding Quadrilaterals
• Chapter 4 Data Handling
• Chapter 5 Squares And Square Roots
• Chapter 6 Cubes And Cube Roots
• Chapter 7 Comparing Quantities
• Chapter 8 Algebraic Expressions And Identities
• Chapter 9 Mensuration
• Chapter 10 Exponents And Powers
• Chapter 11 Direct And Inverse Proportions
• Chapter 12 Factorisation
• Chapter 13 Introduction To Graphs

NCERT Class 8 Maths Chapter 11 Direct And Inverse Proportions Introduction

Question 1. Mohan prepares tea for himself and his sister. He uses 300 mL of water, 2 spoons of sugar, 1 spoon of tea leaves and, 50 mL of milk. How much quantity of each item will he need, if he has to make tea for five persons?
Solution:

Quantity of water required for 2 persons = 300 mL

Quantity of water required for 1 person = \(\frac{300}{2} \mathrm{~mL}\)

Quantity of water required for 5 persons \(\frac{300}{2} \times 5 \mathrm{~mL}\)

Quantity of sugar required for 2 persons = 2 spoons

Quantity of sugar required for 1 person \(\frac{2}{2} \text { spoons }\)

Read and Learn More NCERT Solutions For Class 8 Maths

Quantity of sugar required for 5 persons \(\frac{2}{2} \times 5=5 \text { spoons }\)

Quantity of tea leaves required for 2 persons = 1 spoon

Quantity of tea leaves required for 1 person \(\frac{1}{2} \text { spoons }\)

Quantity of tea leaves required for 5 persons \(=\frac{1}{2} \times 5=2 \frac{1}{2} \text { spoons }\)

Quantity of milk required for 2 persons = 50 mL

Quantity of milk required for 1 person = \(\frac{50}{2} \mathrm{~mL}\)

Quantity of milk required for 5 persons = \(\frac{50}{2} \times 5 \mathrm{~mL}=125 \mathrm{~mL}\)

Thus, Mohan will require 750 mL of water, 5 spoons of sugar, \(2 \frac{1}{2}\) Spoons of tea leaves, and 125 mL of milk for 5 persons.

Question 2. If two students take 20 minutes to arrange chairs for an assembly, then how much time would five students have to do the same job?
Solution:

Time taken by 2 students = 20 minutes

Time taken by 1 student = 20 x 2 minutes

Time taken by 5 students = \(\frac{20 \times 2}{5} \text { minutes }=8 \text { minutes }\)

We come across many such situations in our day-to-day life, where we need to see variation in one quantity bringing in variation in the other quantity. Quantities so related are called variables and the relation is called variation.

For example :

1. If the number of articles purchased increases, the total cost also increases.
2. The more money deposited in a bank, more is the interest earned.
3. As the speed of a vehicle increases, the time taken to cover the same distance decreases.
4. For a given job, the more the number of workers less will be the less time taken to complete the work.

Question 3. Write five more such situations where a change in one quantity leads to  change in another quantity.
Solution:

Five more such situations where a change in one quantity leads to a change in another quantity are as follows :

1. If the amount of the commodity purchased increases, the total cost also increases.
2. If the amount of loan taken from a bank increases, then the interest to be paid also increases.
3. At a particular temperature, if the pressure of gas increases, then the volume decreases.
4. If the number of pipes to fill a tank increases, the time taken to fill the tank decreases.
5. If the number of guests increases, the number of days for which a given amount of food will last decreases.

Question 4. How do we find out the quantity of each item needed by Mohan? or, the time five students take to complete the job?
Solution:

We need to study some concepts of variation as follows :

1. Direct variation
2. Inverse variation

Two quantities may be linked in two ways :

Both increase or decrease together proportionally.

If one increases, the other decreases proportionally, and vice-versa.

The first way is named a direct variation whereas the second way is named as an inverse variation.

1. Direct Proportion

If two quantities are related in such a way that an increase in one leads to a corresponding proportional increase in the other, then such a variation is called direct variation.

Thus, two numbers x and y are said to vary in direct proportion if

\(\frac{x}{y}\)= k or x = ky where k is a constant.

In this case, if ÿ1,y2 are the values ofy corresponding to the values x1, x2 of x respectively, then

⇒  \(\frac{x_1}{y_1}=\frac{x_2}{y_2}\)

Question 5. Let the consumption of petrol be x liters and the corresponding distance traveled by km. Now, complete, the following table:

We find that as the value increases, the value also increases in such a way that the ratio \(\frac{x}{y}\) does not change; it remains constant (say It). In this case, it is……(check it !)
Solution:

In this case, it is \(\frac{1}{15}\)

Question 6. Think of a few more examples for direct proportion.
Solution:

1. A few more examples of direct proportion are as follows :
2. Length of the cloth purchased and its total cost.
3. Number of months and total salary.
4. Number of hours of production and the amount of the commodity produced.

Take a clock and fix its minute hand, at 12.

Record the angle turned through by the minute hand from its original position and the time that has passed, in the following table

What do you observe about T and A ? Do they increase together? Is \(\frac{T}{A}\) same every time ?

Is the angle turned through by the minute hand directly proportional to the time that has passed? Yes; From the above table, you can also see

T₁ : T₂ = A₁ : A₂, because

T₁ : T₂ = 15 : 30 = 1:2

A₁ : A₂ = 90 : 180 = 1:2

Check if T₂ : T₃ = A₂ : A₃ and T₃ : T₄ = A₃ : A₄

You can repeat this activity by choosing your time interval.

We observe about T and A that they increase together and \(\frac{T}{A}\)is the same every time. A Yes, the angle turned by the minute hand is directly proportional to the time that has passed.

On checking, we find that

T₂ : T₃ = A₂ : A₃ = 2:3

T₃ : T₄ = A₃ : A₄ = 3:4

Ask your friend to fill in the following table and find the ratio of his age to the corresponding age of his mother:

What do you observe?

Do F and M increase (or decrease) together? Is — the same every time? No.

You can repeat this activity with other friends and write down your observations.

Solution:

We observe that F and M increase (or decrease) together. But — is not the same every
time.

Note: It is not. necessary that the variables increasing (or decreasing) together are always in direct proportion.

For example :

Height and age of an individual

Height and weight of an individual

Height of mango tree and the number of mango fruits growing on its branches.

Question 7. Observe the following tables and find if x and y are directly proportional:

Solution:

We have \(\frac{20}{40}=\frac{1}{2}\)

⇒ \(\frac{17}{34}=\frac{1}{2}\)

⇒ \(\frac{14}{28}=\frac{1}{2}\)

⇒ \(\frac{11}{22}=\frac{1}{2}\)

⇒ \(\frac{8}{16}=\frac{1}{2}\)

⇒ \(\frac{5}{10}=\frac{1}{2}\)

⇒ \(\frac{2}{4}=\frac{1}{2}\)

We find that each ratio is the same.

Hence, x and y are directly proportional.

We have \(\frac{6}{4}=\frac{3}{2}\)

⇒ \(\frac{10}{8}=\frac{5}{4}\)

⇒ \(\frac{14}{12}=\frac{7}{6}\)

⇒ \(\frac{18}{16}=\frac{9}{8}\)

⇒ \(\frac{22}{20}=\frac{11}{10}\)

⇒ \(\frac{26}{24}=\frac{13}{12}\)

⇒ \(\frac{30}{28}=\frac{15}{14}\)

We find that all the ratios are not the Same.

Hence, x and y are not directly proportional.

⇒ \(\frac{5}{15}=\frac{1}{3}\)

⇒ \(\frac{8}{24}=\frac{1}{3}\)

⇒ \(\frac{12}{36}=\frac{1}{3}\)

⇒ \(\frac{15}{60}=\frac{1}{4}\)

⇒ \(\frac{18}{72}=\frac{1}{4}\)

⇒ \(\frac{20}{100}=\frac{1}{5}\)

We find that all the ratios are not the same.

Hence, x and y are not directly proportional.

Question 8. Principal = f1,000, Rate = 8% per annum. Fill in the following table and find which type of interest (simple or compound) changes in direct proportion with time.

Solution:

1 year \(\frac{\mathrm{P} \times r \times t}{100}\)

⇒ \(₹ \frac{1000 \times 8 \times 1}{100}\)

₹ 80

We have \(\frac{80}{1}=80\)

2 years \(\frac{\mathrm{P} \times r \times t}{100}\)

⇒ \(₹ \frac{1000 \times 8 \times 2}{100}\)

₹ 160

\(\frac{160}{2}=80,\)

3 years \(\frac{\mathrm{P} \times r \times t}{100}\)

⇒\(₹ \frac{1000 \times 8 \times 3}{100}\)

₹ 240

⇒ \(\frac{240}{3}=80\)

We find that each ratio is the same.

Hence, simple interest changes in direct proportion with a period.

Compound Interest (in ?) for different periods

1 year \(\mathrm{P}\left(1+\frac{r}{100}\right)^t-\mathrm{P}\)

⇒ \(1000\left(1+\frac{8}{100}\right)-1000\)

⇒ \(1080-1000=₹ 80\)

We have \(\frac{80}{1}=80\)

2 years \(\mathrm{P}\left(1+\frac{r}{100}\right)^t-\mathrm{P}\)

v\(1000\left(1+\frac{8}{100}\right)^2-1000\)

₹ 166.40

⇒ \(\frac{166.40}{2}=83.20\)

3 years \(\mathrm{P}\left(1+\frac{r}{100}\right)^t-\mathrm{P}\)

⇒ \(1000\left(1+\frac{8}{100}\right)^3-1000\)

₹ 259.712

⇒ \(\frac{259.712}{3}=86.5706\)

We find that all the ratios are not the same.

Hence, compound interest does not change in direct proportion to the period.

Question 9. If we fix the period and the rate of interest, simple interest changes proportionally with the principal. Would there be a similar relationship for compound interest? Why?
Solution:

Case 1

⇒ \(\text { S.I. }=\frac{\mathrm{P} r t}{100}\)

⇒ \(\frac{\text { S.I. }}{\mathrm{P}}=\frac{r t}{100}\)

r and t are fixed

⇒ \(\frac{r t}{100} \text { is constant }\)

⇒ \(\frac{\mathrm{S} . \mathrm{I}}{\mathrm{P}}=\text { constant }\)

S.I. changes proportionally with principal

Case 2

⇒ \(\text { C.I. }=\mathrm{P}\left\{\left(1+\frac{r}{100}\right)^t-1\right\}\)

⇒ \(\frac{\mathrm{C} . \mathrm{I}}{\mathrm{P}}=\left(1+\frac{r}{100}\right)^t-1\)

r and tare fixed

⇒ \(\left(1+\frac{r}{100}\right)^t-1 \text { is constant }\)

⇒ \(\frac{\mathrm{C} . \mathrm{I}}{\mathrm{P}}=\text { constant }\)

C.I. changes proportionally with principal

There is a similar relationship for compound interest

Question 10. Take a map of your State. Note the scale used there. Using a ruler, measure the “map distance’’ between any two cities. Calculate the actual distance between them.
Solution:

Let the scale be 1 cm = 100 km

Let the distance between any two cities on the map be 5 cm.

Then, the actual distance between these two cities = 100 x 5 km = 500 km

Direct And Inverse Proportions Exercise 11.1

Question 1. Following are the car parking charges near a railway station up to

1. 4 hours ₹ 60
2. 8 hours ₹ 100
3. 12 hours ₹ 140
4. 24 hours ₹ 180

Check if the parking charges are in direct proportion to the parking time.
Solution:

We have

⇒ \(\frac{60}{4}=\frac{60 \div 4}{4 \div 4}=\frac{15}{1}\) HCF (60, 4) = 4

⇒ \(\frac{100}{8}=\frac{100 \div 4}{8 \div 4}=\frac{25}{2}\) HCF (100, 8) = 4

⇒ \(\frac{140}{12}=\frac{140 \div 4}{12 \div 4}=\frac{35}{3}\) HCF (140, 12) = 4

⇒ \(\frac{180}{24}=\frac{180 \div 12}{24 \div 12}=\frac{15}{2}\) HCF (180, 24) = 12

Since all the ratios are not the same, therefore, the parking charges are not in direct proportion to the parking time.

Question 2. A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of the base. In the following table, find the parts of the base that need to be added

Solution:

Let the number of parts of red pigments be x and the number of parts of the base is y.

As the number of parts of red pigments increases, several parts of the base also increase in the same ratio. So it is a case of direct proportion.

We make use of the relation of the type \(\frac{x_1}{y_1}=\frac{x_2}{y_2}\)

Here

x₁= 1

y₁ = 8 and x₂ = 4

Therefore \(\frac{x_1}{y_1}=\frac{x_2}{y_2} \text { gives }\)

⇒ \(\frac{1}{8}=\frac{4}{y_2}\)

y₂ = 8 x 4

y₂ = 32

Hence, 32 parts of the base are needed to be added to 4 parts of red pigments.

Here,

x₁ = 1

y₄ = 8 and x3 = 7

Therefore \(\frac{x_1}{y_1}=\frac{x_3}{y_3}\) gives

⇒ \(\frac{1}{8}=\frac{7}{y_3}\)

y₃ = 8 x 7

y₃ = 56

Hence, 56 parts of the base are needed to be added to 7 parts of red pigments.

Here

x₁= 1

y₁= 8

x₄ = 12

Therefore,\(\frac{x_1}{y_1}=\frac{x_4}{y_4}\) gives

⇒ \(\frac{1}{8}=\frac{12}{y_4}\)

y4= 12 x 8

y₄ = 96

Hence, 96 parts of the base are needed to be added to 12 parts of red pigments.

Here,

x₁= 1

y₁ = 8

x₅ = 20

Therefore \(\frac{x_1}{y_1}=\frac{x_5}{y_5}\)gives

⇒ \(\frac{1}{8}=\frac{20}{y_5}\)

y₅ = 8 x 20

y₅ = 160

Hence, 160 parts of the base are needed to be added to 20 parts of red pigments.

Question 3. In Question 2 above, if part of the red pigment requires 75 mL of the base, how much red pigment should we mix with 1800 mL of the base?
Solution:

Let the number of parts of red pigments be and the amount of base be y mL.

As the number of parts of red pigment increases, the amount of base also increases in the same ratio. So it is a case of direct proportion.

We make use of the relation of the type.

⇒ \(\frac{x_1}{y_1}=\frac{x_2}{y_2}\)

Here x1=1

y₂ = 75 and ,y₂ = 1800

⇒ \(\frac{x_1}{y_1}=\frac{x_2}{y_2}\) give

⇒ \(\frac{1}{75}=\frac{x_2}{1800}\)

⇒ \(\frac{1}{75} =\frac{x_2}{1800}\)

⇒ \(75 x_2 =1800 \)

⇒ \(x_2 =\frac{1800}{75}=\frac{600}{25}\)

x₂ = 24

Hence, 24 parts of the red pigment should be mixed with 1800 mL of base.

⇒ \(\frac{x_1}{y_1}=\frac{x_2}{y_2} \Rightarrow x_1 y_2=x_2 y_1\)

⇒ \(\frac{x_1}{x_2}=\frac{y_1}{y_2}\)

Question 4. A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Solution:

Let the machine fill x bottles in five hours. We put the given information in the form of a table as shown below :

The more the number of hours, the more the y2 number of bottles would be filled. So, the number of bottles filled and the number of hours are directly proportional to each other.

⇒ \(\frac{x_1}{x_2}=\frac{y_1}{y_2}\)

⇒ \(\frac{840}{x_2}=\frac{6}{5}\)

6×2 = 840 x 5

⇒ \(x_2=\frac{840 \times 5}{6}=700\)

Hence, 700 bottles will be filled in five hours.

Question 5. A photograph of bacteria enlarged 50,000 limes allies a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what, would be its enlarged length?

Solution:

Actual length of the bacteria \(\frac{5}{50000} \mathrm{~cm}=\frac{1}{10000} \mathrm{~cm}=10^{-4} \mathrm{~cm}\)

Let the enlarged length be cm. We put the given information in the form of a table as shown below :

⇒ \(\frac{x_1}{y_1}=\frac{x_2}{y_2}\)

⇒ \(\frac{50000}{5}=\frac{20000}{y_2}\)

50000 y₂ = 5 x 20000

⇒ \(y_2=\frac{5 \times 20000}{50000}=\frac{5 \times 2}{5}\)

y₂ = 2

Hence, its enlarged length would be 2 cm.

Question 6. In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m. high. If the length of the ship is 28 m, how long is the model ship?

Solution:

Let the length of the model ship be x₂ cm.

We form a table as shown below :

The more the length of the ship, the more would be the length of its mast. Hence, this is a case of direct proportion. That is,

⇒ \(\frac{x_1}{y_1}=\frac{x_2}{y_2}\)

⇒ \(\frac{28}{12}=\frac{x_2}{9}\)

12 x₂ = 28 x 9

⇒ \(x_2=\frac{28 \times 9}{12}=\frac{28 \times 3}{4}\)

x₂ = 21

Hence, the length of the model ship is 21 m

Question 7. Suppose 2 kg of sugar contains 9 Y.106 crystals. How many sugar crystals are there in (i) 5 kg of sugar? (ii) 1.2 kg ofsugar ?
Solution:

Suppose the amount of sugar is x kg and the number of crystals is y. We put the given information in the form of a table as shown below :

As the amount of sugar increases, the number of crystals also increases in the same ratio. So it is an ease of direct proportion. We make use of the relation of the type \(\frac{x_1}{y_1}=\frac{x_2}{y_2}\)

Here, x₁ = 2

y₁ = 9 x 106

x₂= 5

Therefore \(\frac{x_1}{y_1}=\frac{x_2}{y_2}\) gives

⇒ \(\frac{2}{9 \times 10^6}=\frac{5}{y_2}\)

2y₂= 5 x 9 x 106

⇒ \(y_2=\frac{5 \times 9 \times 10^6}{2}\)

y₂ = 22.5 x 106

y₂ = 2.25 x 107

[Standard form]

Hence, there are 2.25 x 107 crystals of sugar in 5 kg of sugar

Here, .x₁ = 2

y₁ =9 X 106

x₃ = 1.2

Therefore \(\frac{x_1}{y_1}=\frac{x_3}{y_3}\) gives

⇒ \(\frac{2}{9 \times 10^6}=\frac{1.2}{y_3}\)

2y₃ = 1.2 x 9 x 106

2y₃= 10.8 x 106

⇒ \(y_3=\frac{10.8 \times 10^6}{2}\)

y₃ = 5.4 x 106

Hence, there are 5.4 x 106 crystals of sugar in 1.2 kg of sugar.

Question 8. Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?
Solution:

Let. the distance covered in the map be x cm. We put the given information in the form of a table as shown below :

The more the actual distance covered on the road, the more the distance on the map. So, it is a case of direct proportion

⇒ \(\frac{x_1}{y_1}=\frac{x_2}{y_2} \Rightarrow \frac{18}{1}=\frac{72}{y_2}\)

⇒ \(y_2=\frac{72}{18}=4\)

Hence, the distance covered on the map would be 4 cm.

Question 9. A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time) the length of the shadow cast by another pole 10 m 50 cm high (ii) the height of a pole which casts a shadow 5 m long.
Solution:

Let the height of the vertical pole be * m and the length of the shadow be y m. We put the given information in the form of a table as shown below :

As the height of the vertical pole increases, the length of the shadow also increases in the same ratio. So it is a case of direct proportion.

We make use of the relation of the type \(\frac{x_1}{y_1}=\frac{x_2}{y_2} \text {. }\)

Here xl = 5 m 60 cm = 5.60 m

yi= 3 m 20 cm = 3.20 m

x2 = 10 m 50 cm = 10.50 m

Therefore \(\frac{x_1}{y_1}=\frac{x_2}{y_2}\) gives

⇒ \(\frac{5.6}{3.2}=\frac{10.5}{y_2}\)

5.6y2 = 3.2 x 10.5

⇒ \(y_{2 .}=\frac{3.2 \times 10.5}{5.6}=\frac{32 \times 105}{560}\)

⇒ \(\frac{4 \times 105}{70}=\frac{4 \times 15}{10}\)

y2= 6

Hence, the length of the shadow is 6 m.

(ii) Here xx = 5 m 60 cm = 560 cm

y1 = 3 m 20 cm = 320 cm

y3 = 5 m = 500 cm

Therefore \( \frac{x_1}{y_1} =\frac{x_3}{y_3} \text { gives } \frac{560}{320}=\frac{x_3}{500}\)

⇒ \(320 x_3 =560 \times 500\)

⇒ \(x_3 =\frac{560 \times 500}{320}=\frac{560 \times 50}{32}\)

⇒ \(\frac{70}{4} \times 50=\frac{3500}{4}\)

x3 = 875

Hence, the height of the pole is 8 m 75 cm.

Question 10. A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Solution:

Two quantities x and y which vary £ in direct proportion have the relation

⇒ \(x=k y \text { or } \frac{x}{y}=k\)

⇒ \(k =\frac{\text { Number of km it can travel }}{\text { time in hours }}\)

⇒ \(\frac{14}{\left(\frac{25}{60}\right)}=\frac{14 \times 60}{25}\)

⇒ \(\frac{14 \times 12}{5}=\frac{168}{5}\)

Now, x is the distance traveled in 5 hours

Using the relation x = ky, obtain

⇒ \(x=\frac{168}{5} \times 5\)

x = 168

Hence, it can travel 168 km in 5 hours.

Question 11. On a squared paper, draw five squares of different sides. Write the following information in a tabular form.

Find whether the length of a side is in direct proportion to the:

1. The perimeter of the square.
2. The area of the square.

Solution:

Since each \(\frac{L}{\mathrm{P}}\) is the same, so we find that the length of a side is in direct proportion to the perimeter of the square.

Since all \(\frac{L}{\mathrm{a}}\) are not the same, therefore, we find that the length of a side is not in direct proportion to the area of the square.

Question 12. The following ingredients are required to make halwa for 5 persons: SujiRawa = 250 g, Sugar = 300 g, Ghee = 200 g, Water = 500 ml. Using the concept of proportion, estimate the changes in the quantity of ingredients, to prepare halwa for your class.
Solution:

Suppose that there are 25 (= 5 x 5) students in the class. Then, this is a case of direct variation. Hence, we shall require five times the quantity of ingredients.

Question 3. Choose a scale and make a map of your classroom., showing windows, doors blackboard, etc. (An example is given here),

Solution:

Please make yourself.

Question 13. Take a few problems discussed so far under ‘direct variation’. Do you think that they can be solved by the ‘unitary method’?
Solution:

Yes, they can be solved by the unitary method

∴  75 mL of the base is required for

= 1 part of red pigment

∴  1 mL of the base is required for

\(\frac{1}{75}\) part of red pigment

∴  1800 mL of the base is required for

⇒ \(=\frac{1}{75} \times 1800=\frac{120}{5}\)

= 24 parts of red pigment

Number of bottles filled in 6 hours = 840

Number of bottles filled in 1 hour \(\)

⇒ \(\frac{840}{6}=140\)

∴  Number of bottles filled in 5 hours = 140 x 5 = 700

∴  Distance travelled in 25 minutes = 14 km

∴  Distance travelled in 1 minute\(=\frac{14}{25} \mathrm{~km}\)

∴  Distance travelled in 5 hours(= 5 x 60 minutes or 300 minutes)

⇒ \(\frac{14}{25} \times 300 \mathrm{~km}\)

14 x 12 = 168 km

Inverse Proportion

If two quantities are related in such a way that an increase in one quantity leads to a corresponding proportional decrease in the other and vice-versa, then such a variation is called inverse proportion.

Thus, two quantities x and y are said to vary in inverse proportion if xy = k where k is a constant of proportionality

In this case, if. y1, y2 are the values of y corresponding to the values xv x2
f x respectively,

⇒ \(x_1 y_1=x_2 y_2 \quad \text { or } \quad \frac{x_1}{x_2}=\frac{y_2}{y_1}\)

Question 14. Think of more such examples of pairs of quantities that vary in inverse proportion.
Solution:

Some more such examples of pairs of quantities that vary in inverse proportion are as follows :

Number of machines to produce a given number of articles and the number of days required for production.

Number of periods a day in a school and the length of the period.

Number of workers to build a wall and The number of hours to build the wall

Question 15. Take a squared paper and arrange 48 counters on it in different numbers of rows shown below

What do you observe? As R increases, C decreases.

1. Is R1: R₂ = C₂: C₁?
2. Is R₃:R₄ = C₄:C₃ ?
3. Are R and C inversely proportional to each other?

Try this activity with 36 counters.

Solution:

So \(\frac{\mathrm{R}_1}{\mathrm{R}_2} =\frac{2}{3}\)

⇒ \( \frac{\mathrm{C}_2}{\mathrm{C}_1} =\frac{16}{24}=\frac{2}{3} \)

⇒ \( \frac{\mathrm{R}_1}{\mathrm{R}_2} =\frac{\mathrm{C}_2}{\mathrm{C}_1} \)

⇒ \(\mathrm{R}_1: \mathrm{R}_2 =\mathrm{C}_2: \mathrm{C}_1\)

R1: R₂ = C₂: C₁

⇒ \(\frac{\mathrm{R}_3}{\mathrm{R}_4}=\frac{4}{6}=\frac{2}{3}\)

⇒ \(\frac{\mathrm{C}_4}{\mathrm{C}_3}=\frac{8}{12}=\frac{2}{3}\)

⇒ \(\frac{\mathrm{R}_3}{\mathrm{R}_4}=\frac{\mathrm{C}_4}{\mathrm{C}_3}\)

R₃:R₄ = C₄:C₃

Yes! R and C are inversely proportional to each other.

Let us arrange 36 counters in different numbers of rows as shown below :

Let us try this activity with 36 counters as follows :

We observe that as R increases, C decreases.

R₁ : R₂ = 2:3

C₂ : C4= 12 : 18 = 2 : 3

R₁:R₂ = C₂ : C₁

R₃ : R₄ = 4:6 = 2:3

C₄ : C₃ = 6:9 = 2:3

R₃:R₄= c₄-c₃

Yes! R and C are inversely proportional to each other.

Question 16. Observe the following tables and find which pair of variables (here x and y) are in inverse proportion.

Solution:

x₁y₁ = 50 x 5 = 250

x₂y₂ = 40 x6 = 240

So, x₁y₁ ≠ x₂y₂

Hence, x and y are not in inverse proportion

x₁y₁= 100 x 60 = 6000

x₂y₂ = 200 x 30 = 6000

x₃y₃ = 300 x 20 = 6000

x₄ÿ₄ = 400 x 15 = 6000

So , x₁y₁ = x₂y₂ = x₃y₃ = x₄y₄

Hence, x and y are in inverse proportion.

x₁y₁= 90 x 10 = 900

x₂y₂ = 60 x 15 = 900

x₃y₃ = 45 x 20 = 900

x₄ÿ₄ = 30 x 25 = 750

As x₁y₁ = x₂y₂ = x₃y₃ ≠ x₄y₄

So, x and y are not in inverse proportion.

Note: When two quantities x and y are in direct proportion (or vary directly) they are also written as x y.

When two quantities x andy are in inverse proportion (or vary inversely) they are also written as x «c \(\frac{1}{y}\)

Direct And Inverse Proportions Exercise 11.2

Question 1. Which of the following are in inverse proportion 1

1. The number of workers on a job and the time to complete the job.
2. The time taken for a journey and the distance traveled at a uniform speed,
3. Area cultivated land and the crop harvested,
4. The time taken for a fixed journey and the speed of the vehicle.
5. The population of a country and the area, of land per person.

Solution:

The number of workers on a job and the time to complete the job are in inverse
proportion, since as the number of workers increases, then the time to complete the job decreases proportionally.

The time taken for a journey and the distance traveled at a uniform speed are not in inverse proportion, since for a longer distance, more time will be required.

The area of cultivated land and the crop harvested is not in inverse proportion, since for more area of cultivated land, more crops will be needed to be harvested.

The time taken for a fixed journey and the speed of the vehicle are in inverse proportion, since the speed of the vehicle, proportionally less would be the time to cover a fixed journey.

The population of the country and the area of land per person are in inverse proportion,
since for more population of the country, the area of land per person would be proportionally less.

Question 2. In a Television game show, the prize money of f1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners.

Solution:

Let the prize (in T) for each winner bey3, y4, y5, y6, and y1 corresponding to the number of winners 4, 5, 8, 10, and 20 respectively. Then,

1 X 1,00,000 = 4 X y3 \(y_3=\frac{1,00,000}{4}=25,000\)

1 x 1,00,000 = 5 x y4 \(y_3=\frac{1,00,000}{5}=20,000\)

1 x 1,00,000 = 8 x y5 \(y_3=\frac{1,00,000}{8}=12,000\)

1 x 1,00,000 = 10 x y6 \(y_3=\frac{1,00,000}{10}=10,000\)

1 x 1,00,000 = 20 x y7 \(y_3=\frac{1,00,000}{20}=5,000\)

As the number of winners increases, amount of prize decreases. So the prize money given to an individual winner is inversely proportional to the number of winners.

Question 3. Rehrnan is making any paira wheel of using spokes in such a way that the angles between any pair of consecutive equal. Help equal him spokes by completing the following table:

1. Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion?
2. Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.
3. How many spokes would be needed, if the angle between a pair of-consecutive spokes is 40°?

Solution:

Let the angles (in degree) between a pair of consecutive spokes be y₃, y₄, and y₅ respectively. Then,

⇒ \(y_3=\frac{360^{\circ}}{8}=45^{\circ}\)

⇒ \(y_4=\frac{360^{\circ}}{10}=36^{\circ}\)

⇒ \(y_5=\frac{360^{\circ}}{12}=30^{\circ}\)

Yes; The number of spokes and the angles formed between the pairs of consecutive spokes are in inverse proportion.

|4 x 90° = 6 x 60° = 8 x 45° = 10 x 36° = 12 x 30°

Let the angle (in degrees) between a pair of consecutive spokes on a wheel with 15 spokes be y₆.

The smaller the number of spikes, the more the angle between a pair of consecutive spokes.

So, this is a case of inverse proportion.

Hence, 4 x 90° = 15 x y

⇒ \(y_6=\frac{4 \times 90^{\circ}}{15}\) x1y1 = x_2y2

y₆ = 24°

Hence, the angle between a pair of consecutive spokes on a wheel with 15 spokes is 24°.

Let x spokes be needed.

The lesser the number of spokes, the more will be the angle between a pair of consecutive spokes, bo, this is a case of inverse proportion.

Hence, 4 x 90° = x x 40°

⇒ \(x=\frac{4 \times 90^{\circ}}{40}\) x1y1 = x₂y₂

X = 9

Hence, 9 spokes would be needed, if the angle between a pair of consecutive spokes is 40″,

Question 4. If a box of sweets is divided among 21 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4?
Solution:

Suppose that each would get y2 sweets

Thus, we have the following table :

The fewer the number of children, the more the number of sweets each would get. So, this is a case of inverse proportion.

Hence 24 x 5 = 20 x y₂

⇒ \(y_2=\frac{24 \times 5}{20}\)

y₂ = 6

Hence, each would get 6 sweets, if the number of children is reduced by 4.

Question 5. A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?
Solution:

Suppose that the food would last for two days. We have the following table :

We note that the more the number of animals, the lesser will be the number of days for which the food will last. Therefore, this is a case of inverse proportion.

So, 20 x 6 = 30 x y₂

⇒ \(y_2=\frac{20 \times 6}{30}\)

y₂ = 4

Hence, the food would last for 4 days, if there were 10 more animals in his cattle.

Question 6. A contractor estimates that 3 persons could rewire Dasminder’s house in 4 days. If, he uses 4 persons instead of three, how long should they take to complete the job?
Solution:

Suppose that they take y₂ days to complete the job. We have the following table :

The more the number of persons, the lesser will be the number of days required to complete the job. So, this is a case of inverse proportion.

Hence, 3 x 4 =4 x y₂

⇒ \(y_2=\frac{3 \times 4}{4}\)

y₂ = 3

Hence, they would take 3 days to complete the job.

Question 7. A batch of bottles was packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?

Solution:

Suppose that y₂ boxes would be filled. We have the following table:

The smaller the number of bottles, the more the number of boxes required to be filled. So, this is a case of inverse proportion.

Hence, 12 x 25 = 20 x y₂

⇒ \(y_2=\frac{12 \times 25}{20}=3 \times 5\)

y₂= 15

Hence, 15 boxes would be filled if the same batch is packed using 20 bottles in each box.

Question 8. A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days?
Solution:

Suppose that x2 machines would be required. We have the following table :

The smaller the number of machines, the more will be the number of days to produce the same number of articles.

So, this is a case of inverse proportion.

Hence, 42 x 63 = x₂ x 54

⇒ \(x_2=\frac{42 \times 63}{54}=\frac{21 \times 7}{3}\)

x₂= 49

Hence, 49 machines would be required to produce the same number of articles in 54 days.

Question 9. A car takes 2 hours to reach a destination by traveling at the speed of 60 km/h. How long will it take when the car travels at the speed of 80 km/h?
Solution:

Let it take two hours. We have the following table :

Lesser the speed, the more the number of hours to reach the destination. So, this is a case of inverse proportion.

Hence, 60 x 2 = 80 x y₂

⇒ \(y_2=\frac{60 \times 2}{80}=\frac{3 \times 2}{4}\)

⇒ \(y_2=\frac{3}{2}=1 \frac{1}{2}\)

Thus, \(1 \frac{1}{2}\) hours would be taken when the car travels at the speed of 80 km/h.

Question 10. Two people could fit new windows in a house in 3 days.

1. One of the people fell ill before the work started. How long would the job take now?
2. How many people would be needed to fit the windows in one day?

Solution:

Let the job would take y2 days.

We have the following table :

The more the number of people, the lesser the number of days to do the job. So, the number of persons and number of days vary in inverse proportion.

So, 2 x 3= 1 x y₂

Thus, the job would now take 6 days when one of the persons fell ill before the work started.

Let y3 persons be needed.

We have the following table :

Clearly, the more the number of people, the lesser the number of days to do the job. So, the number of persons and number of days vary in inverse proportion.

So, 3 x 2 = 1 x y₃

y₃= 6

Thus, 6 people would be needed to fit the windows in one day.

Question 11. A school has 8 periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school hours to be the same?
Solution:

Let each period be two minutes long. We have the following table :

Wo note that the more the number of periods, the lesser would be the length of each period. Therefore, this is a case of inverse proportion.

So, 8 x 45 = 9 x y₂

⇒ \(y_2=\frac{8 \times 45}{9}\)

y₂ = 40

Hence, each period would be 40 minutes long if the school had 9 periods a day.

Question 12. Take a sheet of paper. Fold it as shown in the figure. Count the number of parts and the area of a part in each case.

Tabulate your observations and discuss them with your friends. Is it a case of inverse proportion? Why?

Solution:

We see that

⇒ \(1 \times 1=2 \times \frac{1}{2}=4 \times \frac{1}{4}=8 \times \frac{1}{8}=16 \times \frac{1}{16}=1 \text { (= constant) }\)

So yes it is a case of inverse proportion because the greater the number of parts lesser the area of each part.

Question 13. Take a few containers of different sizes with circular bases, Fill the same amount of water in each container. note the diameter of each container and the respective height at which the water levels stand. Tabulate your observations. Is it a case of inverse proportion?

Solution:

Let the same amount of water in each container be V. Then

⇒ \(V=\pi\left(\frac{d_1}{2}\right)^2 h_1=\pi\left(\frac{d_2}{2}\right)^2 h_2=\pi\left(\frac{d_3}{2}\right)^2 h_3\)

d₁²h₁ = d²2h₂ = d_3²h₃

d₁h₁= d₂h₂ = d₃h3

So, it is not a case of inverse proportion.

It does not imply that

Direct And Inverse Proportions Multiple-Choice Question And Solutions

Question 1. 10 meters of cloth cost? 1000. What will 4 meters cost?

1. ₹ 400
2. ₹ 800
3. ₹ 200
4. ₹ 100

Solution: 1. ₹ 400

⇒ \(\frac{10}{1000}=\frac{4}{?} \quad \Rightarrow ?=400\)

Question 2. 15 books weigh 6 kg. What will 6 books weigh?

1. 1.2 kg
2. 2.4 kg
3. 3.8 kg
4. 3 kg

Solution: 2. 2.4 kg

⇒ \(\frac{15}{6}=\frac{6}{?} \quad \Rightarrow ?=2.4\)

Question 3. Does a horse eat 18 kg of corn in 12 days? How much does he eat in 9 days?

1. 11.5 kg
2. 12.5 kg
3. 13.5 kg
4. 14.5 kg

Solution: 3. 13.5 kg

⇒ \(\frac{12}{18}=\frac{9}{?} \quad \Rightarrow ?=\frac{18 \times 9}{12}\)

= 13.5

Question 4. 8 g of sandalwood cost ₹ 40. What will 10 g cost?

1. ₹ 30
2. ₹ 36
3. ₹ 48
4. ₹ 50

Solution: 4. ₹ 50

⇒ \(\frac{8}{40}=\frac{10}{?} \quad \Rightarrow ?=50\)

Question 5. 20 trucks can hold 150 metric tonnes. How much will 12 trucks hold?

1. 80 metric tonnes
2. 90 metric tonnes
3. 60 metric tonnes
4. 40 metric tonnes.

Solution: 2. 90 metric tonnes

⇒ \(\frac{20}{150}=\frac{12}{?} \quad \Rightarrow ?=90 .\)

Question 6. 120 copies of a book cost ₹ 600. What will 400 copies cost?

1. ₹ 1000
2. ₹ 2000
3. ₹ 3000
4. ₹ 2400

Solution: 2. ₹ 2000

⇒  \(\frac{120}{600}=\frac{400}{?} \Rightarrow ?=2000 .\)

Question 7. The rent of 7 hectares is ₹ 875. What is the rent of 16 hectares?

1. ₹ 2000
2. ₹ 1500
3. ₹ 1600
4. ₹ 1200

Solution: 1. ₹ 2000

⇒ \(\frac{7}{875}=\frac{16}{?} \quad \Rightarrow ?=\frac{875 \times 16}{7}\)

= 2000

Question 8. A boy runs 1 km in 10 minutes. How long will it take to run 600 m?

1. 2 minutes
2. 3 minutes
3. 4 minutes
4. 6 minutes

Solution: 4. 6 minutes

⇒ \(\frac{1000}{10}=\frac{600}{?} \Rightarrow ?=6\)

Question 9. A shot travels 90 m in 1 second. How long will it take to go 225 m?

1. 2 seconds
2. 2.5.seconds
3. 4 seconds
4. 3.5 seconds

Solution: 2. 2.5.seconds

⇒ \(\frac{90}{1}=\frac{225}{?} \Rightarrow ?=2.5 .\)

Question 10. 3 knives cost ₹ 63. What will 17 knives cost?

1. ₹ 357
2. ₹ 375
3. ₹ 537
4. ₹ 573

Solution: 1. ₹ 357

⇒ \(\frac{3}{63}=\frac{17}{?} \quad \Rightarrow \quad ?=357\)

Question 11. 15 men can mow 40 hectares of land in 1 day. How much will 6 men mow in 1 day?

1. 16 hectares
2. 12 hectares
3. 20 hectares
4. 24 hectares

Solution: 1. 16 hectares

⇒ \(\frac{15}{40}=\frac{6}{?} \quad \Rightarrow ?=\frac{40 \times 6}{15}=16 .\)

Question 12. A man walks 20 km in 5 hours. How long would it take to walk 32 km?

1. 3 hours
2. 4 hours
3. 6 hours
4. 8 hours

Solution: 4. 8 hours

⇒ \(\frac{20}{5}=\frac{32}{?} \quad \Rightarrow ?=8\)

Question 13. What is the cost of 50 sticks at ₹ 24 per score?

1. ₹ 30
2. ₹ 40
3. ₹ 50
4. ₹ 60

Solution: 4. ₹ 60

⇒ \(\frac{20}{24}=\frac{50}{?} \quad \Rightarrow ?=60 \text {. }\)

Question 14. A train travels 60 km in 1 hour. How long will it take to go 150 km?

1. 2 hours
2. 3 hours
3. 2.5 hours
4. 4 hours.

Solution: 3. 2.5 hours

⇒ \(\frac{60}{1}=\frac{150}{?} \quad \Rightarrow ?=2.5\)

Question 15. If 3 quintals of coal cost ₹ 6000, what is the cost of 120 kg?

1. ₹ 1200
2. ₹ 2400
3. ₹ 3600
4. ₹ 4800

Solution: 2. ₹ 2400

⇒ \(\frac{300}{6000}=\frac{120}{?} \Rightarrow ?=2400\)

Question 16. If 20 cows eat as much as 15 oxen, how many cows will eat as much as 36 oxen?

1. 40
2. 44
3. 45
4. 48

Solution: 4. 48

⇒ \(\frac{15}{20}=\frac{36}{?} \quad \Rightarrow ?=\frac{20 \times 36}{15}\)= 48

Question 17. The fare for a journey of 40 km is ₹ 25. How much can be traveled for ₹ 40?

1. 32 km
2. 64 km
3. 50 km
4. 60 km

Solution: 2. 64 km

⇒ \(\frac{25}{40}=\frac{40}{?} \Rightarrow ?=\frac{40 \times 40}{25}\)= 74

Question 18. Apala types 200 words in half an hour. How many words will she type in 12 minutes?

1. 80
2. 50
3. 100
4. 60

Solution: 1. 80

⇒ \(\frac{30}{200}=\frac{12}{?} \quad \Rightarrow ?=80 \text {. }\)

Question 19. A labourer is paid? 400 for 2 days of work. If he works for 5 days, how much will he get?

1. ₹ 1000
2. ₹ 800
3. ₹ 750
4. ₹ 900

Solution: 1. ₹ 1000

⇒ \(\frac{2}{400}=\frac{5}{?} \quad \Rightarrow ?=1000 .\)

Question 20. A machine in a soft drink factory fills 600 bottles in 5 hours. How many bottles will it fill in 2 hours?

1. 120
2. 180
3. 150
4. 240

Solution: 4. 240

⇒ \(\frac{5}{600}=\frac{2}{?} \quad \Rightarrow ?=240 .\)

Question 21. If 8 men can do a piece of work in 20 days, in how many days could 20 men do the same work?

1. 6 days
2. 8 days
3. 4 days
4. 10 days

Solution: 2. 8 days

⇒ \(8 \times 20=20 \times ? \quad \Rightarrow ?=8\)

⇒ \(5 \times 3=15,7 \times 3=21 \)

Question 22. If an amount of food last for 40 days for 120 men, how long will it last for 80 men at the same rate?

1. 50 days
2. 60 days
3. 80 days
4. 100 days

Solution: 2. 60 days

⇒ \( 40 \times 120=80 \times ? \quad \Rightarrow ?=60 \)

Question 23. If 18 women can reap a field in 7 days, in what time can 6 women reap the same field?

1. 15 days
2. 21 days
3. 30 days
4. 36 days

Solution: 2. 21 days

⇒ \(18 \times 7=6 \times ? \quad \Rightarrow ?=21 .\)

Question 24. 10 men can dig a trench in 15 days. How long will 3 men take?

1. 50 days
2. 60 days
3. 100 days
4. 75 days

Solution: 1. 50 days

⇒ \( 10 \times 15=3 \times ? \quad \Rightarrow ?=50\)

Question 25. 3 lambs finish eating turnips in 8 days. In how many days will 2 lambs finish them?

1. 6
2. 8
3. 10
4. 12

Solution: 4. 12

\(3 \times 8=2 \times ? \Rightarrow ?=12 .\)

Question 26. 6 pipes are required to fill a tank in 1 hour. How long will it take if only 5 pipes of the same type are used?

1. 75 minutes
2. 72 minutes
3. 80 minutes
4. 90 minutes.

Solution: 2. 72 minutes

⇒ \( 6 \times 60=5 \times? \)

⇒ \(?=72 \text { minutes. }\)

Question 27. 40 cows can graze a field in 16 days. How many cows will graze the same field in 10 days?

1. 60
2. 64
3. 80
4. 75

Solution: 2. 64

⇒ \(40 \times 16=? \times 10 \quad \Rightarrow ?=64 .\)

Question 28. If x = ky and when y = 4, x = 8 then k =

1. 1
2. 2
3. 3
4. 4

Solution: 2. 2

⇒ \(8=4 k \quad \Rightarrow k=2 .\)

Question 29. The constant of variation, if x «= y, from the following table, is

x     6     12     15    21

y     2      4       5      7

1. 1
2. 2
3. 3
4. 4

Solution: 3. 3

⇒ \(2 \times 3=6,4 \times 3=12, \\\)

Question 30. x and y vary inversely with each other. If x = 15 when y- 6, then the value of x when = 15 is

1. 2
2. 4
3. 5
4. 6

Solution: 4. 4

⇒ \( 15 \times 6=? \times 15 \quad \Rightarrow ?=6 \)

Direct And Inverse Proportions True-False

Write whether the following statements are True or False:

1. The distance traveled by a CNG vehicle and the amount of CNG consumed are directly proportional: True

2. If and q are in inverse proportion, then (p+l) and (q- 1) are also in inverse proportion: True

3. If Apala can finish a work in n days, then the amount of work done by her in one day is \(\frac{1}{y}\): True

4. If two quantities p and q vary inversely, then pq remains constant: True

5. If x and y are in direct proportion, then \(\frac{1}{x} \text { and } \frac{1}{y}\) are also in direct proportion: True

Direct And Inverse Proportions Fill in the Blanks

1. If xy = 1, then x and y vary ______ with each other: Inversely

2. The radius and the circumference of a circle vary ______ with each other.
sheets of the same: Directly

3. If the weight of 10 sheets of an A-4 size paper sheet is 30 g, then ______
the paper would weigh 90 g: 30

4. If the amount of work finished by Meenu in one day is \(\frac{1}{m}\) then the whole work will be finished by her in______days: m

5. If the speed remains constant, then the distance traveled is ______ proportional to the time: Directly

6. The area occupied by 10 postal stamps is 40 cm2. Find the area occupied by 100 such stamps: 400cm2

7. In a camp, there is enough flour for 50 persons for 40 days. How long will the food last, if 30 more people join the camp: 25days

8. 20 persons can reap a field in 15 days. How many more persons should be engaged to reap the same field in 10 days: 30

9. A car is traveling 80 km per hour. Find the distance traveled by car in 3 hours, if the speed remains constant: 240km

10. If x varies inversely as y and x = 20 when y = 30. Then, find y when x = 50: 12

Factorization Introduction

In this chapter, we shall learn about the factorization of algebraic expressions, methods of factorization, and division of algebraic expressions.

Factors Of Natural Numbers

A number when written as a product of its prime factors is said to be in the prime factor form. Similarly, we can express algebraic expressions as products of their factors.

Factors Of Algebraic Expressions

A fundamental factor cannot be expressed further as a product of factors.

What Is Factorisation

When we factorize an algebraic expression, we write it as a product of irreducible factors. These factors may be numbers, algebraic variables, or algebraic expressions. This process is called factorization.

Method Of Common Factors

We factorize each term of the given algebraic expression as a product of irreducible factors and separate the common factors. Then, we combine the remaining factors in each term using the distributive law.

Read and Learn More NCERT Solutions For Class 8 Maths

Question 1. Factorise:

12x + 36

22y – 33z

14pq + 35pqr.

Solution:

We have

12i = 2 x 2 x 3 x x

36 = 2 x 2 x 3 x 3

The two terms have 2, 2, and 3 as

common factors.

Therefore, 12x + 36

= (2 x 2 x 3 x x) + (2 x 2 x 3 x 3)

= 2 x 2 x 3 x (x + 3) using distributive law

= 12 x (x + 3)

= 12(x + 3)

which is the required factor form.

We have

22y = 2 x ll x y

33z = 3 x li x z

The two terms have 11 as a common factor.

Therefore,

22y – 33z

= (11 x 2 x y) – (11 x 3 x z)

= 11 X [(2 X y) – (3 X Z)]

I use the distributive law

= 11 x (2y – 3z)

= ll(2y – 3z)

which is the required factor form.

We have

14pq = 2 x 7 x p x q

35pgr = 5×7 x p x g x r

The two terms have 7, p, and q as common factors.

Therefore, 14pq + 35pqr

= 7 X p x q x 2 + 7 x p x q x 5 x r

= 7 x p x q x [2 + (5 x r)]

using distributive law

= 7pq x (2 + 5r)

= 7pq(2 + 5r).

which is the required factor form.

Note:  We notice that the factor form of an expression has only  one term

Factorization By Regrouping Terms

Sometimes it so happens that all the terms in a given algebraic expression do not have a common factor, but the terms can be grouped so that all the terms in each group have a common factor. In doing so, we get a common factor across all the groups formed. This leads to the required factorization of the given algebraic expression.

Factorization Exercise 12.1

Question 1. Find the common factors of the given terms:

1. 12x, 36
2. 2y, 22xy
3. 14pq, 28p²q²
4. 2x, 3x², 4
5. 6abc, 24ab², 12a²b
6. 16x³-4x², 32x
7. 10pq, 20qr, 30rp
8. 3x²y³, 10x³y², 6x²y²z.

Solution:

1.  12x = 2 × 2 × 3 × x

36 = 2 × 2 × 3 × 3

Common prime factors are 2 (occurs twice) and 3.

Required common factor

= 2 × 2 × 3 = 12

2.  2y = 2 x y

22xy = 2 × 11 × x × y

Common factors are 2 and y.

Required common factor = 2 × y = 2y

3.  14pq = 2 x 7 x p x q

28 p²q² = 2 × 2 × 7 × p × p × q × q

Common factors are 2, 7, p, and g.

Required common factor

= 2 × 7 × p × q = 14pg

4.  2x = 1 × 2 × x

3ÿ = 1 × 3 × x × x

4 = 1× 2 × 2

Common factor is 1

Required common factor = 1

5.  6abc = 2 x 3 x a x b x c

24ab² = 2 × 2 × 2 × 3 × a × b × b

12a²b = 2 × 2 × 3 × a × a × b

Common factors are 2, 3, a and b

Required common factor

= 2 × 3 × a × b = 6ab

6. 16x³ = 2 × 2 × 2 × 2 × x × y × x

-4X² =-1 × 2 × 2 × y × X

32r = 2 ×2 × 2 × 2 × 2 × x

Common factors are 2 (occurs twice) and x (occurs once).

Required common factor = 2 × 2 × x = 4x

7.  10pq = 2 ×5 × p × q

20qr = 2 × 2 × 5 × q × r

30rp = 2 x 3 x 5 x r xp

Common factors are 2 and 5.

Required common factor = 2 x 5 = 10

8.  3x²y³ = 3 X x X x X y X y X y

10x³y² = 2 x 5 x x x x x x y x y

6 x³ y² z = 2 x 3 x z x x x y X y X z

Common factors are x (occurs twice) and y (occurs twice)

Required common factor = X X X X y X y = x²y²

Question 2. Factorize the following expressions:

1. 7x – 42
2. 6p – 12q
3. 7a² + 14a
4. -16z + 20z³
5. 20l²m + 30 alm
6. 5x²y – 15xy²
7. 10a² – 15b² + 20c²
8. – 4a² + 4ab – 4ca
9. x²yz + xy²z + xyz²
10. ax²y + bxy² + cxyz.

Solution:

1.  7x = 7 × x

42 = 2 x 3 x 7

7x – 42

= l x x – 2 x 3 x 7

= 7(x-2 x 3)

using distributive law

= 7 (x – 6)

It is the required factor form.

2. 6p – 2 x 3 x p

12g = 2 x 2 x 3 x q

6p – 12q

= 2 x 3 x p – 2 x 2 x 3 x q

= 2 x 3(p – 2 x q)

using distributive law

= 6(p – 2q)

It is the required factor form.

3. 7a² = 7 x a x a

14a = 2 x 7 x a

7a² + 14a

= 7 x a x a + 2 x 7 x a

= 7 x a (a + 2)

I use the distributive law

= 7a (a + 2)

It is the required factor form.

4. -16z = (-1) x 2 x 2 x 2 x 2 x z

20Z³= 2 × 2 × 5 × Z × x × Z

A – 16z + 20 ×3 N

= (-1) x 2 x 2 x 2 x 2 x 2 + 2 x 2 x 5 x z x z x z

= (2 x 2 x z) [(-1) x 2 x 2 + 5 X z X z] 5

I am using the distributive law

= 4z(-4 + 5z²)

It is the required factor form.

5.  20l²m = 2 x 2 x 5 x l x l x m

30alm = 2 x 3 x 6 x a x l x m

A 20l²m + 30alm

= 2 x 2 x 5 x l x l x m

+ 2 x 3 x 5 x a x l x m

= 2 x 5 x l x m

x (2 x l + 3 x a)

= 10lm (2l + 3a)

| Using the distributive law

It is the required factor form.

6.  5x²y = 5 × x × x × y

15xy× = 3 × 5 × x × y × y

A 5x²y – 15xy²

= 5 × x × x × y

-3 × 5 × x × y × y

= (5 × x × y)(x – 3 x y)

| Using the distributive law

= 5xy (x – 3y)

It is the required factor form.

7.  10a² = 2 × 5 × a × a

156² = 3 × 5 × b × b

20c²= 2 × 2 × 5 × c × c

10a² – 15b² + 20c²

= 2 × 5  × a × a – 3 × 5 × b × b +2 × 2 × 5 × c × c

+2 × 2 × 5 × c × c

= 5 (2 × a × a -3 × b × b + 2 × 2 × c × c)

using distributive law

= 5(2a² – 3b² + 4c²)

It is the required factor form.

8. 4a² = 2 x 2 x a x a

4ab = 2 x 2 x a x 6

4ca = 2 x 2 x c x a

– 4a² + 4ab – 4ca

= (-1) × 2 ×2 ×a × a

+ 2 ×  2 × a × b

-2 × 2 × c × a

= 2 × 2 × a [(- 1) × a + 6 – c)]

I use the distributive law

= 4a (- a. + 6 – c)

It is the required factor form.

9.  x²yz = x × x × y × z

xy²z =x × y × y × z

xyz² =x × y × z × z

x²yz + xyz + xyz²

= X × X × y × z

+ X × y × y × Z

+ X × y × Z × z

= X × y × z(x + y + z)

I use the distributive law

= xyz{x + y + z)

It is the required factor form.

10.  ax²y = a × x × x × y

bxy² = b × x × y × y

Xyz = C × X × y × z

ax²y + bxy² + XYZ

= a X × X × X y + b × x × y X y + C × X × y × z

= X X y (a × x+b × y + c × z)

I use the distributive law

= xy (ax + by + cz).

It is the required factor form.

Question 3. Factorise :

1.  x² + xy + 8x + 8y
2. 15xy – 6x + 5y – 2
3. ax + bx – ay – by
4. 15pq + 15 + 9q + 25p
5.  z – 7 + 7xy – xyz.

Soiution: 

x² + xy + 8x + 8y

= x(x + y) + 8(x + y)

Taking x common in the first two terms

and 8 commons in the last two terms.

= (x + y)(x + 8)

Taking (x + y) common

It is the required form.

2. 15xy – 6x + 5y – 2

= 3x(5y – 2) + 1(5y – 2)

Taldng 3x common in first two terms and 1 common in last two terms.

= (5y – 2) (3x + 1)

Taking (5y- 2) common

It is the required factor form.

3.  ax + bx – ay – by

= x(a + b) – y(a + b)

I Taking .v common in the first two terms x and -y common in the last two terms.

= (a + b)(x – y)

| Taking (a + b) common

It is the required factor form.

4. 15pq + 15 + 9g + 25p

= 15pq + 9q + 25p + 15

Arranging the terms

= 3q(5p + 3) + 5(5p + 3)

Taking 3g common in the first two terms and 5 common in the last two terms.

= (5p + 3)(3q + 5)

Taking (5p + 3) common

It is the required factor form.

5.  z-7 + 7xy – xyz

= z – 7 – XYZ + Ixy

Arranging the terms

= 1(z – 7) – xy (z – 7)

Taking 1 common in the first two terms

and- xy common in the last two terms.

= (z – 7) (1 – xy).

Taking (z – 7) common

It is the required factor form.

Factorization Using Identities

The following identities prove to be quite helpful in factorization of an algebraic expression :

(a + b)² = a² + 2ab + b²

(a – b)² = a² – 2ab + b²

(a + b) (a – b) = a² – b²

12.2.4 Factors Of The Form (X + A) {X + B)

(x + a) (x + b) = x² + (a. + b) x + ab

To factorize an algebraic expression of type x² + px + q, we find two factors a and b of q such that

ab = q and a + b-p

Then, the given expression becomes

x² + (a + b) x + ab = x² + ax + bx + ab

= x (x + a) + b (x + a) = (x + a) (x + b)

which are the required factors.

Factorization Exercise 12.2

Question 1. Factorize the following expressions:

1. a² + 8a + 16
2. p² – lOp + 25
3. 25m² + 30m + 9
4. 49y² + 84yz + 36z²
5. 4x² – 8x + 4 K
6. 121b² – 88bc + 16c²
7. (l + m)² – 4lm      [Hint: Expand {l + m)2 first]
8. a⁴ + 2a²b² + b⁴

Solution:

1.  a² + 8a+ 16

= (a)² + 2 (a)(4) + (4)²

= (a + 4)²

Using Identity I

= (a + 4) (a + 4)

It is the required factor form.

2.   p² – 10p + 25

= (p)² – 2(p) (5) + (5)²

= (p-5)²

1 Using Identity II

= (p – 5) (p – 5)

It is the required factor form.

3.  25m² + 30m + 9

= (5m)² + 2(5m) (3) + (3)²

= (5m + 3)²

Using Identity I

= (5m + 3) (5m + 3)

It is the required factor form.

4.  49y² + 84yz + 36z²

= (7y)² + 2(7y) (6z) + (6z²

= (7y + 6z)² I Using Identity I

= (7y + 6z) (7y + 6z)

It is the required factor form.

5.  4x² – 8x + 4

= 4(x² – 2x + 1)

= 4 [(x)2 – 2(x) (1) + (1)²]

= 4(x – 1)²

Using Identity II

= 4(x – 1) (x – 1)

It is the required factor form.

6.  121b² – 88bc + 16c²

= (11b)² – 2(11b) (4c) + (4c)2

= (11b – 4c)²

Using Identity II

= (11b – 4c) (11b – 4c)

It is the required factor form.

7.  (l + m)² – 4lm

= (l² + 2lm + m²) – 4lm

I Using Identity I

= l2 + (2lm – 4lm) + m²

I Combining the like terms

= l² – 2lm + m²

= (l)² – 2(l)(m) + (m²

= (l – m)²

Using Identity II

= (l – m) (l- m)

It is the required factor form.

8.  a⁴ + 2a²b² + b⁴

= (a²)² + 2(a²) (b²) + (b²)²

= (a² + b²)² I Using Identity I

= (a² + b²) (a² + b²)

It is the required factor form.

Question 2. Factorise:

1. 4p² – 9q²
2. 63a² – 112b²
3. 49x² – 36
4. 16X⁵ – 144x³
5. (l + m)² – (l – m)²
6. 9x²y² – 16
7. (x² – 2xy + y²) – z²
8. 25a² – 4b² + 28bc – 49c².

Solution:

1.  4p² – 9q²

= (2p)²– m²

= (2p – 3q) (2p + 3q)

I am Using Identity III

It is the required factor form.

2.  63a² – 112b²

= 7 (9a2 – 1662) | Taking 7 common

= 7 {(3a)² – (46)²}

= 7 (3a – 45) (3a + 45)

Using Identity III

It is the required factor form

3.  49x² – 36

= (7x)² – (6)²

= (7x – 6) (7x + 6)

1 Using Identity III

It is the required factor form.

4. 16x⁵ – 144x³

= 16x³(x² – 9)

I Taking 16x³ common

= 16x³ {(x)² – (3)²}

= 16x³ (x – 3) (x + 3)

Using Identity III

It is the required factor form.

5.  (l + m)² – (l – m)²

= {(l + m) – (l -m)}

x {(l + m) + (l – m)}

Using Identity III

= (l + m – l + m) (l + m + l – m)

= (2m) (2l)

= 4lm

It is the required factor form.

6.  9x²y² – 16

= (3xy)² – (4)²

= (3xy – 4) (3xy + 4)

Using Identity II

It is the required factor form

7. (x² – 2xy + y²) – z²

= (x- y)² – z²

Using Identity II

= (x – y – z) (x – y + z)

Using Identity III

It is the required factor form.

8.  25a² – 4b² + 28b² – 49c²

= 25a² – (4b² – 28bc + 49c²)

= 25a² – {(2b)² – 2(2b) (7c) + (7c)²}

= (5a)² – (2b – 7c)²

Using Identity II

= {5a – (2b – 7c)} {5a + (2b – 7c)}

I Using Identity III

= (5a – 2b + 7c) (5a + 2b – 7c).

It is the required factor form.

Question 3. Factorize the expressions:

1. ax² + bx²
2. 7p² + 21q²
3. 2x³ + 2xy²+ 2xz²
4. am² + bm² + bn² + an²
5. (Im + l) + m + 1
6. y(y + z) + 9(y + z)
7. 5y² – 20y – 8z + 2yz
8. lOab + 4a + 5b + 2
9. 6xy – 4y + 6 – 9x.

Solution:

1.  ax² = a × x × x

bx = b × x

ax² + bx = ax × x × +  x = x (a × x + b)

| Using the distributive law

= x (ax + b)

It is the required factor form

2.  7p² = 7 × p × p

21q² = 3 × 7 × q × q

7p² + 21q² =7 × p × p + 3 × 7 × q × q

= 7(p × p + 3 × q × q)

Using distributive law

= 7(p² + 3g²)

It is the required factor form

3.  2x³ = 2 × x ×  x × x

2xy² = 2 × x × y × y

2xz² = 2 × x × z × z

2X²+ 2y² + 2z²

= 2 × x × x × x + 2 × x × y × y + 2 × x × z × z

= (2 × x) (x  × X + y × y + Z × z)

Using distributive law

= 2x(x² + y² + z²)

It is the required factor form

4.  am² + bm² + bn² + an²

= am²+ bm² + an² + bn²

= m² (a + b) +n² (a + b)

I Taking m² commonly in the first two terms

‘ and n² are common in the last two terms.

= (a + b) (m² + n²)

I Taking (a + b) common

It is the required factor form.

5.  (lm + l) + m + 1

= l(m + 1) + 1 (m + 1)

Taking l common in the first two terms and 1 common in the last two terms.

= (HI + 1) (l+ 1)

Taking (m + 1) common

It is the required factor form.

6.  y(y + z) + 9(y + z)

= (y + z) (y + 9)

Taking (y + z) common

It is the required factor form.

7. 5y² – 20y – 8z + 2yz

= 5y² – 20y + 2yz – 8z

= 5y (y – 4) + 2z(y – 4)

distributive law

= (y – 4) (5y + 2z)

I Taking (y – 4) common

It is the required factor form.

8.  10ab + 4a + 5b+ 2

= (10ab + 4a) + (5b + 2)

I Grouping the terms

= 2a (5b + 2) + 1 (5b + 2)

Using distributive law

= (5b + 2) (2a + 1)

| Taking (56 + 2) common

It is the required factor form.

9.  6xy – 4y + 6 – 9x

= 6xy – 4y – 9x + 6

Grouping the terms

= 2y (3x – 2) – 3(3x – 2)

Using distributive law

= (3x – 2) (2y – 3).

Taking (x – 2) common

It is the required form.

Question 4. Factorise:

1. a⁴ – b⁴
2. p⁴– 81
3. x⁴ – (y + z)⁴
4. x⁴ – (x – z)⁴
5. a⁴– 2a²b² + b⁴

Solution:

1.  a⁴ – b⁴

= (a²)² – (b²)²

= (a² – b²) (a² + b²)

Using Identity III

= (a – b) (a + b) (a² + b²)

| Using Identity III

It is the required factor form.

2.  p⁴– 81= (p²)² – (9)²

= (P² – 9) (p² + 9)

| Using Identity III

= (P)²-(3)²)(p² + 9)

= (P – 3) (p + 3) (p² + 9)

Using Identity III

It is the required factor form.

3.  x⁴ – (y + z)⁴

= (x²)² -{(y + z)²}²

= {x² – (y + 2)²} {x²+ (y + 2)²}

Using Identity III

= {x – (y + z)} {x + (y + z)} {x² + (y + z)²}

Using Identity III

= (X – y – Z)(X + y + z) {x² + (y + z)²}

It is the required factor form.

4.  x⁴ – (x – z)⁴

= (x²)² – {(x – z)}²

= {x² – (x – z)²}{x²+(x – z)²

Using Identity III

= {x – (x – z)}{x +(x- z)} {x² + (x – z)²}

Using Identity III

= (x – x + z) (x + X – z)

{x² + (X – 2)²}

= Z(2X – Z) {X² + (X – 2)²}

= z(2x – z) (x² + X² – 2xz + z²)

Using Identity II

= z(2x – z) (2x² – 2xz + z²)

It is the required factor form.

5.  a⁴ – 2a²b² + b⁴

= (a²)²– 2(a²) (b²) + (b²)²

= (a² – b²)²

| Using Identity II

= {(a – 6) ( a + b)}²

Using Identity III

= (a – b)² (a + b)²

= (a-b) (a – b) (a + b) (a + 6).

It is the required factor form

Question  5. Factorize the following expressions:

1. p² + 6p + 8
2. q²– lOq + 21
3. p² + 6p – 16

Solution:

1.  P² +6p + 8

= p² + 6p + 9 – 1

= {(p)² + 2(p)(3) + (3)²}-(l)²

= (p + 3)² – (l)²

I Using Identity I

= (p + 3 – l)(p + 3 + 1)

Using Identity III

= (p + 2) (p + 4)

It is the required factor form

2.  q²– lOq + 21

= q² – lOq + 25-4

= {(q)²-2(q)(5) + (5)²}-4

= (q – 5)² – (2)²

Using Identity II

= (q – 5 – 2) (q – 5 + 2)

Using Identity III

= (q – 7) (q – 3)

It is the required factor form

3.  P²+ 6p – 16

= p² + 6p + 9 – 25

= (p)² + 2(p)(3) + (3)²-(5)²

= (P + 3)² – (5)²

Using Identity I

= (P + 3 – 5) (p + 3 + 5)

I Applying for Identity III

= (p – 2) (p + 8)

It is the required factor form.

Division Of Algebraic Expressions

Here, we shall divide one algebraic expression by another.

Division Of A Monomial By Another Monomial

We shall factorize the numerator and denominator into irreducible factors and cancel out the common factors from the numerator and the denominator.

Question 1. Divide:

1. 24xy²z³ by 6yz²
2. 63a²b⁴c⁶ by 7a²b²c³

Solution:

1.  24xy²z³+ 6yz2

⇒ \(\frac{2 \times 2 \times 2 \times 3 \times x \times y \times y \times z \times z \times z}{2 \times 3 \times y \times z \times z}\)

⇒ \(\frac{2 \times 2 \times x \times y \times z}{1}=4 x y z\)

2.  63a²b⁴c⁶ + 7a²b²c³

⇒ \(\frac{3 \times 3 \times 7 \times a \times a \times b \times b \times b \times b \times c \times c \times c \times c \times c \times c}{7 \times a \times a \times b \times b \times c \times c \times c}\)

⇒ \(\frac{3 \times 3 \times b \times b \times c \times c \times c}{1}\)

⇒ \(=9 b^2 c^3\)

Division Of A Polynomial By A Monomial

We divide each term of the polynomial in the numerator by the monomial in the denominator.

Division Of Algebraic Expressions Continued (Polynomial + Polynomial)

We factorize the algebraic expressions in the numerator and the denominator into irreducible factors and cancel the common factors from the numerator and the denominator.

Factorization Exercise 12.3

Question 1. Carry out the following divisions:

1. 28x⁴ + 56x
2. – 36y³ 9y²
3. 66pq²r³ + llqr²
4. 34x²y²z² – 51xy²z²
5. 12a⁸b⁸ + (-6a⁶b⁴).

Solution:

1.  28x² + 56x

⇒ \(\frac{28 x^4}{56 x}\)

⇒ \(\frac{2 \times 2 \times 7 \times x \times x \times x \times x}{2 \times 2 \times 2 \times 7 \times x}\)

⇒ \(=\frac{x \times x \times x}{2}=\frac{x^3}{2}\)

2.  – 36y³ + 9y²

⇒ \(\frac{-36 y^{\circ}}{9 y^2}\)

⇒ \(\frac{(-1) \times 2 \times 2 \times 3 \times 3 \times y \times y \times y}{3 \times 3 \times y \times y}\)

⇒ \(\frac{(-1) \times 2 \times 2 \times y}{1}=\frac{-4 y}{1}=-4 y\)

3.  66pq²P + llqr²

⇒ \(\frac{66 p q^2 r^5}{11 q r^2}\)

⇒ \(\frac{2 \times 3 \times 11 \times p \times q \times q \times r \times r \times r}{11 \times q \times r \times r}\)

⇒ \(\frac{2 \times 3 \times p \times q \times r}{1}=\frac{6 p q r}{1}=6 p q r\)

4.   34x³y³z³ + 5lxy²z³

\(\frac{34 x^3 y^3 z^3}{51 x y^2 z^3}\)

⇒ \(\frac{2 \times 17 \times x \times x \times x \times y \times y \times y \times z \times z \times z}{3 \times 17 \times x \times y \times y \times z \times z \times z}\)

⇒ \(\frac{2 \times x \times x \times y}{3}=\frac{2}{3} x^2 y\)

5. 12a⁸b⁸ + (-6a⁶b⁴).

⇒ \(\frac{12 a^8 b^8}{-6 a^6 b^4}\)

⇒ \(\frac{2 \times 2 \times 3 \times a \times a \times a \times a \times a \times a \times a \times a \times b \times b \times b \times b \times b \times b \times b \times b}{(-1) \times 2 \times 3 \times a \times a \times a \times a \times a \times a \times b \times b \times b \times b}\)

⇒ \(\frac{-2 \times a \times a \times b \times b \times b \times b}{1}\)

⇒ \(\frac{-2 a^2 b^4}{1}=-2 a^2 b^4 \)

Question 2. Divide the given polynomial by the given monomial:

1. (5x² – 6x) + 3x
2. (3y⁸ – 4y⁶ + 5y⁴) + y⁴
3. 8(x³y²z² + x²y² + x²y²Z²) 4x²y²z²
4. (x²+ 2x² + 3x) x 2x
5. (P³q² – P⁶q³) +p³q³

Solution:

1.  (5x² – 6x) + 3x

⇒ \(=\frac{5 x^2-6 x}{3 x}=\frac{5 x^2}{3 x}-\frac{6 x}{3 x}=\frac{5}{3} x-2=\frac{1}{3}(5 x-6)\)

2.  (3y⁸ – 4y⁶ + 5y⁴) + y⁴

⇒ \(=\frac{3 y^8-4 y^6+5 y^4}{y^4}=\frac{3 y^8}{y^4}-\frac{4 y^6}{y^4}+\frac{5 y^4}{y^4}\)

– 3y4 – 4y2 + 5

3.  8(x³y²z² + x²y² + x²y²Z²) 4x²y²z²

⇒ \(\frac{8\left(x^3 y^2 z^2+x^2 y^3 z^2+x^2 y^2 z^3\right)}{4 x^2 y^2 z^2}\)

⇒ \(\frac{8 x^2 y^2 z^2(x+y+z)}{4 x^2 y^2 z^2}\) Taking (x? y2 z2) common

⇒ \(\frac{2(x+y+z)}{1}=2(x+y+z)\)

4.  (x²+ 2x² + 3x) x 2x

⇒ \(\frac{x^3+2 x^2+3 x}{2 x}=\frac{x \times\left(x^2+2 x+3\right)}{2 \times x}\)

⇒ \(\frac{1}{2}\left(x^2+2 x+3\right)\)

5.  (P³q² – P⁶q³) +p³q³

⇒ \(\frac{p^3 q^6-p^6 q^3}{p^3 q^3}=\frac{p^3 q^3\left(q^3-p^3\right)}{p^3 q^3}\)

⇒ \(\frac{q^3-p^3}{1}=q^3-p^3\)

Question 3. Work out the following divisions:

1. (10x – 25) 5
2. (10x – 25) 4- (2x – 5)
3. 10y(6y + 21) + 5(2y + 7)
4. 9x2y2(3z – 24) + 27xy(z – 8)
5. 96abc(3a – 12) (5b – 30) + 144(a – 4) (b – 6).

Solution:

1. (10x – 25) – 5

⇒ \(\frac{10 x-25}{5}=\frac{5(2 x-5)}{5}\)

⇒ \(\frac{2 x-5}{1}=2 x-5\)

⇒ \((10 x-25) \div(2 x-5)\)

⇒ \(\frac{10 x-25}{2 x-5}=\frac{5(2 x-5)}{2 x-5}=\frac{5}{1}=5\)

2. 10 y(6 y+21)+5(2 y+7)

\(\frac{10 y(6 y+21)}{5(2 y+7)}=\frac{10 y \times 3(2 y+7)}{5(2 y+7)}\)

⇒ \(\frac{2 y \times 3}{1}=\frac{6 y}{1}=6 y\)

⇒ \(9 x^2 y^2(3 z-24)+27 x y(z-8)\)

⇒ \(\frac{9 x^2 y^2(3 z-24)}{27 x y(z-8)}=\frac{9 x^2 y^2 \times 3(z-8)}{27 x y(z-8)}\)

⇒ \( \frac{x y}{1}=x y\)

3. 96abc(3a – 12) (5b – 30) + 144(a – 4) (b – 6)

⇒ \(\frac{96 a b c(3 a-12)(5 b-30)}{144(a-4)(b-6)}\)

⇒ \(\frac{2 \times 2 \times 2 \times 2 \times 2 \times 3 \times a b c \times 3(a-4) \times 5(b-6)}{2 \times 2 \times 2 \times 2 \times 3 \times 3 \times(a-4)(b-6)}\)

|Taking 3 common in (3 a-12) and 5 common in $(5 b-30)

⇒ \(\frac{2 \times 5 \times a b c}{1}=\frac{10 a b c}{1}\)

Question 4. Divide as directed.

1. 5(2x + 1) (3x + 5) -5- (2x + 1)
2. 26xy(x + 5) (y – 4) * 13x(y – 4)
3. 52pqr (p + q)(q + r)(r + p) + 104pq(q + r)(r + p)
4. 20(y + 4) (y2 + 5y + 3) + 5(y + 4)
5. x(x + l)(x + 2) (x + 3) + x(x + 1).

Solution:

1.  5(2x + 1) (3x + 5) + (2x + 1)

⇒ \(=\frac{5(2 x+1)(3 x+5)}{2 x+1}=\frac{5(3 x+5)}{1}\)

= 5(3x + 5)

2. 26xy(x + 5) (y – 4) + 13x(y – 4)

⇒ \(=\frac{26 x y(x+5)(y-4)}{13 x(y-4)}=\frac{2 y(x+5)}{1}\)

= 2y(x + 5)

3.  52pqr (p + q) (q + r) (r + p) -s- 104pq(q + r) (r + p)

⇒ \(\frac{52 p q r(p+q)(q+r)(r+p)}{104 p q(q+r)(r+p)}\)

⇒ \(\frac{2 \times 2 \times 13 \times p q r(p+q)(q+r)(r+p)}{2 \times 2 \times 2 \times 13 \times p q(q+r)(r+p)}=\frac{1}{2} r(p+q)\)

4.  20(y + 4) (y2 + 5y + 3) + 5(y + 4)

⇒ \(\frac{20(y+4)\left(y^2+5 y+3\right)}{5(y+4)}\)

⇒ \(\frac{2 \times 2 \times 5 \times(y+4)\left(y^2+5 y+3\right)}{5 \times(y+4)}=\frac{2 \times 2 \times\left(y^2+5 y+3\right)}{1}\)

= 4(y2 + 5y + 3)

5.  x(x + 1) (x + 2) (x + 3) + x(x + 1)

⇒ \(=\frac{x(x+1)(x+2)(x+3)}{x(x+1)}=\frac{(x+2)(x+3)}{1}\)

= (x + 2) (x + 3)

Question 5. Factorize the expressions and divide them as directed,

1. ( y² + 7y + 10) + (y + 5)
2. (m² – 14m – 32) + (m + 2)
3. (5p² – 25p + 20) + (p – 1)
4. 4yz(z² + 6z – 16) + 2y(z + 8)
5. 5pq(p² – q2) + 2p(p + q)
6. 12xy(9x² – 16y²) + 4xy(3x + 4y)
7. 39y³(50y² – 98) + 26y²(5y + 7)

Solution:

1.  ( y² + 7y + 10) + (y + 5)

⇒ \(\frac{y^2+7 y+10}{y+5}=\frac{y^2+2 y+5 y+10}{y+5}\)

Using Identity IV; 2y + 5y = 7y; (2y) (5y) = 10/

⇒ \(\frac{y(y+2)+5(y+2)}{y+5}=\frac{(y+2)(y+5)}{y+5}\)

Taking (y + 2) common

⇒ \(\frac{y+2}{1}=y+2\)

2.  (m² – 14m – 32) + (m + 2)

⇒ \(\frac{m^2-14 m-32}{m+2}=\frac{m^2-16 m+2 m-32}{m+2}\)

⇒ \(\frac{m(m-16)+2(m-16)}{m+2}=\frac{(m-16)(m+2)}{m+2} \quad \text { |Taking }(m-16) \text { common }\)

⇒ \(\frac{m-16 m+2 m=-14 m ; \quad(-16 m)(2 m)=-32 m^2}{}=m-16\)

3.  (5p² – 25p + 20) + (p – 1)

⇒ \(\frac{5\left(p^2-5 p+4\right)}{p-1}=\frac{5\left(p^2-p-4 p+4\right)}{p-1}\)

\(\frac{5\{p(p-1)-4(p-1)\}}{p-1}=\frac{5(p-1)(p-4)}{p-1}\)

⇒ \(\frac{5(p-4)}{1}=5(p-4)\)

4.  4yz(z² + 6z – 16) + 2y(z + 8)

⇒ \(\frac{4 y z\left(z^2+6 z-16\right)}{2 y(z+8)}=\frac{2 \times 2 \times y z\left(z^2+6 z-16\right)}{2 \times y(z+8)}\)

⇒ \(\frac{2 z\left(z^2+6 z-16\right)}{z+8}=\frac{2 z\left(z^2+8 z-2 z-16\right)}{z+8}\)

Using Identity IV: 8z – 2z = 6z; (8z) (- 2z) = – 16z2

⇒ \(\frac{2 z[z(z+8)-2(z+8)]}{z+8}\) Taking (z + 8) common

⇒ \(\frac{2 z(z+8)(z-2)}{z+8}=\frac{2 z(z-2)}{1}\)

= 2z (z – 2)

5.  5pq(p² – q2) + 2p(p + q)

⇒ \(\frac{5 p q\left(p^2-q^2\right)}{2 p(p+q)}=\frac{5 p q(p+q)(p-q)}{2 p(p+q)}\) Using Identity III

⇒ \(\frac{5}{2} q(p-q)\)

6.  12xy(9x² – 16y²) + 4xy(3x + 4y)

⇒ \(\frac{12 x y\left(9 x^2-16 y^2\right)}{4 x y(3 x+4 y)}=\frac{3\left(9 x^2-16 y^2\right)}{3 x+4 y}\)

⇒ \(\frac{3\left\{(3 x)^2-\left(4 y^2\right)\right\}}{3 x+4 y}\)

⇒ \(\frac{3(3 x+4 y)(3 x-4 y)}{3 x+4 y}\)

⇒ \(\frac{3(3 x-4 y)}{1}=3(3 x-4 y)\)

7.  39y³(50y² – 98) + 26y²(5y + 7)

⇒ \(\frac{39 y^3\left(50 y^2-98\right)}{26 y^2(5 y+7)}\)

⇒ \(\frac{3 \times 13 \times y^3 \times 2 \times\left(25 y^2-49\right)}{2 \times 13 \times y^2(5 y+7)}\) I Taking 2 common

⇒ \(\frac{\left.3 y(5 y)^2-(7)^2\right)}{(5 y+7)}=\frac{3 y(5 y+7)(5 y-7)}{(5 y+7)}\) Using Identity III

⇒ \(\frac{3 y(5 y-7)}{1}=3 y(5 y-7)\)

Factorization Multiple-Choice Questions and Solutions

Question 1. The common factor of x²y² and x³y³ is

1. x²Y²
2. x³y3
3. x²y³
4. x³y²

Solution: 1. x2Y2

x²y²= X X X X y X y

x³y³ = X X X X x X y X y X y

Question 2. The common factor of x²y² and x⁴y is

1. x⁴y²
2. x²y²
3. x³y²
4. x³y

Solution: 4. x³y

x³y²= x x x x x y x y

x⁴y = x X x X X x x x y

Question 3. The common factor of a²m⁴ and a⁴m² is

1. a⁴m⁴
2. a²m²
3. a²m⁴
4. a⁴m²

Solution: 2. a²m

a²m⁴ = a x a x m x m x m x m

a⁴m² = a x a x a x a x m x m

Question 4. The common factor of p³q⁴ and p⁴q³ is

1. p⁴q⁴
2. p⁴q³
3. p³q⁴
4. p³q³

Solution: 3. p³q⁴

p³q⁴=p x p x p x q x q x qx q

p⁴q³ = p x p x p x p x q x q x q,

Question 5. The common factor 12y and 30 is

1. 6
2. 12
3. 30
4. 6y

Solution: 1. 6

12y = 2 x 2 x 3 x y

30 = 2 x 3 x 5

Question 6. The common factor of 2x, 3x³, 4 is

1. 1
2. 2
3. 3
4. 4

Solution: 1. 1

2x = 2 x x

3x³ = 3 x T x x x x

4 = 2 x 2.

Question 7. The common factor of 10ab, 30bc, 50ca is

1. 10
2. 30
3. 50
4. abc

Solution: 1. 10

10a6 = 2 x 5 x a x b

306c = 2 x 3 x 5 x b x c

50ca = 2 x 5 x 5 x c x a

Question 8. The common factor of 14a²b and 35a⁴b²

1. a4b²
2. 35a⁴b²
3. 14a²b
4. 7a²b

Solution: 4. 7a²b

14a²b = 2 x 7 x a x a x b

35a⁴b² = 5 x 7 x a x a x a x a x a x b

Question 9. The common factor of 8a²b⁴c², 12a⁴bc⁴ and 20a³b⁴ is

1. a⁴b⁴
2. a2b²
3. 4a²b²
4. 4a²b.

Solution: 4. 4a²b.

8a²b⁴c² = 2 x 2 x 2 x a x a x b x b x b x b x c x c

12a⁴bc² = 2 x 2 x 3 x a X a X a x a x b x c x c

20a³b⁴ = 2 x 2 x 5 x a x a x a x b x b x b x b

Question 10. The common factor of 6 a³b⁴c², 21a²b and 15a³ is

1. 3a²
2. 3a³
3. 6a³
4. 6a³

Solution: 1. 3a²

6a³b⁴c³ = 2 x 3 x a’x o x a x b x b x t x 6 x c x c

21a²b = 3 x 7 x o x a x 6

15a³ = 3 x 5 x a x a x a

Question 11. The common factor of 2a²b⁴c², 8a⁴b³c4 and 6a³b⁴c² is

1. 2a²b³c²
2. 6a²b³c²
3. 8a²b³c²
4. a⁴b⁴c⁴

Solution: 1. 2a²b³c²

2a²b⁴c² = 2 x a x a X b X b X b X b X c x c

8a⁴b³c² = 2 x 2 x 2 x 0 x a x n x a x b x b x b X c X c X c X c

6a³b⁴c² = 2 X 3 X a x a X a X b X b X b X b X c X C

Question 12. The common factor of 3a²b⁴c², 12b²c⁴ and 15a³b⁴c⁴ is

1. b⁴c⁴
2. 3b²c²
3. 15b²c²
4. 12b²c⁴

Solution: 2. 3b²c²

3a³b⁴c³ = 3 x o x a x b x b x 6 x 6 x c x c

12b³c⁴ = 2 x 2 x 3 x b x. b xcxcxcxc

15a³b⁴c⁴ = 3 x 5 x a x a x a x b x b x b x b X c x c x c x c.

Question 13. The common factor of 3a²b⁴c², 12b³c⁴ and 15a³b⁴b⁴ is

1. 12x³
2. 24x³
3. 36x³
4. 48x³

Solution: 1. 12x³

243y⁴ = 2 x 2 x 2 x 3

x x x x x x x y x y x y x y

36x⁴z⁴ = 2 x 2 x 3 x 3 X X X X X X X X X z X z X z X z

48x³y²z = 2 x 2 x 2 x 2 x 3 x x x x x x x y x y x z.

Question 14. The common factor of 72x³y⁴z⁴, 120z²d⁴x⁴ and 96y³z⁴d⁴ is

1. 96z³
2. 120z³
3. 72z³
4. 24z³

Solution: 4. 24z³

72x³y⁴z⁴ = 2 x 2 x 2 x 3 x 3 x x x x x x X y X y X y X y x z x z x z x z

120z²d⁴x⁴ = 2 x 2 x 2 x 3 x 5 x z x z x d x d x d x d x x x x x x x x

96y³z⁴d⁴ = 2 x 2 x 2 x 2 x 2 x 3 x y X y X y x z X z x z X z x d x d x d x d,

Question 15. The common factor of 36p²q³x⁴, 48pq³x², and 54p³q³x⁴ is

1. 6pq³x²
2. 36pq³x²
3. 54pq³x²
4. 48pq³x²

Solution: 1. 6pq³x²

36p²q³x⁴ = 2 x 2 x 3 x 3 x p x p x q x q x q x x x x x x x x

48pq³x²= 2 X 2 X 2 X 2 X 3 x p x q x q x q x x x x

P³q⁴x⁴ = p x p x p x q x q x q x x x x x x x x.

Question 16. The factorisation of 12a²b + 15ab² is

1. 3ab(4a + 5b)
2. 3a²b(4a + 5b)
3. 3ab²(4a + 5b)
4. 3a²b²(4a + 5b).

Solution: 1. 3ab(4a+ 5b)

12a²b + 16ab² = 3ab(4a + 5b)

Question 17. The factorisation of 10x² – 18x³ + 14x⁴ is

1. 2×2(7x² – 9.x + 5)
2. 2x(7x² – 9x + 5)
3. 2(7x² – 9x + 5)
4. 2×3(7x² – 9x + 5).

Solution: 1. 2x²(7x² – 9x + 5)

10x² – 18x³+ 14x⁴ = 2x² (6 – 9x + 7x²).

Question 18. The factorisation of 6x – 42 is

1. 6(x – 7)
2. 3(x – 7)
3. 2(x – 7)
4. 6(x + 7)

Solution: 1. 6(x – 7)

6x – 42 = 6(x – 7)

Question 19. The factorisation of 6x + 12y is

1. 14a³b³(2b² – 3a²)
2. 3(3x + 4y)
3. 2(3x + 12y)
4. none of these.

Solution: 1. 14a³b³(2b² – 3a²)

6x + 12y = 6(x + 2y)

Question 20. The factorization of 28a³b² – 42a²b³ is

1. 14a³b³(2b² – 3a²)
2. 14a²b³(2b² – 3a²)
3. 14a³b²(2b² – 3a²)
4. none of these.

Solution: 1. 14a³b²(2b² – 3a²)

28a³b³ – 42a²b³ = 14a³b³(2b² – 3a²)

Question 21. The factorisation of a³ + a²b + ab² is

1. a(a² + ab + b²)
2. b(a² + ab + b²)
3. ab(a² + ab + b²)
4. none of these.

Solution: 1. a(a² + ab + b²)

a³ + a²b + ab² = a(a² + ab+ b²)

Question 22. The factorisation of x²yz + xy²z + xyz² is

1. XYZ(x + y + z)
2. x²yz(x + y + z)
3. xy²z(x + y + z)
4. xyz²(x + y + z).

Solution: 1. xyz(x + y + z)

x²yz + xy²z + xyz² = xyz (x+ y + z)

Question 23. The factorisation of ax²y + bxy² + cxyz is

1. xy(ax + by + cz)
2. axy(ax + by + cz)
3. bxy(ax + by + cz)
4. cxy(ax + by + cz).

Solution: 1. xy(ax + by + cz)

ax²y + bxy² + XYZ

= xy (ax + by + cz)

Question 24. The factorisation of a (x + y + z) + b(x + y + z) + c(x + y + z) is

1. (a + b + c)(x + y + z)
2. (ab + be + ca)(x + y + z)
3. (xy + yz + zx)(a + b + c)
4. none of these.

Solution: 1. (a + b + c)(x + y + z)

a(x + y + z) + b(x + y + z) + c(x + y + z)

= (x + y + z) (a + b + c)

Question 25. The factorisation of 6xy – 4y + 6 – 9xis

1. (3x – 2)(2y – 3)
2. (3x + 2)(2y – 3)
3. (3x – 2)(2y + 3)
4. (3x + 2)(2y + 3).

Solution: 1. (3x – 2)(2y – 3)

6xy – 4y + 6 – 9x

= 2y(3x – 2) – 3(-2 + 3x)

= (3x – 2)(2y – 3)

Question 26. The factorisation of x² + xy + 2x + 2y is

1. (x + 2)(x + y)
2. (x + 2)(x – y)
3. (x – 2)(x + y)
4. (x – 2)(x – y).

Solution: 1. (x + 2)(x + y)

x² + xy + 2x + 2y

= x(x + y) + 2(x + y)

= (x + 2) (x + y)

Question 27. The factorization of ax + bx – ay – by is

1. (x – y)(a + b)
2. (x + y)(o + b)
3. (x – y)(a – b)
4. (x + y)(a – b).

Solution: 1. (x – y)(a + b)

ax + bx – ay – by

= x(a. + b) – y(a + b)

= (x – y)(a + b)

Question 28. The factorization of ab – a – b + 1 is

1. (a – 1)(b – 1)
2. (a + 1)(6 + 1)
3. (a – 1)(6 + 1)
4. (a + 1)(6 – 1)

Solution: 1. (a – 1)(b – 1)

ab – a – b + 1

= a(b – 1) – 1(b – 1)

= (a – l)(b – 1).

Question 29. The factorisation of x² + x + xy + y + zx + z is

1. (x + y + z)(x + 1)
2. (x + y + z)(x + y)
3. (x + y + z)(y + z)
4. (x + y + z)(z + x).

Solution: 1. (x + y + z)(x + 1)

x²+ x + xy + y + zx + z

= x(x + 1) + y(x + 1) + z(x + 1)

= (X + 1)(x + y + z)

Question 30. The factorisation of x²y² + xy + xy²z + yz + x²yz + xz is

1. (xy + yz + zx)(xy + 1)
2. (xy + yz + zx)(yz + 1)
3. (xy + yz + zx)(zx + 1)
4. none of these.

Solution: 1. (xy + yz + zx)(xy + 1)

x²y² + xy + xy²z + yz + x²yz + xz

= xy(xy + 1) + yz(xy + 1) + Zx(xy + 1)

= (xy + yz + zx)(xy + 1).

Question 31. The factorisation of x² + 8x + 16 is

1. (x + 2)²
2. (x + 4)²
3. (x – 2)²
4. (x – 4)²

Solution: 2. (x + 4)²

x² + 8x + 16

= (x)2 + 2 (x)(4) + (4)2 = (x+ 4)2.

Question 32. The factorisation of 4y² – 12y + 9 is

1. (2y + 3)²
2. (2y – 3)²
3. (3y + 2)²
4. (3y- 2)²

Solution: 2. (2y – 3)²

4y² – 12y + 9

= (2y)² – 2(2y)(3) + (3)²

= (2y – 3)²

Question 33. The factorisation of 49p² – 36 is

1. (7p + 6)(7p – 6)
2. (6p + 7)(6p – 7)
3. (7p + 6)²
4. (Ip – 6)²

Solution: 1. (Ip + Q)(lp – 6)

49p² – 36

= (7p)² – (6)2 = (7p – 6)(7p + 6)

Question 34. The factorisation of y² – 7y + 12 is

1. (y + 3)(y + 4)
2. (y + 3)(y – 4)
3. (y – 3)(y + 4)
4. (y – 3)(y – 4).

Solution: 4. (y – 3)(y – 4).

y²– 7y + 12

= y2 – 3y – 4y + 12

= y(y – 3) – 4(y – 3)

= (y – 3)(y – 4)

Question 35. The factorisation of z² -4z – 12 is

1. (z + 6)(z + 2)
2. (z – 6)(z – 2)
3. (z – 6)(z + 2)
4. (z + 6)(z – 2).

Solution: 3. (z – 6)(z + 2)

z2 – 4z – 12

= z2 – 6z + 2z – 12

= z(z – 6) + 2(z – 6)

= (z – 6)(z + 2).

Question 36. The factorisation of am²+ bm² + bn² + an² is

1. (a + b)(m² – n²)
2. (a + b)(m² + n²)
3. (a – b)(m² + n²)
4. (a – b)(m² – n²).

Solution: 2. (a + b)(m² + n²)

am² + bm² + bn² + an²

= m²(a + b) + n²(b + a)

= (a + b)(m² + n²).

Question 37. The factorisation of (Im + t) + m + 1 is X

1. (l + l)(m + 1)
2. (l – 1)(m – 1)
3. (l + l)(m – 1)
4. (l – 1)(m + 1).

Solution: 1. (l + 1)(m + 1)

Im + l + m + 1

= l(m + 1) + l(m + 1)

= (l + 1)(m + 1)

Question 38. The factorization of (I + m)² – 4lm is

1. (l- m)²
2. (l+m-2)²
3. (l + m + 2)²
4. none of these

Solution: 1. (l- m)²

(I + m)2 – 4Im

= l² + m²+ 2Im – 4lm

= l²– 2Im + m² = (l- m)²

Question 39. The factorisation of 1 + p + q + r + pq + qr + pr + pqr is

1. (1 +p)(1 + q)(1 + r)
2. (1 – p)1- q)(1- r)
3. (1 – p)(1 – q)(1 + r)
4. (1 + p)(1 – q)(1 – r).

Solution: 1.(l +p)(1 + q)(1 + r)

(1 + p) + q + r + pq + qr + pr + pqr = 1 + p + q + pq + r (1 + p + q + pq)

= (1 + r)(1 + p + q + pq)

= (1 + r)(1 + p)(l + q).

Question 40. The value of 0.645 x 0.645 + 2 x 0.645 x 0.355 + 0.355 x 0.355 is

1. l
2. 0
3. -l
4. 2

Solution: 1. 1

Value = (0.645 + 0.355)² = (l)² = 1.

Question 41. The factorisation of 1 + 16x + 64×2 is

1. (l – 8x)2
2. (l + 8x)2
3. (8 – x)2
4. (8 + x)2

Solution: 2. (1 + 8x)²

1 + 16x + 64X²

= (l)² + 2(1) (8x) + (8x)² = (1 + 8x)²

Question 42. The factorisation of a:2 + x + \(\frac{1}{4}\) is

1. \(\left(\frac{x}{2}-1\right)^2\)
2. \(\left(\frac{x}{2}+1\right)^2\)
3. \(\left(x+\frac{1}{2}\right)^2\)
4. \(\left(x-\frac{1}{2}\right)^2.\)

Solution: 3. \(\left(x+\frac{1}{2}\right)^2\)

⇒ \( x^2+x+\frac{1}{4}=x^2+2(x)\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2\)

⇒ \(\left(x+\frac{1}{2}\right)^2\)

Question 43. The value of 992 is

1. (90)2 + 2(90)(9) + (9)²
2. (90)² _ 2(90)(9) + (9)²
3. (90)² + (9)²
4. None of these

Solution: 1. (90)² + 2(90)(9) + (9)²

992 = (90 + 9)²

= (90)² + 2(90)(9) + (9)²

Question 44. The value of 492 is

1. (50)²– 2(50)(1) + (l)²
2. (50)² + 2 (50) (1) + (l)²
3. (50)² -(1)²
4. (50)²+ (1)²

Solution: 1. (50)2- 2(50)(1) + (1)2

492 = (50 – 1)²

= (50)² _ 2(50)(l) + (1)²

Question 45. The factorisation of\(\left(\frac{x^2}{y^2}-2+\frac{y^2}{x^2}\right)\) x – 0, y * 0 is

1. \(\left(\frac{x}{y}+\frac{y}{x}\right)^2\)
2. \(\left(\frac{x}{y}-\frac{y}{x}\right)^2\)
3. \(\left(\frac{x}{y}-1\right)^2\)
4. \(\left(\frac{x}{y}+1\right)^2.\)

Solution: 2. \(\left(\frac{x}{y}-\frac{y}{x}\right)^2\)

⇒ \(\frac{x^2}{y^2}-2+\frac{y^2}{x^2}\)

⇒ \(\left(\frac{x}{y}\right)^2-2\left(\frac{x}{y}\right)\left(\frac{y}{x}\right)+\left(\frac{y}{x}\right)^2 \)

⇒ \(\left(\frac{x}{y}-\frac{y}{x}\right)^2\)

Question 46. The value of \( \frac{0.73 \times 0.73-0.27 \times 0.27}{0.73-0.27} is\)

1. 1
2. 0
3. 0.73
4. 0.27

Solution: 1. 1

Value= \(\frac{(0.73+0.27)(0.73-0.27)}{0.73-0.27}\) = 1

Question 47. The factorisation of x² – 9 is

1. (x – 3)²
2. (x + 3)²
3. (a + 3)(a – 3)
4. None of these

Solution: 3. (a + 3)(a – 3)

x² – 9 = (x)² – (3)² = (x – 3) (x + 3).

Question 48. The factorisation of 36xy – 1 is

1. (6xy – 1)(6xy + 1)
2. (6xy – 1)²
3. (6xy + 1)²
4. ( 6 + xy)²

Solution: 1. (6xy – 1)(6xy + 1)

36x²y² – 1 = (6xy)² – (1)²

= (6xy – 1)(6xy + 1).

Question 49. The value of \(\frac{0.564 \times 0.564-0.436 \times 0.436}{0.564-0.436} \text { is }\)

1. 0
2. 1
3. -1
4. None of these

Solution: 2. 1

Value = \(\frac{(0.564+0.436)(0.564-0.436)}{0.564-0.436}\) = 1

Question 50. The value of (0.68)² – (0.32)² is

1. -l
2. 0
3. 1
4. 0.36.

Solution:  4. 0

Value = (0.68 + 0.32) (0.68 – 0.32) = 0.36.

Question 51. The factorisation of 3x² + 10x + 8 is

1. (3x + 4)x + 2)
2. (3x – 4)(x – 2)
3. (3x + 4)(x – 2)
4. (3x – 4)(x + 2).

Solution: 1. (3x + 4)(x + 2)

3X² + lOx + 8

= 3X² + 6x + 4x + 8

= 3x(x + 2) + 4(x + 2)

= (x + 2)(3x + 4).

Question 52. The factorisation of 3a² – 16a + 16 is

1. (x – 4)(3a – 4
2. (a + 4)(3a + 4)
3. (a – 4)(3a + 4)
4. (a + 4)(3a – 4).

Solution: 1. (x – 4)(3a – 4)

3×2 – 16x + 16

= 3s2 – 12x – 4x + 16

= 3x(x – 4) – 4(x – 4)

= (x – 4) (3x – 4).

Question 53. The factorisation of 6a² – 5a – 6 is

1. (2a – 3)(3a + 2)
2. (2a + 3)(3a + 2)
3. (2a – 3)(3a – 2)
4. (2a + 3)(3a – 2).

Solution: 1. (2a – 3)(3a + 2)

6X² – 5x – 6

= 6×2 – 9x + 4x – 6

= 3x(2x – 3) + 2(2x – 3)

= (2x – 3)(3x + 2).

Question 54. The factorisation of 6 – a – 2a² is

1. (2 + a)(3 – 2a)
2. (2 + a)(3 + 2a)
3. (2 – a)(3 – 2a)
4. (2 – a)(3 + 2a).

Solution: 1. (2 + a)(3 – 2a)

6 – x – 2x²

= 6 + 3x – 4x – 2x²

= 3(2 + x) – 2x (2 + x)

= (2 + x)(3 – 2x)

Question 55. If x² – x – 42 = (x + k)(x + 6), then k =

1. 6
2. -6
3. 7
4. -7

Solution: 4. -7

x² – x – 42

= x2 – 7x + 6x – 42

= x(x — 7) + 6(x — 7)

= (x – 7)(x + 6)

= (x + k) (x + 6)

k = -7

Question 56. The value of 3.5 x 3.5 – 2.5 x 2.5 is

1. -6
2. 6
3. 60
4. 1

Solution: 2. 6

Value = (3.5 + 2.5)(3.5 – 2.5) = 6

Question 57. If \(\left(x-\frac{1}{x}\right)^2=x^2+a+\frac{1}{x^2}\) then a=

1. -2
2. 2
3. 2x
4. -2x

Solution: 1. -2

⇒ \(\left(x-\frac{1}{x}\right)^2 =x^2-2+\frac{1}{x^2}\)

⇒ \(x^2+a+\frac{1}{x^2} a=-2 .\)

Question 58. If x = 2, y = -1, then the value of x² + 4xy + 4y² is

1. 0
2. 1
3. -1
4. 2

Solution: 1. 0

x² + 4xy + 4y²

= (x)2 + 2(x)(2y) + (2y)²

– (x + 2y)² = {2 + 2(- l)}² = 0

Question 59. The quotient of 28x² ÷ 14r is

1. 2
2. 2x
3. x
4. x2

Solution: 2. 2x

⇒ \(\frac{28 x^2}{14 x}=\frac{2 \times 2 \times 7 \times x \times x}{2 \times 7 \times x}=2 x .\)

Question 60. The quotient of 12a⁸b⁸ ÷ (- 4a⁶b⁶) is

1. 3a²b²
2. 3a²b
3. 3ab²
4. -3a²b²

Solution: 4. -3a²b²

⇒  \(-\frac{12 a^8 b^8}{4 a^6 b^6}\)

⇒  \(2 \times 2 \times 3 \times a \times a \times a \times a \times a \times a \times a \times a \)

⇒  \(-\frac{\times b \times b \times b \times b \times b \times b \times b \times b}{2 \times 2 \times a \times a \times a \times a \times a \times a} \times b \times b \times b \times b \times b \times b\)

= – 3a²b²

Factorization True-False

Write whether the following statements are True or False:

1. An equation is true for all values of its variables: False

2. The difference of the square of two consecutive natural numbers is equal to their sum: True

3. An identity is true for all values of its variables: True

4. (x + 1) (x – 1) (x² + 1) = x4 + 1: False

5. The difference between the areas of the two squares with sides 5o and 5b is 25 (a – b) (a + b): True

Factorization Fill In The Blanks

1. The name of the property a (b + c) = ab + ac is: Distributive Property

2. The greatest common factor of 5a. and 151) is: 5

3. The common factor of 2xy and 3zt is: 1

4. The quotient obtained on dividing (x² – 1) (x – 2) by – (x – 2) is:- (x² – 1)

5. On dividing (x⁴+ y⁴) (x – y) by (x – y), the remainder is: 0

6. Find the value of an in 9a = 502 – 412: 91

7. Ifa + b = 10 and a² + b² = 44, then find ab: 28

8. Factorise x³ – 64x: x (x – 8) (x + 8)

9. x – \(\frac{1}{x}\) = 5, then find the value of \(x^2+\frac{1}{x^2}\): 27

10. Simplify : (a + b)² + (a – b)²: . 2 (a² + b²)

Introduction To Graphs Introduction

The purpose of the graph is to show numerical facts in visual form for their better and quicker understanding. It is especially very useful when there is a trend or comparison to be shown.

A Line Graph

It is a graph which displays data that changes continuously over periods.

Question What is the information that you gather from the given histogram? Try to list them out.
Solution:

This histogram illustrates the distribution of weights (in kg) of 40 persons in a locality. The information that we gather from this histogram is as follows :

A maximum number of persons in the locality have their weights (in kg) in the interval 50-55.

The minimum number of persons in the locality have their weights (in kg) in the interval 40- 45

Read and Learn More NCERT Solutions For Class 8 Maths

Introduction To Graphs Exercise 13.1

Question 1. The following graph shows a patient’s temperature in a hospital, recorded every hour.

1. What was the patient’s temperature at 1 p.m.?
2. When was the patient’s temperature 38.5°C?
3. The patient’s temperature was the same two times during the period given. What were these two times?
4. What was the temperature at 1.30 p.m.? How did you arrive at your answer?
5. During which period did the patient’s temperature show an upward trend?

Solution:

The patient’s temperature at 1 p.m. was 36.5°C.

The patient’s temperature was 38.5°C at 12 noon

The two times when the patient’s temperature was the same were 1 p.m. and 2 p.m.

The temperature at 1.30 p.m. was 36.5°C.

From the graph, we see that the temperature was constant from 1 p.m. to 2 p.m. Since 1.30 p.m. comes in between 1 p.m. and 2 p.m., therefore we arrived at our answer.

The patient’s temperature showed an upward trend during the periods 9 a.m. to 10 a.m., 10 a.m. to 11 a.m. and 2 p.m. to 3 p.m.

Question 2. The following line graph shows the yearly sales figures for a manufacturing company.

1. What were the sales in
   1. 2002
   2. 2006?
2. What were the sales in
   1. 2003
   2. 2005?
3. Compute the difference between the y sales in 2002and 2006.
4. In which year was the greatest difference between the sales compared to the previous year?

Solution:

The sales in

1. 2002 were ₹ 4 crores and in
2. 2006 were ₹ 8 crores.

The sales in

1. 2003 were ₹ 7 crores and in
2. 2005 were ₹ 10 crores.

The difference between the sales in 2002 and 2006 = ₹ 8 crores – ₹ 4 crores = ₹ 4 crores

The difference between the sales in 2002 and 2003 =₹ 7 crores – ? 4 crores =₹ 3 crores

The difference between the sales in 2003 and 2004=₹ 7 crores – ₹ 6 crores = ₹ 1 crore

The difference between the sales in 2004 and 2005 =₹ crores – ₹ 6 crores = ₹ 4 crores

The difference between the sales in 2005 and 2006

= ₹ 10 crores – ₹ 8 crores = ₹ 2 crores

Therefore, in the year 2005, the difference between the sales as compared to the previous year was the greatest.

Question 3. For an experiment in Botany, two different plants, plant A and plant B were grown under similar laboratory conditions. Their heights were measured at the end of each week for 3 weeks. The results are shown in the following graph.

1. How high was Plant A after
   1. 2 weeks
   2. 3 weeks?
2. (b) How high was Plant B after
   1. 2 weeks
   2. 3 weeks?
3. How much did Plant A grow during the 3rd week?
4. How much did Plant B grow from the end of the 2nd week to the end of the 3rd?
5. During which week did Plant. A grow most?
6. During which week did the Plant grow the least?
7. Were the two plants of the same height during any week shown here? Specify.

Solution:

1.  The height of Plant A

1. after 2 weeks was 7 cm high and
2. after 3 weeks it was 9 cm high.

2.  The height of Plant B

1. after 2 weeks it was 7 cm high and
2. after 3 weeks it was 10 cm high.

3. Plant A grew 9 cm – 7 cm = 2 cm during the 3rd week.

4.  From the end of the 2nd week to the end of the 3rd week, Plant B grew = 10 cm – 7 cm = 3 cm.

5.  The Plant A grew in 1st week = 2 cm – 0 cm = 2 cm

The Plant A grew in 2nd week = 7 cm – 2 cm = 5 cm

The Plant A grew in 3rd week = 9 cm – 7 cm = 2 cm

Therefore, the Plant A grew mostly in Weeks second week

6.  The Plant B grew in 1st week = 1 cm – 0 cm = 1cm

The Plant B grew in 2nd week = 7 cm – 1 cm = 6 cm

The Plant B grew in 3rd week = 10 cm – 7 cm = 3 cm

Therefore, Plant B grew the least in the first week.

7.  At the end of the week, the two plants shown here were of the same height.

Question 4. The following shows the temperature forecast and the actual temperature for each day of the week:

1. On which days was the forecast temperature the same as the actual temperature?
2. What was the maximum forecast temperature during the week?
3. What was the minimum actual temperature during the week?
4. On which day did the actual temperature differ the most from the forecast temperature?

Solution:

1. The forecast temperature was the same as the actual temperature on Tuesday, Friday and Sunday. This is indicated by the points where the two graphs intersect.
2. The maximum forecast temperature during the week was 35°C.
3. The minimum actual temperature during the week was 15°C.

Since the maximum temperature difference is 7.5°C, therefore, the actual temperature filtered the most from the forecast temperature on Thursday.

Question 5. Use the tables below to draw linear graphs.

The number of days a hillside city received snow in different years

Population (in thousands) of men and women in a village in different years

Solution:

Question 6. A courier person cycles from a town to a neighbouring suburban area to deliver a parcel to a merchant. His distance from the town at different times is shown in the following graph.

1. What is the scale taken for the time xis?
2. How much time did the person take 4 for the trave?
3. Flow far is the place of the merchant from the tow?
4. Did the person slop on his way? Explain.
5. During which period did he ride fast?

Solution:

1.  The scale taken for the time axis (i.e., along the x-axis) is 4 units = 1 hour.

2.  Time of the start of the travel = 8 a.m.

Time of the end of the travel = 11.30 a.m.

The time taken by the person for the travel = Difference between 8 a.m. and 11.30 a.m. = 31/2 hours.

3.  The place of the merchant from the town is 22 km.

4.  Yes. This is indicated by the horizontal part of the graph (10 a.m. -10.30 a.m.)

Distance covered by the person between 8 a.m. and 9 a.m. = (10 – 0) km = 10 km

Distance covered by the person between 9 a.m. and 10 a.m. = (16- 10)=6 km

covered by km the person between 10 a.m. and 10: 30 a.m. = (16 – 16) km = 0 km

Distance covered by the person between 10.30 a.m. and 11.30 a.m. = (22 – 16) km = 6 km

He rides fastest between 8 a.m. and 9 a.m

Question 7. Can there be a time-temperature graph as follows? Justify your answer.

Solution:

1. Yes, it can be It is a time-temperature graph because it shows an increase in temperature with an increase in time. Yes, it can be
2. it is a time-temperature graph because it shows a decrease in temperature with an increase in time.
3. It cannot be a time-temperature graph because it shows infinitely many different temperatures at one particular time which is not possible.
4. Yes, it can be It is a time-temperature graph because it shows a fixed temperature at different times (or increasing times)

Some Applications

In our daily life, we observe two quantities that are interrelated i.e., the change in one quantity is accompanied by a change in the other quantity, for example: the more the number of days a labourer works, the more the wages he gets. More the sugar we purchase, more the amount we have to pay.

In the first, case, the number of days is called the independent variable (or control variable) and the wages are called the dependent variable. Similarly, in the second case, the amount of sugar is the independent variable whereas the money paid is the dependent variable.

Thus, an independent variable is a variable which can take values freely i.e., whose value is not dependent on any other variable whereas a dependent variable is a variable where value depends upon the other variable, i.e., independent variable.

The relation between the independent and dependent variables can be shown by a graph.

Question 1. The number of litres of petrol you buy to fill a car’s petrol tank will decide the amount you have to pay. Which is the independent variable here? Think about it
Solution:

The independent variable here is the amount of petrol we buy to fill a car’s petrol tank.

Question 2. In the given example use the graph to find how much petrol can be purchased for 800.
Solution:

Mark the point representing 800 the on y-axis. From this point, draw a line, parallel to the line graph at point, draw a line parallel to OY to ow cut OX at a point This point represents litres on the x-axis Hence, 16 litres of petrol can be purchased for 800

Question Is Example a case of direct variation?
Solution:

It is a case of direct variation as

⇒  \(\frac{100}{10}=\frac{200}{20}=\frac{300}{30}=\frac{500}{50}=\frac{1000}{100}\) = 10 (constant)

Exercise 13.2

Question 1. Draw the graph for the following tables of values, with suitable scales on the axes.

1.  Cost of apples

2.  Distance travelled by car

1. How much distance did the car cover during the period 7.30 a.m. to 8 a.m.?
2. What was the time when the car had covered a distance of 1000 km since its star?

3.  Interest on deposits for a year

1. Does the graph pass through the origin?
2. Use the graph to find the interest on 72500 for a year.
3. To get an interest of f280 per year, how much money should be deposited?

Solution:

Scale:

On the horizontal axis: 2 units = 1 apple

On the  vertical axis: 1 unit = ? 5

Mark the number of apples on the horizontal axis.

Mark cost (in ?)the on vertical axis.

Plot, the points: (1, 5), (2, 10), (3, 15), (4, 20) and (5, 25).

Join the points.

We get a linear graph.

Scale :

On the  horizontal axis: 2 units = 1 hour

On the  vertical axis: 2 units = 40 km

Mark time (in hours)the on horizontal axis.

Mark distances (in km)the on vertical axis.

Plot the points (6 a.m., 40), (7 a.m., 80), (8 a.m., 120) and (9 a.m. 160).

Join the points.

We get a linear graph.

Distance tance covered during 7.30 a.m. to 8 a.m. = 120 km – 100 km = 20 km

The time when the car had covered a distance of 100 k Since its start was 7.30 a.m.

Scale :

On the horizontal axis: 2 units? 1000

On the vertical axis: 2 units  80

Mark deposit (in ?) on the horizontal axis.

Mark simple interest (in ?) on the vertical axis.

Plot the points (1000, 80), (2000, 160), (3000, 240), (4000, 320) and (5000, 400).

Join the points.

We get a linear graph.

Yes, The graph passes through the origin.

Interest o ? 2500 for a year = ? 200

To get an interest of? 280 per year? 3500 should be deposited.

Question 2. Draw a graph for the following:

Is it a linear graph?

Is it a linear graph?

Solution:

Scale:

On the  horizontal axis : 1 unit = 1 cmOn thee  vertical axi : 1 unit = 4 cm

Mark the  side of the square (in cm) on the  horizontal axis

Mark the  perimeter (in cm) on vertical

Plot the points (2, 8), (3,12), (3.5, 14), (5, 20) and (6, 24).

Join the points.

Yes, It is a linear graph.

Scale :

On the horizontal axis: 2 units = 2 cm

On the vertical axis: 1 unit = 2 cm

Mark the side of the square (in cm) on the horizontal axis.

Mark the area (in cm2) on the vertical axis.

Plot the points (2, 4), (3, 9), (4, 16), (5, 25) and (6, 36).

Join the points.

The graph we get is not linear.

Multiple-Choice Questions and Solutions

Observe the following bar graph and answer the related questions:

Question 1. On which head, is the expenditure maximum?

1. Travelling allowance
2. Rent
3. Appliances
4. Salary of employees.

Solution: 4. Salary of employees.

Question 2. On which head or head, is the expenditure minimum?

1. Travelling allowance or rent
2. Appliances
3. Salary of employees
4. Others.

Solution: 1. Travelling allowance or rent

Question 3. On which two heads, is the expenditure the same?

1. Salary of employees and others
2. Travelling allowance and rent
3. Appliances and rent
4. Appliances and others

Solution: 2. Travelling allowance and rent

Question 4. What is the difference in expenditure (in thousands of rupees) between the salary of employees and rent?

1. 100
2. 200
3. 300
4. 400

Solution: 3. 300

Question 5. What is the sum of the expenditures (in thousands of rupees) on travelling allowance and rent?

1. 100
2. 200
3. 300
4. 400

Solution: 2. 200

Observe the following circle graph and answer the related questions:

Question 6. On which head is the expenditure maximum?

1. Food
2. Clothes
3. House rent
4. Education.

Solution: 1. Food

Question 7. On which head is the expenditure minimum

1. Education
2. Food
3. House rent
4. Clothes.

Solution: 1. Education

Question 8. If the budget of the family is ₹ 10800, what is the savings?

1. ₹ 1050
2. ₹ 950
3. ₹ 1000
4. ₹ 1200.

Solution: 1. ₹ 1050

Question 9. What is the difference in spending on clothes and education the budget of the family is ₹ 1000.

1. ₹ 1200
2. ₹ 800
3. ₹ 1000
4. ₹ 1500.

Solution: 1. ₹ 1200

Question 10. What is the sum of the expenditures on food and education if the budget of the family is ₹ 1080?

1. ₹ 5000
2. ₹ 5400
3. ₹ 8000
4. ₹ 6000

Solution: 3. ₹ 8000

Observe the following circle graph and answer the related questions:

Question 11. In which class intervals, are the maximum number of students?

1. 0-5
2. 5-10
3. 20-25
4. 15-20

Solution: 3. 20-25

Question 12. In which class intervals, are the minimum number of students?

1. 0-5
2. 5-10
3. 10-15
4. 15-20

Solution: 1. 0-5

Question 13. In which class intervals, is the number of students 200?

1. 0-5
2. 5-10
3. 20-25
4. 15-20

Solution: 1. 0-5

Question 14. The difference in the number of students in class intervals 0-5 and 5- 10 is

1. 100
2. 200
3. 300
4. 400

Solution: 1. 100

Question 15. The sum of the number of students in the class intervals 10-15 and 20-25 is

1. 800
2. 900
3. 600
4. 400

Solution: 1. 800

Observe the following temperature-timeline graph and answer the related questions:

Question 16. At what time is the temperature maximum?

1. 13 hours
2. 15 hours
3. 11 hours
4. 19 hours

Solution: 1. 13 hours

Question 17. At what time(s) is the temperature minimum?

1. 7 hours and 21 hours
2. 9 hours
3. 11 hours
4. 13 hours.

Solution: 1. 7 hours and 21 hours

Question 18. 103°F temperature is the time

1. 11 hours
2. 13 hours
3. 15 hours
4. 21 hours.

Solution: 1. 11 hours

Question 19. What is the difference of temperatures at 7 hours and 21 hours?

1. 0°F
2. 1°F
3. 2°F
4. 3°F

Solution: 1. 0°F

Question 20. What is the rise in temperature from 11 hours to 13 hours?

1. 1°F
2. 2°F
3. 3°F
4. 4°F

Solution: 1. 1°F

Question 21. What is the fall in temperature from 13 hours to 21 hours?

1. 2°F
2. 3°F
3. 4°F
4. 6°F

Solution: 1. 2°F

Question 22. The coordinates of the origin are

1. (0,0)
2. (1,0)
3. (0,1)
4. (1,1)

Solution: 1. (0,0)

Question 23. What are the coordinates of a point whose .v-coordinate is 3 andy-coordinate is 4?

1. (3,3)
2. (3,4)
3. (4,3)
4. (4,3)

Solution: 2. (3,4)

Question 24. What are the coordinates of a point whose x-coordinate is 1 and y-coordinate is 0?

1. (1,0)
2. (0,0)
3. (0,1)
4. (1,1)

Solution: 1. (1,0)

Question 25. What are the coordinates of a point whose v-coordinate is 0 and whose y-coordinate is?

1. (0,1)
2. (0,0)
3. (1,0)
4. (1,1).

Solution: 1. (0,1)

Observe the following velocity-lime graph and answer the related questions:

Question 26. At what time is the velocity maximum?

1. 7
2. 8
3. 9
4. 10

Solution: 1. 6

Question 27. At what time is the velocity minimum?

1. 8
2. 9
3. 10
4. 11

Solution: 4. 11

Question 28. At what times are the velocities equal?

1. 8 and 12
2. 9 and 11
3. 7 and 12
4. 11 and 13.

Solution: 1. 8 and 12

Question 29. What is the fall in velocity from 7 to 1 1?

1. 80 km/hour
2. 90 km/hour
3. 100 km/hour
4. 20 km/hour.

Solution: 1. 80 km/hour

Question 30. What is the rise in velocity from 11 to 12?

1. 10 km/hour
2. 20 km/hour
3. 30 km/hour
4. 60 km/hour.

Solution: 4. 60 km/hour.

Observe the following runs-over graph and answer the related questions:

Question 31. In which over is the maximum scoreboard scored?

1. 2
2. 4
3. 5
4. 6

Solution: 4. 6

Question 32. In which over are the maximum runs scored?

1. 10
2. 11
3. 12
4. 9

Solution: 3. 12

Question 33. What is the difference in runs scored in IV and V overs?

1. 1
2. 2
3. 3
4. 4

Solution: 1. 1

Question 34. What is the sum of runs scored in I and XII overs?

1. 1
2. 2
3. 3
4. 4

Solution: 1. 1

Question 35. 3 runs are scored in which over?

1. 2 and 10
2. 1 and 5
3. 7 and 8
4. 10 and 12

Solution: 3. 7 and 8

Read the graph and answer the related questions:

Question 36. In which year was the rate of interest maximum?

1. 2005
2. 2003
3. 2006
4. 2002.

Solution: 1. 2005

Question 37. In which year was the rate of interest minimum 

1. 2006
2. 2004
3. 2001
4. 2005.

Solution: 1. 2006

Question 38. The difference in the maximum and minimum rates of interests

1. 2%
2. 4%
3. 6%
4. 8%.

Solution: 4. 8%.

Question 39. The rise in interest from 2004 to 2005 was

1. 2%
2. 4%
3. 6%
4. 8%.

Solution: 2. 4%

Question 40. The fall in interest from 2001 to 2002 was

1. l%
2. 2%
3. 3%
4. 4%.

Solution: 2. 2%

Read the graph and answer the related questions :

Question 41. How many students appeared in the year 2000?

1. 200
2. 250
3. 300
4. 350.

Solution: 2. 250

Question 42. In which year did 50 students appear? 

1. 1998
2. 2001
3. 1997
4. 1996.

Solution: 4. 1996.

Question 43. In which year did the maximum number of students appear?

1. 2002
2. 2000
3. 2001
4. 1999.

Solution: 1. 2002

Question 44. What is the maximum number of students that appeared in any year?

1. 350
2. 300
3. 250
4. 300.

Solution: 1. 350

Question 45. In which two years was the number of students appearing the same?

1. 1997 and 1998
2. 1998 and 1999
3. 1999 and 2000
4. 2000 and 2001.

Solution: 1. 1997 and 1998

Read the graph and answer the related questions:

Question 46. The maximum books are of which subject?

1. Hindi
2. Science
3. English
4. Sanskrit.

Solution: 1. Hindi

Question 47. The minimum books are of which subject

1. Home Science
2. Sanskrit
3. Science
4. English.

Solution: 1. Home Science

Question 18. 500 books are of which subject 

1. English
2. Science
3. Hindi
4. Maths.

Solution: 4. Maths.

Question 49. How many books are on the subject of Home Science?

1. 100
2. 200
3. 300
4. 400

Solution: 1. 100

Question 50. How many books are there in Sanskrit and Home Science taken together?

1. 100
2. 200
3. 300
4. 400

Solution: 3. 300

Read the graph and answer the related questions:

Question 51. In which year was the number of labourers maximum?

1. 2001
2. 2002
3. 2003
4. 2004

Solution: 4. 2004

Question 52. In which year was the number of labourers minimum?

1. 2003
2. 2004
3. 2005
4. 2006

Solution: 4. 2006

Question 53. What was the difference in the number of labourers in the years 2002 and 203?

1. 100
2. 200
3. 300
4. 400

Solution: 2. 200

Question 54. Find the rise in the number of labourers from 2001 to 2004.

1. 200
2. 300
3. 400
4. 500

Solution: 3. 400

Question 55. Find the sum of the number of labourers in the years 2004 and 2006.

1. 700
2. 600
3. 200
4. 500

Solution: 1. 700

Read the circle graph and answer the related questions:

Question 56. There are in all 101000 students a school. The number of students in the class I are

1. 500
2. 250
3. 125
4. None of these

Solution: 1. 500

Question 57. The number of students in class II is

1. 500
2. 250
3. 125
4. 100

Solution: 2. 250

Question 58. In which two classes is the number of students the sesame 

1. 1 and 2
2. 1 and 3
3. 3 and 4
4. 1 and 4

Solution: 3. 3 and 4

Question 59. The minimum number of students in any class is

1. 125
2. 250
3. 500
4. 1000

Solution: 1. 125

Question 60. The sum of the number of students in class 3 and class 4 is

1. 500
2. 1000
3. 50
4. 250

Solution: 4. 250

True-False

Write whether the following statements are ‘D ue or False:

1. The coordinates of the origin are (0, 0): True

2. The .v-coordinate of a point is its distance from r-the axis: False

3. For fixing a point on the graph sheet, we need two coordinates: True

4. The relation between the pendent variable and the dependent variable is shown through a graph: True

Fill in the Blanks

1. Comparison of parts of a whole can be done by_____: pie chart

2. A point which lies on both these is_____: (0,0)

3. The r-coordinate of every point lying on the y-axis is_____: 0

4. The process of fixing a point with the help of x and y coordinates is called_____of the point: Plotting

5. What will you get after joining the points (-1, -1), (0, 0) and (3, 3): A straight line passing through the origin

6. What will you get after joining the points (1, 0), (1, 1) and (1, -1): A straight line not passing through the origin

7. What is the distance of the point (3, 4) from s-axis: 4

8. In the following graph, what are the coordinates of P: (3,2)

9. In the following graph, which letter indicates (the point (3, 0): A

NCERT Solutions For Class 8 Maths Chapter 1 Rational Numbers Introduction

• The counting numbers 1, 2, 3, 4,………… are called natural numbers.
• The smallest natural number is 1. There is no greatest natural number.
• If we add 0 to the collection of natural numbers, what we get is called the collection of whole numbers. are called natural numbers. are whole numbers.
• Every natural number is a whole number but every whole number is not a natural number, e.g., 0 is a whole number but not a natural number.
• The positive integers are the same as natural numbers.
• We get negative integers if we put a negative sign before each positive integer. Thus, – 1, – 2, – 3, – 4,
• The smallest positive integer is 1. The greatest negative integer is -1.
• The number ‘O’ is neither positive nor negative. It is greater than every negative integer and smaller than every positive integer. It is simply an integer.
• Thus, 0, 1, 2, 3, and 4, are negative integers.÷
• Several p/q where p and q are integers and q ≠ 0 is called a rational number.
• All the above types of numbers are needed to solve various kinds of simple algebraic equations  

Read and Learn More NCERT Solutions For Class 8 Maths

Properties Of Rational Numbers

The list of properties of rational Numbers can be given as follows :

1. Closure
2. Commutativity
3. Associativity
4. The role of zero
5. The role of 1
6. Negative (or additive inverse) of a number
7. Reciprocal (or multiplicative inverse)
8. Distributivity of multiplication over addition for rational numbers

Closure

Whole numbers

Question 1. 4+7 =________________ Is it a whole number
Solution:

4 + 7=11

Yes, It is a whole number.

Question 2. 3×7 = _______________Is it a whole number?
Solution:

3 × 7 = 21

Yes, It is a whole number

Question 3. Check for closure property under all four operations for natural numbers
Solution:

Integers

Question 1. Is – 7 + (- 5) an integer?
Solution:

– 7 + (- 5)

– 7 + (- 5) = – 12

Yes,  Which is an integer.

Question 2. Is 8 + 5 an integer?
Solution:

8 + 5

8 + 5 = 13

Yes,  Which is an integer

Question 3. Is 5 – 7 an integer?
Solution:

5 – 7

5 – 7 = -2

Yes, Which is an integer.

Question 4. Is 8 – (- 6) an integer?
Solution:

8 – (- 6)

8 – (- 6) = 8 + 6

= 14,

Yes, Which is an integer.

Question 5. Is – 5 × 8= do integer?
Solution:

– 5 × 8

– 5 × 8= – 40

Yes,  Which is an integer

Rational numbers

Question 1.\(\frac{-3}{8}+\frac{(-4)}{5}\)= ________ Is it a rational number ?
Solution:

⇒ \(\frac{-3}{8}+\frac{(-4)}{5}\)

=\(\frac{-15+(-32)}{40}\)

= \(\frac{-47}{40}\)

Yes, It is a rational number

Question 2. \(\frac{5}{8}-\frac{4}{5}=\frac{25-32}{40}\) = _________________ it is a ralional number?
Solution:

⇒ \(\frac{5}{8}-\frac{4}{5}\)

= \(\frac{25-32}{40}\)

= \(\frac{-7}{40}\)

Yes, It is a rational number.

Question 3. \(\frac{-5}{9}+\frac{(-7)}{6}\) _______________ Is it a rational number?
Solution:

⇒ \(\frac{3}{7}-\left(\frac{-8}{5}\right)=\frac{3}{7}+\frac{8}{5}\)

= \(\frac{15+56}{35}\)

=\(\frac{71}{35}\) .

Yes, It is a rational number.

Question 4. \(\frac{4}{7}+\frac{6}{11}\) Is it a rational number ?
Solution:

⇒  \(\frac{4}{7}+\frac{6}{11}\)

= \(\frac{44+42}{77}=\frac{86}{77}\)

Yes, It is a rational number

Question 5. \({4}{5} \times \frac{-6}{11}\)___________Is it a rational number?
Solution:

⇒ \(\frac{4}{5} \times \frac{-6}{11}\)

= \(\frac{24}{55}\)

Yes, It is a rational number.

Question 6. \(\frac{2}{7}+\frac{5}{3}\)= _______________Is it a rational number?
Solution:

⇒ \(\frac{2}{7} \div \frac{5}{3}\)

=\(\frac{2}{7} \times \frac{3}{5}\)

= \(\frac{6}{35}\)

Yes, It is a rational number

Question 7. \(\frac{-3}{8}+\frac{-2}{9}\)= ______________ It is a rational numbers
Solution:

⇒ \(\frac{-3}{8} \div \frac{-2}{9}\)

= \(\frac{-3}{8} \times \frac{-9}{2}\)

= \(\frac{27}{16}\)

Yes, it is a rational number.

Question 8. Check it for a few more pairs of rational numbers

1. \(\frac{3}{5}+\frac{7}{13}\)
Solution: 

⇒ \(\frac{3}{5}+\frac{7}{13}\)

= \(\frac{39+35}{65}\)

= \(\frac{74}{65}\)

Rational number

2. \(\frac{3}{7}+\frac{(-4)}{9}\)
Solution:

⇒ \(\frac{3}{7}+\frac{(-4)}{9}\)

= \(\frac{27+(-28)}{63}\)

Rational number

3. \(\frac{-5}{9}+\frac{(-7)}{6}\)
Solution:

⇒ \(\frac{-5}{9}+\frac{(-7)}{6}\)

= \(\frac{-10+(-21)}{18}\)

Rational number

Question 9. Try this for some more pairs of rational numbers

1. \(\frac{-3}{8}-\frac{4}{5}\)
Solution:

⇒ \(\frac{-3}{8}-\frac{4}{5}\)

= \(\frac{-15-32}{40}\)

=\(\frac{-47}{40}\)

Rational number

2. \(\frac{2}{9}-\frac{3}{7}\)
Solution:

⇒ \(\frac{2}{9}-\frac{3}{7}\)

= \(\frac{14-27}{63}\)

= –\(\frac{13}{63}\)

Rational number

3. \(\frac{-3}{4} \times \frac{7}{8}\)
Solution:

⇒ \(\frac{-3}{4} \times \frac{7}{8}\)

= –\(\frac{21}{32}\)

Rational number

4. \(\frac{5}{8} \times \frac{3}{7}\)
Solution:

⇒ \(\frac{5}{8} \times \frac{3}{7}\)

= \(\frac{15}{56}\)

Rational number

Question 10. Fill in the blanks in the following table

Solution:

Commutativity

Whole numbers

Question 1. Recall the commutativity of different operations for whole numbers by filling the following table.

Check whether the commutativity of the. operations hold for natural numbers also.
Solution:

Commutativity of the operations for natural numbers also.

Integers

Question 1.  Fill in the following table and check the commutativity of different operations for integers:

Solution:

Rational numbers

Question 1. \(\frac{-6}{5}+\left(\frac{-8}{3}\right)\) = \(\ldots \ldots \text { and } \frac{-8}{3}+\left(\frac{-6}{5}\right)=\ldots\) Is \(\frac{-6}{5}+\left(\frac{-8}{3}\right) \)= \(\left(\frac{-8}{3}\right)+\left(\frac{-6}{5}\right) \)
Solution:

⇒ \(\frac{-6}{5}+\left(\frac{-8}{3}\right)\)

= \( \frac{-18+(-40)}{15}=\frac{-58}{15}\)

= \(\frac{-8}{3}+\left(\frac{-6}{5}\right)\)

= \(\frac{-40+(-18)}{15}=\frac{-58}{15}\)

So, \(\frac{-6}{5}+\left(\frac{-8}{3}\right)=\left(\frac{-8}{3}\right)+\left(\frac{-6}{5}\right)\)

Question 2. Is \(\frac{3}{8}+\frac{1}{7}=\frac{1}{7}+\left(\frac{-3}{8}\right)\)?
Solution.

⇒ \(\frac{-3}{8}+\frac{1}{7}=\frac{-21+8}{56} \)

= \(\frac{-13}{56}\)

= \(\frac{1}{7}+\left(\frac{-3}{8}\right)=\frac{8+(-21)}{56}\)

= \(\frac{-13}{56}\)

So, \(\frac{-3}{8}+\frac{1}{7}=\frac{1}{7}+\left(\frac{-3}{8}\right)\).

Question 3. Is \(\frac{2}{3}-\frac{5}{4}=\frac{5}{4}-\frac{2}{3}\)
Solution:

⇒ \(\frac{2}{3}-\frac{5}{4}=\frac{8-15}{12}=\frac{-7}{12} \)

⇒ \(\frac{5}{4}-\frac{2}{3}=\frac{15-8}{12}=\frac{7}{12}\)

⇒  \(\frac{-7}{12} \neq \frac{7}{12} \)

∴ \(\frac{2}{3}-\frac{5}{4} \neq \frac{5}{4}-\frac{2}{3} \)

⇒  \(\frac{1}{2}-\frac{3}{5}=\frac{3}{5}-\frac{1}{2} \)

⇒  \(\frac{1}{2}-\frac{3}{5}=\frac{5-6}{10}=\frac{-1}{10} \)

⇒  \(\frac{3}{5}-\frac{1}{2}=\frac{6-5}{10}=\frac{1}{10} \)

⇒  \(\frac{-1}{10} \neq \frac{1}{10} \)

⇒  \(\frac{1}{2}-\frac{3}{5} \neq \frac{3}{5}-\frac{1}{2}\)

Question 4. \(\frac{-8}{9} \times\left(\frac{-4}{7}\right)=\frac{32}{63} \)
Solution:

⇒ \(\frac{-4}{7} \times\left(\frac{-8}{9}\right)=\frac{32}{63}\)

So, \(\frac{-8}{9} \times\left(\frac{-4}{7}\right)=\frac{-4}{7} \times\left(\frac{-8}{9}\right)\)

Question 5. \(\frac{-5}{9} \times \frac{7}{8}\)
Solution:

= \(\frac{7}{8} \times\left(\frac{-5}{9}\right)\)

= \(\frac{-35}{72}\)

Question 6. \(\frac{-6}{11} \times\left(\frac{-3}{5}\right)\)
Solution:

⇒ \(\frac{-6}{11} \times\left(\frac{-3}{5}\right)\)

=\(\frac{-3}{5} \times\left(\frac{-6}{11}\right)\)

= \(\frac{18}{55}\)

⇒ \(\frac{2}{3} \times \frac{4}{7}=\frac{4}{7} \times \frac{2}{3}=\frac{8}{21}\)

Question 7. Is \(\frac{-5}{4} \div \frac{3}{7}=\frac{3}{7}\div\left(\frac{-5}{4}\right)\)
Solution:

⇒ \(\frac{-5}{4} \div \frac{3}{7}=\frac{3}{7} \div\left(\frac{-5}{4}\right)\)

= \(\frac{-35}{12}\)

And \(\frac{3}{7} \div \frac{-5}{4}=\frac{3}{7} \times\left(\frac{-4}{5}\right) \)

=  \(\frac{-12}{35}\)

Since \(\frac{-35}{12} \neq \frac{-12}{35} \)

∴ \(\frac{-5}{4} \div \frac{3}{7} \neq \frac{3}{7} \div\left(\frac{-5}{4}\right)\)

Question 8. Complete the following table:

Solution: 

Associativity

Whole numbers

Question 1. Recall the associativity of the four operations for whole numbers through this table

Fill in this table and. verify the remarks given in the last column. Check for yourself the associativity of different operations for natural numbers
Solution:

Integers

Question 1. Is (-2) + [3 + (-4)] = [(-2) + 3] + (-4)?
Solution:

Yes; (-2) + [3 + (-4)]

= [(-2) + 3] + (-4)

= -3.

Addition is associative

Question 2. -Is (-6) +[(-4) +(-5)] = (-6) + (-4)] + (-5)?
Solution:

Yes; (-6) + [(-4) + (-5)]

= [(-6) + (-4)] + (-5)

= -15

Addition is associative

Question 3. Is 5 – (7 – 3) = (5 – 7) – 3?
Solution:

5 – (7 – 3) = 5 – 4

= 1

(5 – 7) – 3 = -2 – 3

= -5

1 ≠ – 5

∴  No; 5 – (7 – 3) ≠ (5 – 7) – 3.

Subtraction is not associative

Question 4. Is 5× [(- 7) × (-8)] = [5× (- 7)] × (- 8) ?
Solution:

Yes; 5 × [(-7) × (-8)]

= [5× (-7)] ×(-8)

= 280.

Multiplication is associative

Question 5. Is (- 4) × [(- 8)×(- 5)]= (-4) x (-8)]× (-5)
Solution:

Yes; (-4) × [(-8) × (-5)]

= [(-4) × (-8)] × (-5)

= -160

Multiplication is associative

Question 6.  Is [(-10) ÷ 2] ÷ (- 5)= (-10) ÷[2 ÷ (-5)]?
Solution:

[(-10) ÷ 2] ÷ (-5)= (-5) ÷ (-5) = 1

(-10) ÷ [2 ÷ (-5)]= (-10) ÷ \(\frac{-2}{5}\)

= (-10) ×\(\frac{-5}{2}\)

= 25

∴ 1 ≠ 25

∴ No; [(-10) ÷ 2] ÷ (-5)≠ (-10) ÷ [2 ÷ (-5)]

The division is not associative

Rational numbers

Question 1. Find \(\frac{-1}{2}+\left[\frac{3}{7}+\left(\frac{-4}{3}\right)\right]\) and \(\left[\frac{-1}{2}+\frac{3}{7}\right]+\left(\frac{-4}{3}\right) \text {. }\) Are the two sums equal?
Solution:

⇒ \(\frac{-1}{2}+\left[\frac{3}{7}+\left(\frac{-4}{3}\right)\right]\)

= \(\frac{-1}{2}+\frac{9+(-28)}{21}=\frac{-1}{2}+\left(\frac{-19}{21}\right)\)

= \(\frac{-21+(-38)}{42}=\frac{-59}{42}\)

And \( {\left[\frac{-1}{2}+\frac{3}{7}\right]+\left(\frac{-4}{3}\right)}\)

= \(\frac{-7+6}{14}+\left(\frac{-4}{3}\right)=\frac{-1}{14}+\left(\frac{-4}{3}\right)\)

= \(\frac{-3+(-56)}{42}=\frac{-59}{42}\)

= \(\frac{-1}{2}+\left[\frac{3}{7}+\left(\frac{-4}{3}\right)\right] \)

= \(\left[\frac{-1}{2}+\frac{3}{7}\right]+\left(\frac{-4}{3}\right)\)

= \(\left[\frac{-1}{2}+\frac{3}{7}\right]+\left(\frac{-4}{3}\right)\)

Question 2. \(\frac{-3}{4}+\left[\frac{2}{3}+\left(\frac{-6}{7}\right)\right]\)
Solution:

⇒ \(\frac{-3}{4}+\left[\frac{2}{3}+\left(\frac{-6}{7}\right)\right]\)

= \(\frac{-3}{4}+\frac{14+(-18)}{21}=\frac{-3}{4}+\left(\frac{-4}{21}\right)\)

= \(\frac{-63+(-16)}{84}=\frac{-79}{84} \)

= \(\left[\frac{-3}{4}+\frac{2}{3}\right]+\left(\frac{-6}{7}\right) \)

= \(\frac{-9+(8)}{12}+\left(\frac{-6}{7}\right)=\frac{-1}{12}+\left(\frac{-6}{7}\right) \)

= \(\frac{-7+(-72)}{84}=\frac{-79}{84}\)

So,yes \(\frac{-3}{4}+\left[\frac{2}{3}+\left(\frac{-6}{7}\right)\right]\)

= \(\left[\frac{-3}{4}+\frac{2}{3}\right]+\left(\frac{-6}{7}\right) \)

Question 3. \( \frac{-1}{4}+\left[\frac{2}{9}\right.\left.+\left(\frac{-5}{11}\right)\right] \)
Solution:

⇒  \( \frac{-1}{4}+\left[\frac{2}{9}\right.\left.+\left(\frac{-5}{11}\right)\right] \)

= \(\frac{-1}{4}+\left(\frac{22-(45)}{99}\right) \)

= \(\frac{-1}{4}+\left(\frac{-23}{99}\right) \)

⇒ \(\left[\frac{-1}{4}\right.\left.+\frac{2}{9}\right]+\left(\frac{-5}{11}\right)\)

= \(\frac{-99+(-92)}{396}=\frac{-191}{396}\)

= \(\frac{-9+8}{36}+\left(\frac{-5}{11}\right)\)

= \(\frac{-1}{36}+\left(\frac{-5}{11}\right)=\frac{-11+(-180)}{396}\)

= \(\frac{-191}{396} \)

⇒ \(\frac{-1}{4}+\left[\frac{2}{9}+\left(\frac{-5}{11}\right)\right] \)

= \(\left[\frac{-1}{4}+\frac{2}{9}\right]+\left(\frac{-5}{11}\right)\)

Question 4. \(\frac{-2}{3}-\left[\frac{-4}{5}-\frac{1}{2}\right]\) Check for yourself
Solution:

= \(\left[\frac{-2}{3}-\left(\frac{-4}{5}\right)\right]-\frac{1}{2}\)

= \(\frac{-2}{3}-\left[\frac{-4}{5}-\frac{1}{2}\right]\)

= \(\frac{-2}{3}-\left(\frac{-8-5}{10}\right)=\frac{-2}{3}-\left(\frac{-13}{10}\right)\)

= \(\frac{-2}{3}+\frac{13}{10}=\frac{-20+39}{30}=\frac{19}{30}\)

And \({\left[\frac{-2}{3}-\left(\frac{-4}{5}\right)\right]-\frac{1}{2}=\left[\frac{-2}{3}+\frac{4}{5}\right]-\frac{1}{2}} \)

= \(\left[\frac{-10+12}{15}\right]-\frac{1}{2}=\frac{2}{15}-\frac{1}{2} \)

= \(\frac{4-15}{30}=\frac{-11}{30}\)

So, No ; \(\frac{-2}{3}-\left[\frac{-4}{5}-\frac{1}{2}\right]\)≠ \(\frac{-2}{3}\)– (\(\frac{-4}{5}\)) – \(\frac{1}{2}\)

Question 5. \(\left(\frac{-7}{3} \times \frac{5}{4}\right) \times \frac{2}{9}\)___________
Solution:

⇒\(\left(\frac{-7}{3} \times \frac{5}{4}\right) \times \frac{2}{9}=\frac{-35}{12} \times \frac{2}{9}\)

=\(\frac{-35}{54}\)

So, Yes ;

Question 6. Is \(\frac{2}{3} \times\left(\frac{-6}{7} \times \frac{4}{5}\right)=\left(\frac{2}{3} \times \frac{-6}{7}\right) \times \frac{4}{5} \)
Solution:

⇒ \(\frac{2}{3} \times\left(\frac{-6}{7} \times \frac{4}{5}\right)\)

= \(\frac{2}{3} \times \frac{-24}{35}=\frac{-48}{105}\)________ (1)

=  \(\frac{2}{3}\)× \(\frac{-6}{7}\)× \(\frac{4}{5}\)

= \(\frac{-12}{21} \times \frac{4}{5}=\frac{-48}{105}\)________ (2)

So, Yes; A = B

Question 7. \(\frac{-6}{5} \times\left(\frac{2}{3} \times \frac{1}{8}\right)\)
Solution:

⇒ \(\frac{-6}{5} \times\left(\frac{2}{3} \times \frac{1}{8}\right)=\frac{-6}{5} \times \frac{1}{12}\)

=\(\frac{-1}{10}\)

And  \(\left(\frac{-6}{5} \times \frac{2}{3}\right) \times \frac{1}{8}=\frac{-4}{5} \times \frac{1}{8}=\frac{-1}{10}\)

So, Yes

⇒ \(\frac{-6}{5} \times\left(\frac{2}{3} \times \frac{1}{8}\right)=\left(\frac{-6}{5} \times \frac{2}{3}\right) \times \frac{1}{8}\)

Question 8. \(\frac{2}{7} \times\left(\frac{-5}{9} \times \frac{2}{3}\right) \)
Solution:

⇒ \(\frac{2}{7} \times\left(\frac{-5}{9} \times \frac{2}{3}\right) \)

=\(\frac{2}{7} \times\left(\frac{-10}{27}\right)=\frac{-20}{189} \)

= \(\left(\frac{2}{7} \times \frac{-5}{9}\right) \times \frac{2}{3}=\frac{-10}{63} \times \frac{2}{3}\)

=\(\frac{-20}{189}\)

So, Yes; A=B

Question 9. \(\frac{1}{2}+\left[\frac{-1}{3}+\frac{2}{5}\right]=\left[\frac{1}{2}+\left(\frac{-1}{3}\right)\right]+\frac{2}{5}\),Is L.H.S. =R.H.S. ? Check for yourself
Solution:

L.H.S.\(\frac{1}{2}+\left(\frac{-1}{3}+\frac{2}{5}\right)\)

= \(\frac{1}{2} \div\left(\frac{-1}{3} \times \frac{5}{2}\right)\)

= \(\frac{2}{5} is \frac{5}{2}\)

= \(\frac{1}{2}+\left(-\frac{5}{6}\right) \)

= \(\frac{1}{2} \times \frac{-6}{5}=\frac{-3}{5} \)

R.H.S= \(\left[\frac{1}{2}+\left(\frac{-1}{3}\right)\right]+\frac{2}{5}\)

= \(\left(\frac{1}{2} \times \frac{-3}{1}\right) \div \frac{2}{5}=\frac{-3}{2}+\frac{2}{5} \)

= \(\frac{-3}{2} \times \frac{5}{2}=\frac{-15}{4}\)

So, No; L.H.S ≠R. H.S.

Question 10. Complete the following table:

Solution:

Question 11. Do you think the properties commutativity and associativity made the calculations
Sol.

Yes! The properties of commutativity and associativity made the calculations easier

The Rolie Of Zero (0)

Question 1. \(-5+0=\ldots \ldots. \ldots \ldots .=-5\)
Solution:

-5+0=0+(-5)=-5

Question 2.  \(\frac{-2}{7}+\ldots \ldots .=0+\left(\frac{-2}{7}\right)=\frac{-2}{7}\)

⇒ \(\frac{-2}{7}+0=0+\left(\frac{-2}{7}\right)=\frac{-2}{7}\)

Question 3. Do a few more such additions. What do you observe?
Solution:

1. 3 + 0 = 0 + 3 = 3

2. -9 +0 = 0 + (-9) = -9

3.\(\frac{-3}{4}+0=0+\left(\frac{-3}{4}\right)=\frac{-3}{4}\)

We observe that when we add 0 to a whole number, the sum is again that whole number. This happens for integers and rational numbers also.

The Role Of 1

Question 1. \(\frac{-2}{7} \times 1=\ldots \ldots . . \times \ldots . .=\frac{-2}{7}\)
Solution:

– \(\frac{2}{7} \times 1=1 \times \frac{-2}{7}=\frac{-2}{7}\)

Question 2. \(\frac{3}{8} \times \ldots \ldots=1 \times \frac{3}{8}=\frac{3}{8}\)
Solution:

⇒ \(\frac{3}{8} \times 1=1 \times \frac{3}{8}\)

= \(\frac{3}{8}\)

Question 3.  What do you find?
Solution:

We find that when we multiply any rational number by 1, we get back the same rational number as the product

Question 4. Check for a few more rational numbers

1. 7 × 1
Solution:

7 × 1 = 7 = 1 × 7

2.\(\frac{-3}{5} \times 1\)
Solution:

⇒ \(\frac{-3}{5} \times 1\)

=\(1 \times \frac{-3}{5}=\frac{-3}{5}\)

3. \(\frac{7}{9} \times 1\)
Solution:

⇒ \(\frac{7}{9} \times 1\)

= \(1 \times \frac{7}{9}=\frac{7}{9}\)

Question 5. Is 1 the multiplicative identity for integers? For whole numbers?
Solution:

Yes, 1 is the multiplicative identity for integers as well as for whole numbers.

Question 6. If a property holds for rational numbers, will it also hold for integers?

1. For whole numbers?
2. Which will?
3. Which will not?

Solution:

1. Except for the following property, the properties of the rational numbers will also hold good for integers :

If a and b are rational numbers, then a÷b is also a rational number if b ≠ 0 but a÷ b is not necessarily an integer where a and b are integers

Eample: \(\frac{4}{5}\)is a rational number. Here 4 and 5 are integers, while \(\frac{4}{5}\) is not an integer

2. Except for the following properties, the properties of the rational numbers will also hold good for whole numbers:

If a and b are rational numbers, then, a- b is also a rational number. But if a and b are whole numbers, then A b is not necessarily a whole number

Let 4 and 5 be whole numbers, but 4 – 5 = -1 is not a whole number.

2. If a and b are rational numbers, then, a + b is a rational number if b  but if a and b are whole numbers, then a + b is not necessarily a whole number.

Example: 4 and 5 are whole numbers but \(\frac{4}{5}\) is not a whole number

Disberstivity Of Multiplication Over Addition For Rational

For all rational numbers a, b, and c,

a(b + c) = ab + ac.

a(b-c) = ab – ac

Question 1. \(\left\{\frac{7}{5} \times\left(\frac{-3}{12}\right)\right\}+\left\{\frac{7}{5} \times \frac{5}{12}\right\}\)
Solution:

⇒ \(\left\{\frac{7}{\mathbf{5}} \times\left(\frac{-3}{12}\right)\right\}+\left\{\frac{7}{5} \times \frac{\mathbf{5}}{\mathbf{1 2}}\right\}\)

= \(\frac{7}{5} \times\left\{\left(\frac{-3}{12}\right)+\frac{5}{12}\right\} \)

= \(\frac{7}{5} \times\left\{\frac{(-3)+5}{12}\right\}=\frac{7}{5} \times \frac{2}{12} \)

= \(\frac{7}{5} \times \frac{1}{6}=\frac{7}{30} \)

Question 2. \(\left\{\frac{9}{16} \times \frac{4}{12}\right\}+\left\{\frac{9}{16} \times \frac{-3}{9}\right\}\)
Solution:

⇒ \(\left\{\frac{9}{16} \times \frac{4}{12}\right\}+\left\{\frac{9}{16} \times \frac{-3}{9}\right\} \)

= \(\frac{9}{16} \times\left\{\frac{4}{12}+\left(\frac{-3}{9}\right)\right\}\)

= \(\frac{9}{16} \times\left\{\frac{1}{3}+\left(\frac{-1}{3}\right)\right\} \)

= \(\frac{9}{16} \times\left\{\frac{1-1}{3}\right\}=\frac{9}{16} \times 0=0 .\)

NCERT Solutions For Class 8 Maths Chapter 1 Rational Numbers Exercise 1.1

Question 1.  Name the property under multiplication used in each of the following

1. \( \frac{-4}{5} \times 1=1 \times \frac{-4}{5}=-\frac{4}{5} \)

2. \(\frac{13}{17} \times \frac{-2}{7}=\frac{-2}{7} \times \frac{-13}{17}\)

3. \(\frac{-19}{29} \times \frac{29}{-19}=1\)

Solution:

1.1 is the multiplicative identity

2. Commutativity of multiplication

3. Multiplicative inverse.

Question 2. Tell what property allows you tocompute :\(\frac{1}{3} \times\left(6 \times \frac{4}{3}\right) \text { as }\left(\frac{1}{3} \times 6\right) \times \frac{4}{3} \text {. }\)
Solution:

Associativity of multiplication

Question 3. The product of two rational numbers is always a
Solution:

The product of two rational numbers is always a rational number.

NCERT Solutions For Class 8 Maths Chapter 1 Rational Numbers Multiple Choice Questions

Question 1. Which of the following statements is false?

1. Natural numbers are closed under addition
2. Whole numbers are closed under the addition
3. Integers are closed under the addition
4. Rational numbers are not closed under addition.

Solution: 4. Rational numbers are not closed under addition.

Question 2. Which of the following statements is false?

1. Natural numbers are closed under subtraction
2. Whole numbers are not closed under subtraction
3. Integers are closed under subtraction
4. Rational numbers are closed under subtraction.

Solution: 1. Natural numbers are closed under subtraction

Question 3. Which of the following statements is true?

1. Natural numbers are closed under multiplication
2. Whole numbers are not closed under multiplication
3. Integers are not closed under multiplication
4. Rational numbers are not closed under multiplication.

Solution: 1. Natural numbers are closed under multiplication

Question 4. Which of the following statements is true?

1. Natural numbers are closed under the division
2. Whole numbers are not closed under the division
3. Integers are closed under the division
4. Rational numbers are closed under division.

Solution: 2.  Whole numbers are not closed under the division

Question 5. Which of the following statements is false?

1. Natural numbers are commutative for addition
2. Whole numbers are commutative for addition
3. Integers are not commutative for addition
4. Rational numbers are commutative for addition.

Solution: 3. Integers are not commutative for addition

Question 6. Which of the following statements is true?

1. Natural numbers are commutative for subtraction
2. Whole numbers are commutative for subtraction
3. Integers are commutative for subtraction
4. Rational numbers are not commutative for subtraction.

Solution: 4. Rational numbers are not commutative for subtraction.

Question 7. Which of the following statements is false?

1. Natural numbers are commutative for multiplication
2. Whole numbers are commutative for multiplication
3. Integers are not commutative for multiplication
4. Rational numbers are commutative for multiplication.

Solution: 3. Integers are not commutative for multiplication

Question 8. Which of the following statements is true?

1. Natural numbers are commutative for division
2. Whole numbers are not commutative for division
3. Integers are commutative for division
4. Rational numbers are commutative for division.

Solution: 2. Whole numbers are not commutative for division

Question 9. Which of the following statements is true?

1. Natural numbers are associative for addition
2. Whole numbers are not associative for addition
3. Integers are not associative for addition
4. Rational numbers are not associative for addition.

Solution: 1. Natural numbers are associative for addition

Question 10. Which of the following statements is true?

1. Natural numbers are associative for subtraction
2. Whole numbers are not associative for subtraction
3. Integers are associative for subtraction
4. Rational numbers are associative for subtraction

Solution: 2. Whole numbers are not associative for subtraction

Question 11. Which of the following statements is true?

1. Natural numbers are not associative for multiplication
2. Whole numbers are not associative for multiplication
3. Integers are associative for multiplication
4. Rational numbers are not associative for multiplication.

Solution: 3. Integers are associative for multiplication

Question 12. Which of the following statements is true?

1. Natural numbers are associative for division
2. Whole numbers are associative for division
3. Integers are associative for division
4. Rational numbers are not associative for division

Solution: 4. Rational numbers are not associative for division

Question 13. 0 is not

1. A natural number
2. A whole number
3. An integer
4. A rational number.

Solution: 1. A natural number

Question 14. ½ is

1. A natural number
2. A whole number
3. A n integer
4. A rational number.

Solution: 4.  A rational number.

Question 15. a + 6 = 6 + a is called

1. Commutative law of addition
2. Associative law of addition
3. Distributive law of addition
4. None of these.

Solution: 1. Commutative law of addition

Question 16. a × b = b × a is called

1. Commutative law for addition
2. Commutative law for multiplication
3. Associative law for addition
4. Associative law for multiplication.

Solution: 2. Commutative law for multiplication

Question 17. (a + b) + c = a + (b + c) is called

1. Commutative law for multiplication
2. Commutative law for
3. Additionassociative law for addition
4. Associative law for multiplication.

Solution: 3. Additionassociative law for addition

Question 18. a × (b × c) = (a × b) × c is called

1. Associative law for addition
2. Associative law for multiplication
3. Commutative law for addition
4. Commutative law for multiplication.

Solution: 2. Associative law for multiplication

Question 19. a (6 + c) = ab + ac is called

1. Commutative law
2. Associative law
3. Distributive law
4. None of these

Solution: 3. Distributive law

Question 20. The additive identity for rational numbers is

1. 1
2. -1
3. 0
4. None of these.

Solution: 3. 0

Question 21. The multiplicative identity for rational numbers is

1. -1
2. 1
3. 0
4. None of these

Solution: 2. 1

Question 22. How many rational numbers are there between any two given rational numbers?

1. Only one
2. Only two
3. Countless
4. Nothing can be said.

Solution: 3. Countless

Question 23. If a and 6 are two continuous rational numbers, then

1. \(\frac{a+b}{2}<a\)
2. \(\frac{a+b}{2}<b\)
3. \(\frac{a+b}{2}=a\)
4. \(\frac{a+b}{2}>b\)

Solution: 2.  \(\frac{a+b}{2}<b\)

NCERT Solutions For Class 8 Maths Chapter 1 Rational Numbers True Or False

1. a – 0 and 0 – a represent the same rational number where a ≠ 0. – False

2. 0 is a rational number – True

3. Between any two given rational numbers, there are countless rational numbers – True

NCERT Solutions For Class 8 Maths Chapter 1 Rational Numbers Fill In The Blanks

1. A number of the form p/q, where p and q are integers and q ≠0 is called a → Rational number

2. Write the number 0 reduced by 1 → (-1)

3. Are the two rational numbers p/q and q≠0 equivalent    →Yes

4. In the rational number p/q, why q≠0  → Since division by zero is not defined

5. Which rational number on the number line is equidistant from 0 and -1  → (\(\frac{-1}{2}\)).

NCERT Solutions For Class 8 Maths Chapter 6 Cubes And Cube Roots Introduction

Hardy – Ramanujan Number

Look at the following relations :

1729 = 1728 + 1 = 12³ + 1³

1729 = 1000 + 729 = 10³ + 9³

1729 is the smallest Hardy – Ramanujan Number (a number expressed as a sum of two cubes in two different ways is known as a Hardy- Ramanujan Number). There are infinitely many such numbers.

Few are 4104 (2, 16; 9, 15), 13832 (18, 20; 2, 24

Q. Check it with the numbers given in the brackets. 4104 (2, 16; 9, 15) and 13832 (18, 20; 2, 24)

2³ + 16³  = 8 + 4096 = 4104

9³ + 15³ = 729 + 3375 = 4104

18³ + 20³ = 5832 + 8 000 = 13832

2³ + 24³ = 8 + 13824 = 13832

Read and Learn More NCERT Solutions For Class 8 Maths

Cubes

Numbers obtained when a number is multiplied by itself three times are known as cube numbers or perfect cubes.

For example: 1, 8, 27,

The cube of a natural number m is denoted by m3 and is expressed as

m³ = m x m x m.

Thus, 1³ = 1 x 1 x 1 = 1

2³ = 2 × 2× 2 = 8

3³ = 3 ×  3 × 3 = 27, and so on

Question 1. How many cubes of side? 1 cm will \ make a cube of side 2 cm?
Solution:

2 × 2 × 2 = 8 cubes of side 1 cm will make a cube of side 2 cm.

Question 2. How many cubes of side 1 cm will make a cube of side 3 cm?
Solution:

3 × 3 × 3 = 27 cubes of side 1 cm will make a cube of side 3 cm

Question 3. Following are the cubes of the numbers from 11 to 20
Solution:

Question 4. Complete the following cubes of numbers  From 1 to 10

Solution:

Question 5. Are there only ten perfect cubes from 1 to 1000?
Solution:
Yes :

There are only ten perfect cubes from 1 to 1000.

These are: 1, 8, 27, 64, 126, 216, 343, 512, 729 and 1000

Question 6. How many perfect cubes are there from 1 to 100?
Solution:

There are only four perfect cubes from 1 to 100.

 These are: 1, 8, 27, and 64.

Question 7. Observe the cubes of even numbers. Are they all even? What can you say about the cubes of odd numbers?
Solution:

1. The cubes of even numbers are all even.
2. The cubes of odd numbers are all odd.

Question 8. Consider a few numbers having 1 as the one’s digit (or unit’s). Find the cube of each of them. What can you say about the one digit of the cube of a number having 1 as the one’s digit? Similarly, explore the one’s digit of cubes of numbers ending in 2, 3, 4,
Solution:

We observe that the one digit of the cube of a number having 1 as the one’s digit is 1. Similarly, the one’s digit of cubes of numbers ending in 0, 2, 3, 4, 5, 6, 7, 8, 9 are 0, 8, 7, 4, 5, 6, 3, 2, 9 respectively

Question 9. Find the one digit ofthe cube ofeach of the following numbers.

1. 3331
2. 8888
3. 149
4. 1005
5. 1024
6. 77
7. 5022
8. 53

Solution:

1. 3331

Unit digit of the number = 1

Unit digit of the cube of the number = 1

∴  1 × 1 × 1 = \(\underline{1}\)

2.  8888

Unit digit of the number = 8

Unit digit of the cube of the number = 2.

∴  8 × 8 × 8 = 51\(\underline{2}\)

3.  149

Unit digit of the number = 9

Unit digit of the cube of the number = 9

∴  9 × 9 × 9 = 72\(\underline{9}\)

4.  1005

Unit digit of the number = 5

Unit digit of the cube of the number = 5.

∴  5 × 5 × 5 = 12\(\underline{5}\)

5. 1024

Unit digit of the number = 4

Unit digit of the cube of the number = 4

∴  4 × 4 × 4 × 4 = 6\(\underline{4}\)

4. 77

Unit digit of the number = 7

Unit digit of the cube of the number = 3.

∴  7 × 7 × 7 = 34\(\underline{3}\)

8. 5022

Unit digit of the number = 2

Unit digit of the cube of the number = 8

∴  2 × 2 × 2 = \(\underline{8}\)

8. 53

Unit digit of the number = 3

Unit digit of the cube of the number = 7

∴  3 × 3 × 3 = 2\(\underline{7}\)

Some Interesting Patterns

If in the prime factorization of any number, each prime factor appears three times, then the number is a perfect cube.

For example, 216 =\(\underline{2 \times 2 \times 2} \times \underline{3 \times 3 \times 3} \times \underline{3 \times 3} \times \underline{5 \times 5}\)

= 2³ ×  3³ = (2× 3)³ = 6³ which is a perfect cube.

Question 1. How many consecutive odd numbers will be needed to obtain the sum as 10³?
Solution:  Ten.

Question 2. Express the following numbers as the sum of odd numbers using the above pattern.

1. 6³
2. 8³
3. 7³

Solution: 

1. 6³

Here, n = 6

∴ (n – 1) = 5

We start with n × (n – 1) + 1

= 6× 5 +1

= 31

6³ = 31 + 33 + 35 + 37+ 39 + 41

2. 8³

Here, n = 8

∴ (n – 1) = 7

We start with n × (n – 1) + 1

= 8 × 7  + 1

=  56+1

=  57

8³ = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71

3. 7³

Here, n = 7

(n -1) = 6

We start with n × (n – 1) + 1,

7 × 6 + 1,

= 42+1

= 43

7³ = 43 + 45 + 47 + 49 + 51 + 53 + 55

Question 3. Consider the following pattern.

1. 2³ – 1³ = 1 + 2 × 1 × 3
2. 3³ – 2³ = 1 + 3 × 2 × 3
3. 4³ – 3³= 1 + 4 × 3 × 3

Using the above pattern, find the value of the following:

1. 7³ – 6³
2. 12³ – 11³
3. 20³ – 19³
4. 51³  – 50³ 

Sol.

1. 7³ – 6³ 

= 1 + 7 × 6 ×3

=1+126

= 127

2. 12³ – 11³

= 1 + 12 × 11 × 3

=1 + 396

= 397

3.  20³- 19³

T = 1 + 20 × 19 × 3

= 1 + 1140

= 1141

4.  51³ – 50³

1 + 51 × 50 × 3

= 1 + 7650

= 7651

Question 4. Which of the. following are perfect cubes

1. 400
2. 3375
3. 8000
4. 15625
5. 90000
6. 6859
7. 2025
8. 10648

1.  400

By prime factorisation,

Grouping the factors in triplets

400 = \(\underline{2 \times 2 \times 2} \times\)  2× 5 × 5.

= 2³ × 2× 5× 5

Bylaws of exponents

In the above factorization, 2 and 5 × 5 remain after grouping 2’s in triplets.

∴ 400 is not a perfect cube

2.33375

Therefore, 400 is not a perfect cube

By prime factorisation,

3375 = \(\underline{3 \times 3 \times 3} \times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 3³× 5³

Bylaws of exponents

= (3 × 5)³

= 15³, which is a perfect cube.

∴  3375 is a perfect cube.

3. 80000

By prime factorisation,

8000 =  \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 2³× 2³ × 5³

Bylaws of exponents

= (2 × 2 ×5)³

= 20³, which is a perfect cube.

∴   8000 is a perfect cube

4.15625

By prime factorisation,

15625 = \(\underline{5 \times 5 \times 5} \times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 5³× 5³

Bylaws of exponents

= (5 × 5)³

Bylaws of exponents

25³, which is a perfect cube.

∴  15625 is a perfect cube

5. 9000

By prime factorisation,

9000 = \(\underline{2 \times 2 \times 2} \times\) 3× 3 × \(\underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

In the above factorization, 3×3 remains after grouping 2’s and 5’s in triplets.

∴  9000 is not a perfect cube.

6. 6859

By prime factorisation

6859 = \(\underline{19 \times 19 \times 19}\)

Grouping the factors in triplets

= 19³

Bylaws of exponents which is a perfect cube

∴  6859 is a perfect cube

7. 2025

By prime factorisation,

2025  = \(\underline{3 \times 3 \times 3} \times\) 3 × 5 × 5

Grouping the factors in triplets

= 3³ × 3 × 5 × 5

Bylaws of exponents

In the above factorization, 3 and 5 ×  5 remain after grouping 3’s in triplets.

∴ 2025 is not a perfect cube

8. 10648

By prime factorisation

10648 = \(\underline{2 \times 2 \times 2} \times \underline{11 \times 11 \times 11} \times \)

Grouping the factors in triplets

= 2³ × 11³

Bylaws of exponents

= (2 × 11)³

Bylaws of exponents

= 22³, which is a perfect cube.

∴  10648 is a perfect cube

Smallest Multiple That Is A Perfect Cube

Sometimes we have to find the smallest natural number by which a number is multiplied or divided to make it a perfect cube

Question 1. Check which of the following are perfect cubes

1. 2700
2. 6000
3. 64000
4. 500
5. 125000
6. 50000
7. 21600
8. 10000
9. 27000000
10. 1000

What pattern do you observe in these perfect cubes
Solution:

1. 2700

By prime factorisation,

2700 = 2× 2× \(\underline{3 \times 3 \times 3}\)×  5 ×  5

Grouping the factors in triplets

= 2  × 2 × 3³× 5× 5

Bylaws of exponents

In the above factorization, 2 × 2 and

5× 5 remain after grouping 3’s in triplets.

∴ 2700 is not a perfect cube

2. 16000

By prime factorisation,

16000 = \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2}\)  × 2 × \(\underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 2³ × 2³ × 2 ×  5³

Bylaws of exponents

In the above factorization, 2 remains after grouping 2’s and 5’s in triplets.

∴ 16000 is not a perfect cube

3. 64000

By prime factorisation,

64000 = \(\underline{2 \times 2 \times 2 } \times \underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 2³ ×2³ × 2³× 5³

By laws of exponents

= (2 × 2 × 2 ×5)³

= 40³, which is a perfect cube.

∴  64000 is a perfect cube

4.  900

By prime factorisation,

900 = 2 ×2 × 3 × 3 × 5 × 5

In the above factorization, 2 × 2, 3 × 3, and 5 × 5 remain when we try to group the factors in triplets.

∴ 900 is not a perfect cube.

5. 125000

By prime factorisation,

125000 = \(\underline{2 \times 2 \times 2 } \times \underline{5 \times 5 \times 5} \times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 2³ × 5³ × 5³

Bylaws of exponents

= (2 × 5 × 5)³

= 50³, which is a perfect cube.

∴  125000 is a perfect cube.

6. 36000

By prime factorisation,

3600 = \(\underline{2 \times 2 \times 2 } \times \underline{2 \times 2 } \times \underline{3 \times 3} \times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 2³ × 2 × 2 × 3 × 3 × 5³

Bylaws of exponents

In the above factorization, 2 × 2 and 3 ×3 remain after grouping 2’s and 5’s in  triplets

∴  36000 is not a perfect cube.

7. 21600

By prime factorisation

21600 = \(\underline{2 \times 2 \times 2 }\) × 2 × 2 ×  \(\underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 2³ × 2 × 2×3³×5 × 5

Bylaws of exponents

In the above factorization, 2×2 and 5×5 remain after grouping 2’s and 3’s in triplets

∴  21600 is not a perfect cube.

8. 10000

By prime factorisation,

1000 =\(\underline{2 \times 2 \times 2}\) ×  2 × \(\underline{5 \times 5 \times 5}\) × 5

Grouping the factors in triplets

= 2³ × 2 × 5³ × 5

Bylaws of exponents

In the above factorization, 2 and 5 remain after grouping 2’s and 5’s in triplets.

∴ 10000 is not a perfect cube

9. 27000000

By prime factorisation,

27000000

=\(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times \underline{3 \times 3 \times 3} \times \underline{5 \times 5 \times 5}\times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplet

= 2³×2³ × 3³× 5³ × 5³

Bylaws of exponents

= 300³, which is a perfect cube.

∴  27000000 is a perfect cube.

10. 1000

By prime factorization,

1000 = \( \underline{2 \times 2 \times 2} \times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 2³ × 5³

Bylaws of exponents

= (2 × 5)³ I By laws of exponents

= 10³, which is a perfect cube.

Therefore, 1000 is a perfect cube.

We observe the following pattern in these perfect cubes:

1. If in the end of a number, the number of zeros is not 3 or a multiple of 3, then that number cannot be a perfect cube
2. If in the end of a number, the number £ of zeros is 3 or a multiple of 3, then JJJ that number may be a perfect cube.
3. Thus, the number of zero at the end of a perfect cube (if it is so) must essentially be 3 or a multiple of3, failing which the number cannot be a perfect cube.

NCERT Solutions For Class 8 Maths Chapter 6 Cubes And Cube Roots Exercise 6.1

Question 1. Which of the following numbers are not perfect cubes?

1. 216
2. 129
3. 1000
4. 100
5. 46656

Solution:

1. 216

By prime factorisation

216 =  \(\underline{2 \times 2 \times 2}\times \underline{3 \times 3 \times 3}\)

Grouping the factors in triplets

= 2³ × 3³

Bylaws of exponents

= (2 × 3)³

= 6³ which is a perfect cube.

∴  216 is a perfect cube

2. 128

By prime factorisation

128 = \(\underline{2 \times 2 \times 2}\times \underline{2 \times 2 \times 2}\) × 2

Grouping the factors in triplets

= 2³ × 2³ × 2

Bylaws of exponents

In the above factorization,2remainsafter grouping the 2 ‘s in triplets

∴ 128 is not a perfect cube

3. 1000

By prime factorisation

= \(\underline{2 \times 2 \times 2}\times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 2³× 5³

Bylaws of exponents

= (2 × 5)³ = 10³

Bylaws of exponents

Which is a perfect cube.

∴ 1000 is a perfect cube

4. 100

By prime factorisation

100 =  \(\underline{2 \times 2 \times }\times \underline{5 \times 5}\)

In (he above factorization, 2 × 2,5  × 5 remains when we try to group the factors in triplets

Therefore, 100 is not a perfect, cube.

5.46656

By prime factorisation,

46656 = \(\underline{2 \times 2 \times 2 \times 2 \times 2 \times 2}\) × \(\underline{3 \times 3 \times 3} \times \underline{3 \times 3 \times 3}\).

Grouping the factors in triplets

= 2³ × 2³×3³ × 3³

Bylaws of exponents

= (2 ×2 × 3 × 3)³

Bylaws of exponents

= 36³, which is a perfect cube.

∴  46656 is a perfect cube

Question 2. Find the smallest number by which each of the following numbers must be multi¬ plied to obtain, a perfect cube

1. 243
2. 256
3. 72
4. 675
5. 100

Solution:

1. 243

By prime factorisation

243 =\(\underline{3 \times 3 \times 3}\) × 3 × 3

Grouping the factors in triplets

The prime factor 3 does not appear in a group of three.

Therefore, 243 is not a perfect cube.   To make it a cube, we need one more 3. In that case

243 = \(\underline{3 \times 3 \times 3} \times \underline{3 \times 3 \times 3}\)

= \(3^3 \times 3^3\)

= (3×3)³

729, which is a perfect cube.

Hence, the smallest number by which 243 should be multiplied to obtain a perfect cube is 3.

The resulting perfect cube is 3.

2. 256

By prime factorisation

256 =\(\underline{2 \times 2 \times 2}\underline{2 \times 2 \times 2}\) 2 x 2

Grouping the factors in triplets

In the above factorisation 2× 2  after grouping 2‘s in Iriplols. Therefore, 128 is not a perfect cube. To make it a perfect, cube, we need one 2 more. In that, case

256× 2 = \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times 2 \times 2\)

= 2³×2³× 2³

Bylaws of exponents

= (2×2×2)³

Bylaws of exponents

8³ = 512, which is a perfect cube.

Hence, the smallest number by which 256 must be multiplied to obtain a perfect cube is 2

The resulting perfect cube is 512 (= 8³)

3. 72

By prime factorisation,

72 = 2 × 2 × 2 × 3 × 3

Grouping the factors in triplets

The prime factor 3 does not appear in a group of three.

Therefore, 72 is not a perfect cube. To make it a perfect cube, we need one more 3. In that case,

72 × 3 = \(\underline{2 \times 2 \times 2} \times \underline{3 \times 3 \times 3}\)

= 2³ ×3³

Bylaws of exponent

= (2 × 3)³

Bylaws of exponents

6³ = 216 which is a perfect cube.

Hence, the smallest number by which 72 must be multiplied to obtain a perfect cube is 3.

The resulting perfect cube is 216 (= 6³)

4. 675

By prime factorisation,

675 = 5 × 3 × 3 × 5 × 5

Grouping the factors in triplets

The prime factor 5 does not appear in a group of three. Therefore, 675 is not a perfect cube.

To make it a perfect cube, we need one more 5. In that case

675 ×  5 = \(\underline{3 \times 3 \times 3} \times \underline{5 \times 5 \times 5} \)

= 3³× 5³

Bylaws of exponents

= (3×5)³

Bylaws of exponents

= 15³

∴ 3375 which is a perfect cube.

Hence, the smallest number by which 675 must be multiplied to obtain a perfect cube is 5.

The resulting perfect cube is 3375 (= 15³)

5. 100

By prime factorisation,

100 = 2 × 2 × 5 × 5

Grouping the factors in triplets

The prime factors 2 and 5 do not appear in a group of three. Therefore, 100 is not a perfect cube.

To make it a perfect cube, we need one 2 and one 5 more. In that case,

100 × 2 × 5 = \(\underline{2 \times 2 \times 2} \times \underline{5 \times 5 \times 5} \)

= 2³ × 5³

Bylaws of exponents 2

= (2 × 5)³

Bylaws of exponents

= 10³

∴ 1000 which is a perfect cube.

Hence, the smallest number by which 100 must be multiplied to obtain a perfect cube is 2× 5 = 10

The resulting perfect is 1000 (= 10³ )

Question 3. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.

1. 81
2. 128
3. 135
4. 192
5. 704

Solution:

1. 81

By prime factorisation

81 =\(\underline{3 \times 3 \times 3}\times\) 3

Grouping the factors in triplets

In the above factorization, 3 remains after grouping the 3’s in triplets. Therefore,

81 is not a perfect cube. If we divide the number by 3, then in the prime factorization of the quotient, this 3 will not remain. In that case,

81 ÷ 3  = \(\underline{3 \times 3 \times 3}\)

= 3³

Bylaws of exponents which is a perfect cube

Hence, the smallest whole number by which 81 must be divided to obtain a perfect cube is 3.

The resulting ideal cube is 27 (= 3³)

2. 128

By prime factorisation,

128 = \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2}\)

Grouping the factors in triplets

In the above factorization, 2 remains after grouping’ the 2’s in triplets.

Therefore, 128 is not a perfect cube. If we divide the number by 2, then in the prime factorization ofthe quotient, this 2 will not remain. In that case,

128 ÷ 2 = \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2}\)

= 2³ × 2³

= (2 × 2)³

Bylaws of exponents

= 4³, which is a perfect cube.

Hence, the smallest whole number by  which 128 must be divided to obtain a perfect cube is 2

The resulting ideal cube is 64 (= 4³)

3. 135

By prime factorisation,

135 = \(\underline{3 \times 3 \times 3} \times 5\)

Grouping the factors in triplets.

The prime factor 5 does not appear in a group of three. So 135 is not a perfect cube

In the factorisation 5 appears only once. If we (u) 704 divide 135 by 5, then the prime factorization of the quotient, will not contain 5.

In that case

135 ÷ 5 = \(\underline{3 \times 3 \times 3}\)

Grouping the factors in triplets

= 27

Bylaws of exponents, which is a perfect cube.

Hence, the smallest whole number by  which 135 must be divided to obtain a perfect cube is 5

The resulting perfect cube is 27 (= 33)

4. 192

By prime factorisation,

192  = \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \) × 3

Grouping the factors in triplets

The prime factor 3 does not appear in a group of three. So 192 is not a perfect cube. In the factorization of 192, 3 appears only once. So if we divide the number by 3, then the prime factorization of the quotient will not contain 3.

.In that case,

192 ÷ 3 = \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \) × 3

= 2³ × 2³

= (2 × 2)³

Bylaws of exponents

= 4³ = 64

Which is a perfect cube.

Hence, the smallest whole number by which 192 must be divided to obtain a perfect cube is 3.

The resulting perfect cube is 64 (= 4³)

5. 704

By prime factorisation,

704  =  \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \) × 11

Grouping the factors in triplets

The prime factor 11 does not appear in the group of three.

So, 704 is not a perfect cube.

In the factorization 11 appears only one time.

So if we divide 704 by 11, then the prime factorization of the quotient will not contain 11. In that case

704 ÷ 11 = 2 × 2 × 2 × 2 × 2 × 2

Grouping the factors in triplets

2³ × 2³

Bylaws of exponents

= (2 × 2)³

= 4³

= 64

Which is a perfect cube.

Hence, the smallest whole number by which 704 must be divided to obtain a perfect cube is 11.

The resulting perfect cube is 64 (= 4³).

Question 4. Parikshit makes a cuboid of plasticine of sides 5 cm, 2  How many such cuboids will he need to form a cube? cm, 5 cm.
Solution:

Volume of a cuboid

= 5 × 2 × 5 cm³

Since there is only one 2 and only two 5’s in the prime factorization, so, we need 2 × 2 × 5, i.e.t 20 to make a perfect cube.

Therefore, we need 20 such cuboids to make a cube.

Cube Roots

The cube root is the inverse operation of finding a cube.

2³ = 8 ⇒  2 is the cube root of 8.

The symbol denotes the cube root. Thus,\(\sqrt[3]{8}=2\)

Cube Root Through Prime Factorisation Method

We express the given number into product of its prime factors and make triplets (groups of three) of similar factors. Then, we take one factor from each triplet and multiply. The product so obtained gives the cube root of the given number.

Question 1. State true or false: for any integer m, m²< m³. Why?
Solution:

False, if m is a negative integer.

True, if m is a positive integer (natural number).

Verification. Let us take a negative integer m =-1. Then,

m²= (- 1)2 = (- 1)² ×  (-1) = 1

m³ = (_ 1)3 = (- 1)³ × (-1) × (-1)

= – 1

Clearly, m² > m³

Hence, the given statement is false if m is a negative integer.

We get the same inference for m = – 2, -3,-4, etc.

Again, let us take a positive integer(natural number) m = 2. Then,

m² = (2)² = 2 × 2 = 4

m³ = (2)³ = 2 × 2 × 2 = 8

Clearly, m² < m³

Hence, the given statement is true if m is a positive integer (natural number).

We get the same inference for m = 3, 4,5… etc

Cube Root of a Cube Number Steps

1. Obtain the given number. Start making groups of three digits starting from the rightmost digit of the number.
2. The first group will give one’s (unit’s) digit of the required cube root.
3.  Then, take another group. Find the two closest cube numbers between which this group lies. Take the one’s place of the smaller number as the ten’s place of the required cube root

NCERT Solutions For Class 8 Maths Chapter 6 Cubes And Cube Roots Exercise 6.2

Question 1. Find the cubit root of each of the foUouiu# numbers by prime factorization method :

1. 64
2. 512
3. 10648
4. 27000
5. 15625
6. 13824
7. 110592
8. 46656
9. 175616
10. 91125

Solution:

1. 64

The prime factorization of 64 is

= \(\underline{2 \times 2 \times 2} \times \underline {2\times 2 \times 2}\)

Grouping the factors in triplets

= 2³× 2³

= (2 × 2)³

= 4³

Bylaws of exponents

∴ \(\sqrt[3]{64}\) = 2 × 2

2.512

The prime factorisation of 512 is

⇒ \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2}\)

Grouping the factors in triplets

= 2³× 2³ × 2³

= (2 × 2 ×2)³

= 8³

Bylaws of exponents

∴ \(\sqrt[3]{512}\) = 2 × 2 ×2

= 8³

3. 10648

Prime factorization of 10648 is

⇒ [late]\underline{2 \times 2 \times 2} \times \underline{11 \times 11 \times 11} [/latex]

Grouping the factors in triplets

= 2³ x 11³

= (2 × 11)³

∴ \(\sqrt[3]{10648}\) = 2 × 11

= 22³

4. 27000

Prime factorisation of 27000 is

⇒ \(\underline{2 \times 2 \times 2} \times \underline{3 \times 3  \times 3 } \times \underline{5 \times 5 \times 5 } \times \underline{5 \times 5}\)

Grouping the factors in triplets

= 2³ × 3³ × 5³

= (2 × 3 × 5)³

Bylaws of exponents

= 30³

∴ \(\sqrt[3]{27000}\) = 2×3 × 5

= 30

5. 15625

The prime factorization of 15625 is

⇒ \(\underline{5 \times 5 \times 5} \times \underline{5 \times 5  \times 5 }\)

Grouping the factors in triplets

= 5³ × 5³

= (5 × 5)³

Bylaws of  exponents

= 25³

∴ \(\sqrt[3]{15625}\) = 5 × 5

= 25

6. 13824

Prime factorization of 13824 is

⇒  \(\underline{2 \times 2 \times 2}\times \underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times \underline{3 \times 3 \times 3}\)

Grouping the factors in triplets

= 2³ × 2³× 2³ × 3³

=(2× 2 × 2 × 3)³

Bylaws of exponents

= 24³

∴ \(\sqrt[3]{13824}\)  = 2× 2 × 2 × 3

= 24

7. 1110592

Prime factorisation of 110592 is

⇒  \(\underline{2 \times 2 \times 2}\)× \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{3 \times 3 \times 3}\)

Grouping the factors into triplets

= 2³ × 2³ × 2³ × 2³ × 3

= (2 × 2 × 2 × 2 ×3)³

Bylaws of exponents

= 48³

∴ \(\sqrt[3]{110592}\) =2 × 2 × 2 × 2 ×3

8. 46656

Prime factorization of 46656 is

⇒ \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times \underline{3 \times 3 \times 3} \times \underline{3 \times 3 \times 3}\)

Grouping the factors in triplets

= 2³× 2³ × 3³ × 3³

= (2 × 2 × 3 × 3)³

Bylaws of exponents

= 36³

∴ \(\sqrt[3]{46656}\) = 2 × 2 × 3 × 3

=36

9. 175616

Prime factorisation of 175616 is

⇒  \(\underline{2 \times 2 \times 2 \times} \times \underline{2 \times 2 \times 2 \times} \times \underline{2 \times 2 \times 2 \times} \times \underline{7 \times 7 \times 7}\)

Grouping the factors in triplets

= 2³ × 2³ × 2³ × 7³

= (2 × 2 × 2 × 7)³

Bylaws of exponents

= 56³

∴ \(\sqrt[3]{175616}\) = 2 × 2 × 2 × 7

= 56

10. 91125

Prime factorization of 91125 is

⇒ \(\underline{3 \times 3 \times 3 \times} \times \underline{3 \times 3 \times 3 \times} \times \underline{5 \times 5 \times 5}\)

Grouping the factors in triplets

= 3³ ×3³ × 5³

= (3 × 3× 5)³

Bylaws of exponents

= 45³

∴ \(\sqrt[3]{91125}\) =3 × 3× 5

= 45

NCERT Solutions For Class 8 Maths Chapter 6 Cubes And Cube Roots Multiple Choice Questions

Question 1. Which of the following numbers is a perfect cube?

1. 125
2. 36
3. 75
4. 100

Solution: 1. 125

125 = 5 × 5 × 5 = 5³

Question 2. Which of the following numbers is a cube number?

1. 1000
2. 400
3. 100
4. 600.

Solution: 1.1000

1000 = 10 × 10 × 10 = 10³

Question 3. Which of the following numbers is not a perfect cube?

1. 1331
2. 512
3. 343
4. 100

Solution: 4. 100

100 = 2 ×2 × 5 ×5

= 2² × 5².

Question 4. Which of the following numbers is not a cube number?

1. 10000
2. 3375
3. 64
4. 729.

Solution: 1. 10000

10000 = 2 × 2 ×2 × 2 ×5 ×5× 5 ×5

= 2⁴ × 5⁴

= 2³× 2 × 5³ × 5

Question 5. The cube of an odd natural number is

1. Even
2. Odd
3. Maybe even, maybe odd
4. Prime number.

Solution: 2. Odd

3 × 3 × 3 = 27 (odd)

Question 6. The cube of an even natural number is

1. Even
2. Odd
3. Maybe even, maybe odd
4. Prime number.
5. Solution: 1. Even

6 × 6 × 6 = 216 (even)

Question Question 7. The one digit of the cube of the number 111 is

1. 1
2. 2
3. 3
4. 9.

Solution: 1. 1

1 × 1 × 1 =  1

Question 8. The one digit of the cube of the number 242 is

1. 2
2. 4
3. 6
4. 8.

Solution: 4. 8

2 × 2 × 2= 8.

Question 9. The one digit of the cube of the number 123 is

1. 3
2. 6
3. 9
4. 7

Solution: 4. 7

3 × 3 × 3 =2\(\underline{7}\)

Question 10. The one’s digit of the cube of the number 144 is

1. 1
2. 2
3. 3
4. 4.

Solution: 4. 4

4 × 4 ×4 = 6\(\underline{4}\)

Question 11. The one’s digit of the cube of the number 50 is

1. 1
2. 0
3. 5
4. 4

Solution: 2. 0

0 × 0 × 0 = 0

Question 12. The one digit of the cube of the number 326 is

1. 2
2. 3
3. 6
4. 4

Solution: 3. 3

6 × 6 × 6 = 21\(\underline{6}\)

Question 13. The one digit of the cube of the number 325 is

1. 2
2. 5
3. 3
4. 6.

Solution: 2. 5

5× 5× 5= 12\(\underline{5}\)

Question 14. The one digit of the cube of the number 347 is

1. 3
2. 4
3. 7
4. 1

Solution: 1. 3

7×7×7 = 34\(\underline{3}\)

Question 15. The one digit of the cube of the number 68 is

1. 1
2. 2
3. 6
4. 8

Solution: 2. 2

8× 8 ×8 = 51\(\underline{2}\)

Question 16. The one digit of the cube of the number 249 is

1. 2
2. 4
3. 9
4. 1

Solution: 3. 9

9 × 9 × 9 = 72\(\underline{9}\)

Question 17. What is the one’s digit in the cube root of the cube number 1331?

1. 1
2. 2
3. 3
4. 4

Solution: 1. 1

1 × 1 × 1 = 1

Question 18. What is the one digit in the cube root of the cube number 1000000?

1. 0
2. 1
3. 2
4. 9

Solution: 1. 0

0 ×0 × 0 = 0

Question 19. What is the one’s digit in the cube root of the cube number 1728?

1. 1
2. 2
3. 3
4. 9

Solution: 2. 2

2 × 2 × 2 = \(\underline{8}\)

Question 20. What is the one’s digit in the cube root, of the cube number 2197′?

1. 1
2. 2
3. 3
4. 7

Solution: 3. 3

3× 3 × 3 = 2\(\underline{7}\)

Question 21. What is t lie one’s digit, in the cube root, of the cube number 2744?

1. 1
2. 2
3. 3
4. 4

Solution: 4. 4

4 × 4 × 4 = 6\(\underline{4}\)

Question 22. What is the one’s digit in the cube root of the cube number 3375?

1. 2
2. 3
3. 5
4. 4

Solution: 3. 5

5 × 5 × 5 = 12\(\underline{5}\)

Question 23. What is the one’s digit in the cube root of the cube number 4096? 

1. 2
2. 6
3. 4
4. 9.

Solution: 2. 6

6 × 6 × 6 = 21\(\underline{6}\).

Question 24. What is the one’s digit in the cube root of the cube number 4913?

1. 7
2. 9
3. 3
4. 6

Solution: 1. 7

7 × 7 ×7 = 34\(\underline{3}\)

Question 25. What is the one’s digit in the cube root of the cube number 5832?

1. 2
2. 4
3. 6
4. 8

Solution: 4. 8

8 × 8 × 8 = 51\(\underline{2}\)

Question 26. What is the one’s digit in the cube root of the cube number 6859?

1. 7
2. 8
3. 9
4. 6

Solution: 3. 9

8 × 8 × 8 = 51\(\underline{2}\)

Question 27. What, is the one’s digit in the cube root of the cube number 8000 ?

1. 0
2. 2
3. 4
4. 8

Solution: 1. 0

0 × 0 × 0 = 0.

Question 28. The number of zeroes at the end of the cube of the number 20 is

1. 1
2. 2
3. 3
4. 6

Solution: 3. 3

Number of zeroes at the end of the number 20 = 1

Number of zeroes at the end of its cube = 3 × 1 = 3

Question 29. The number of zeroes at the end of the cube root of the cube number 1000 is

1. 1
2. 2
3. 3
4. 4

Solution: 1. 1

The number of zeroes at the end of the cube = 3

Number of zeroes at the end of the cube root = \(\frac{3}{3}\)

= 1

Question 30. The number of zeroes at the end of the cube of the number 100 is

1. 1
2. 2
3. 4
4. 6

Solution: 4. 6

Number of zeroes at the end of the number 100 = 2

Number of zeroes at the end of its cube = 3 × 2

= 6

Question 31. The number of zeroes at the end of the cube root Of the cube number 8000000 in

1. 1
2. 2
3. 3
4. 6

Solution: 2. 2

The number of zeroes at the end of the cube = 6

Number of zeroes at the end of the cube root =\(\frac{6}{3}\)

= 2

Question 32. Find the smallest number by which the number 108 must be multiplied to obtain a perfect, cube. ,

1. 2
2. 3
3. 4
4. 5

Solution: 1. 2

108 = 2 × 2 × 3 × 3 × 3

= 2 × 2 × 3³

Question 33. Find the smallest number by which the number 250 must be divided to obtain a perfect cube.

1. 2
2. 3
3. 4
4. 5

Solution: 1. 3

250 = 5 ×5 × 5 × 2 = 5³ × 2

Question 34. Find the smallest number by which the number 72 must be multiplied to obtain a perfect cube.

1. 2
2. 3
3. 4
4. 6

Solution: 2. 3

72 = 2 × 2 × 2 × 3 × 3

= 2³ × 3 × 3

Question 35. Find the smallest number by which the number 375 must be divided to obtain a perfect cube.

1. 2
2. 3
3. 5
4. 4.

Solution: 2. 3

375 = 3 × 5 × 5 × 5

= 3× 5³.

Question 36. Find the smallest number by which the number 100 must be multiplied to obtain a perfect cube.

1. 5
2. 2
3. 4
4. 10

Solution: 4. 10

100 = 2 × 2 ×5 × 5.

Question 37. Find the smallest number by which the number 10000 must be divided to obtain a perfect cube.

1. 2
2. 5
3. 10
4. 100.

Solution: 3. 10

10000 = 2 × 2 × 2 × 2 ×5×5 × 5 × 5

= 2³ × 2 × 5³× 5

Question 38. Find the smallest number by which the number 200 must be multiplied to obtain a perfect cube.

1. 2
2. 10
3. 5
4. 100

Solution: 3. 5

200 =2 × 2 × 2 × 5 × 5

= 2³ ×5 × 5

Question 39. Find the smallest number by which the number 625 must be divided to obtain a perfect cube.

1. 3
2. 5
3. 25
4. 125

Solution: 2. 5

625 = 5 × 5 × 5 ×5

= 5³ × 5.

Question 40. Kind the smallest number by which (bo number 128 must the multiplied to obtain a perfect cube.

1. 2
2. 4
3. 3
4. 8

Solution: 2. 4

128 =2 × 2 × 2 × 2 × 2 × 2 × 2

= 2³ ×2³ × 2

Question 41. Find the smallest number bv which the number 250 must be divided to obtain a perfect cube.

1. 2
2. 4
3. 8
4. 16

Solution: 2. 4

256 = 2 × 2× 2× 2 × 2× 2 ×  2×2

= 2³ × 2³ × 2 × 2.

Question 42. Find the smallest number by which the number 30 must be multiplied to obtain a perfect cube.

1. 6
2. 2
3. 3
4. 4

Solution: 1. 6

36 = 2 × 2 × 3 × 3

Question 43. Find the smallest number by which the number 1290 must be divided to obtain a perfect cube.

1. 6
2. 2
3. 4
4. 3

Solution: 1. 6

1296 = 2 × 2× 2 × 2 × 3 × 3

=  2³× 2 ×3³× 3

Question 44. Find the smallest number by which the number 392 must be multiplied to obtain a perfect cube.

1. 3
2. 5
3.  7
4. 6

Solution: 3.  7

392 = 2 × 2 × 2 × 7× 7

Question 45. Find the smallest number by which the number 2401 must be divided to obtain a perfect cube.

1. 1
2. 0
3. 5
4. 9.

Solution: 1. 1

2401 = 7 × 7 × 7 × 7

= 7³ × 7

Question 46. Find the smallest number by which the number 121 must be multiplied to obtain a perfect cube.

1. 1
2. 9
3. 11
4. 5

Solution: 3.11

121 = 11× 11

Question 47. Find the smallest number by which the number 88 must be divided to obtain a perfect cube.

1. 11
2. 5
3. 7
4. 9

Solution: 1.11

88 = 2 × 2 ×2 × 11

= 2³ ×11

Question 48. The volume of a cube is 64 cm3. The edge of the cube is

1. 4 cm
2. 8 cm
3. 16 cm
4. 6 cm

Solution: 1. 4 cm

Edge = \(\sqrt[3]{64}=\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2}\)

= \(\sqrt[3]{2^3 \times 2^3}\)

= 2 × 2

= 4.

Question 49. Apala makes a cuboid of plasticine with sides 5 cm, 4 cm, 2 cm. How many such cuboids will be needed to form a cube?

1. 20
2. 25
3. 10
4. 16

Solution: 2. 25

Volume = 5 × 4 × 2

= 5 × 2× 2 × 2

= 5 ×2³

Question 50. Which of the following is false?

1. A Cube of any odd number is odd
2. A perfect cube does not end with two zeroes
3. The cube of a single-digit number may be a single-digit number
4. There is no perfect cube that ends with 8.

Solution: 4.   There is no perfect cube that ends with 8.

1728 = 12 × 12  × 12

= 12³

NCERT Solutions For Class 8 Maths Chapter 6 Cubes And Cube RootsTrue Or False

There are four perfect cubes between 1 and 100 – False

2. Ifa2 ends in 9, then a3 ends in 9  – False

3. Ifa2 ends in 5, then a3 ends in 25  – False

4. 999 is not a perfect cube  – True

5. Cube roots of 1 are + 1 and -1   – False

6. Cube of any odd number is even – False

7. A perfect cube does not end with two zeros – True

8. If the square of a number ends with 5, then its cube ends with 25. – False

⇒  15² = 225, 15³= 3375

9. There is no perfect cube that ends – False

⇒   12³ = 1728

10. The cube of a two-digit number may be a three-digit number – False 

⇒   10³ = 1000, 99³= 970299

11. The cube of a two-digit number may have seven or more digits -False 

⇒  10³ = 1000, 99³ = 970299

12. The cube of a single-digit number may be a single-digit number – True

⇒  1³ = 1; 2³ = 8

NCERT Solutions For Class 8 Maths Chapter 6 Cubes And Cube Roots Fill In The Blanks

1. \(\sqrt[3]{x}\)represents the Cube  root of the number x

2. 1 cm³  10³ mm³

3. The cube of a number ending in 9 will end in the digit → 9

4. The cube of 100 will have zeros 6 at the end

5. There are  8 perfect cubes between 1 and 1000.

6. Find the value of \(\sqrt[3]{125} \times \sqrt[3]{-64}\) → 20

7. Find the least number by which 72 should be multiplied to make it a perfect cube → 3

8. Find the least number by which 72 should be divided to make it a perfect cube → 9

9. Find(1.1)³→ 1.331

10. Find the number whose cube is 64000 →  40

NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities Recalling Ratios And Percentages Important Points

1. We usually compare two quantities by division, i.e., by using fractions. Comparison by division is called ratio.

Note that two quantities can be compared only when they have the same units. A ratio has no unit.

However, if the two quantities are not in the same units, we convert them into the same units before comparing.

2. Two quantities can also be compared using percentages. By percentage, we mean a fraction where the denominator is 100. The numerator of the fraction is called the rate percent.

For example:  \(\frac{1}{5}\) means 5%. The symbol % is often used to express ‘percent’ (p. c.).

3. To convert the ratio into a percentage, we convert it into a fraction whose denominator is 100. [or we multiply by 100 and employ the % sign.]

4. To convert percentages in \(\frac{5}{100}}\) to fractions, we divide the numerator by 100 and express it in the lowest form.

For example : 5% =\(\frac{1}{20}\)

Read and Learn More NCERT Solutions For Class 8 Maths

.
5. In the unitary method, we find the value of one unit from the given value of some units and then we find the value of the required number of units.

Question 1. In primary school, the parents were asked about the number of hours they spend per day helping their children to do homework. 90 parents helped for hour to \(\frac{1}{2}\) hours to 1 \(\frac{1}{2}\) . The distribution of parents according to the time for which, they said they helped is given in the adjoining figure; 20% helped for more than 1\(\frac{1}{2}\) ho per day; 30% helped for \(\frac{1}{2}\)  an hour to 1 \(\frac{1}{2}\) hours,  50% did not help at all.

Using this, answer the following:

1. How many parents were surveyed?
2. How many said that they did not help
3. How many said that they helped for more than 1\(\frac{1}{2}\) hour?

Solution:

1. According to the question

⇒ \(\frac{30}{100}\) of number of parents surveyed = 90

Number of parents surveyed = 90

Number of parents surveyed

= \(\frac{90 \times 100}{30}\)

= 300

Hence, 300 parents were surveyed.

2. A number of those parents said that they did not help

= 50% of 300

= \(300 \times \frac{50}{100}\)

= 150

Hence, 150 parents said that they did 30% not help.

3. A number of those parents said that they helped for more than 1\(\frac{1}{2}\) hour

= 20% of 300 = \(300 \times \frac{20}{100}\)

= 60

Hence, 60 parents said that they helped for more than 1/[latxe]\frac{1}{2}[/latex] hours

Note: ‘Of’ means multiplication

NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities Exercise 7.1

Question 1. Find the ratio of the following:

1. The speed of a cycle 15 km per hour to the speed of a scooter 30 km per hour.
2. 5 m to 10 km
3. 50 paise to 5

Solution:

1. The speed ofa a cycle 15 km per hour to the speed of a scooter 30 km per hour

= 15 km per hour: 30 km per hour

= 15: 30

= \(\frac{15}{30}=\frac{1}{2}\) Or 1:2

2. 5 m to 10 km

10 km = 10 × 1000 m = 10000 m

∴Ratio of 5 m to 10 km

= 5 m: 10000 m

= 5: 10000

⇒ \(\frac{5}{10000}=\frac{1}{2000}\)

= 1:20000

3. 50 paise to ₹ 5

₹ 5 = 5× 100 = 500 paise

A Ratio of 50 paise to ₹ 5

= 50 paise : 500 paise

= 50: 500

= \(\frac{50}{500}=\frac{1}{10}\) Or 1:10

Question 2. Convert the following ratios to percentages

1. 3:4
2. 2: 3.

Solution:

1. 3: 4

= \(\frac{3}{4}=\frac{3}{4} \times \frac{25}{25}\)

Making denominator 100

= \(\frac{75}{100}=75 \%\)

Aliter:

⇒  \(3: 4=\frac{3}{4} \times 100 \%\)

= 75 %

2. 2:3

= \(\frac{2}{3}=\frac{2}{3} \times \frac{100}{100}\)

= \(66 \frac{2}{3} \%\)

Aliter:

2: 3 = \(\frac{2}{3} \times 100 \%\)

= \(\frac{200}{3} \%=66 \frac{2}{3} \%\)

Question 3. 72% of 25 students are interested in mathematics. How many are not interested in mathematics?
Solution:

Total number of students = 25

Percentage of students interested in Mathematics s = 72%

Percentage of students who are not interested in Mathematics

= (100 – 72)% = 28%

Number of those students who are not interested in Mathematics

= 28% of 25

= \(\frac{28}{100} \times 25\)

= 7

Hence, 7 students are not interested in or 1: 10 Mathematics.

Question 4. A football team won 10 matches out ofthe total number of matches they played. Iftheir win percentage was 40, then how many matches did they play in all?
Solution:

If 40 matches were won, then the total number of matches played = 100

∴ If 1 match was won, then the total number of matches played

=\(\frac{100}{40}\)

If 10 matches were won, then the total number of matches played

= \(\frac{100}{40} \times 10\)

= 25

Hence, they played 25 matches in all.

Aliter:

According to the question,

40% of (total number of matches) = 10

⇒  \(\frac{40}{100} \times \text { (total number of matches) }\)=10

Total number of matches = \(\frac{10 \times 100}{40}\) = 25

Hence, they played 25 matches in all.

Question 5. If Chameli had ₹ 600 left after spending 75% of her money, how much did she have in the beginning ?’
Solution:

Percentage of money spent = 75%

Percentage of money left = (100- 75)% = 25%

∴ If Chameli had ₹ 25 left, then the money she had in the beginning

= 100

∴  If Chameli had ₹1 left, then the money she had in the beginning

= \(₹ \frac{100}{25}\)

∴ If Chameli has v600 left, then the money she had in the beginning

= \(₹ \frac{100}{25} \times 600\)

=  ₹ 24000

Hence, the money she had in the beginning was ₹ 2400.

Aliter :

According to this question,

25% of total money = ₹ 600

⇒ \(\frac{25}{100}\) of total money = ₹ 600

⇒  \(₹ \frac{600 \times 100}{25}\)

= ₹ 2400.

Hence, the money she had in the beginning was? 2400

Question 6. If 60% of people in a city like cricket, 30% like football and, the remaining like other games, then what percent of the people like other games? If the total number of people is 50 lakh, find the exact number who like each type of game.
Solution:

Percentage of people who like cricket = 60%

Percentage of people who like football = 30%

∴Percentage of people who like other games

= [100 -(60 +30)]%

= (100-90)% = 10%.

Total number of people = 50 lakh

= 50,00,000

∴ Number of people who like cricket

= 60% of 50,00,000

= \(50,00,000 \times \frac{60}{100}\)

= 30,00,000 = 30 lakh

Number of people who like football

= 30% of 50,00,000

= \(50,00,000 \times \frac{30}{100}\)

Number of people who like the other games

= 10% of 50,00,000

= \(50,00,000 \times \frac{10}{100}\)

= 5,00,000 = 5 lakh

Finding Discounts

Discount = Marked Price – Sale Price

Discount per cent = \(\frac{\text { Discount }}{\text { Marked Price }} \times 100 \%\)

Question 1. 4 shop gives 20% discount. What Would the sale price ofeach of these be?

1. A dress marked at.₹ 120
2. A pair of shoes marked at ₹ 750
3. A bag marked at₹ 250.

Solution:

1. A dress marked at ₹120

Marked price ofthe dress = ₹120

Discount rate= 20%
,
∴ Discount = 20% of ₹ 120

20 MarkedDiscountprice × 100

=\(₹ \frac{20}{100} \times 120\)

= ₹ 24

∴ Sale price ofthe dress x 100 % = 4%.

= Marked price – Discount

= ₹ 120 – ₹ 24 = ₹ 96

2. A pair of shoes marked at ₹ 750

Marked price ofthe pair ofshoes= ₹ 750

Discount rate = 20%

Discounts 20% of ₹ 750

= \(₹ \frac{20}{100} \times 750\)

= ₹150

∴ Sale price ofthe pair of shoes

= Marked price – Discount

= ₹ 750 -₹ 150

= ₹600

3. A bag marked at ₹ 250

The marked price of the bag =₹ 250

Discount rate = 20%

Discount = 20% of  ₹ 250

= \(₹ \frac{20}{100} \times 250\)

= ₹50

∴ The sale price of the bag

= Marked price – Discount

=₹250 – ₹ 50 =₹200.

Question 2. If the table marked at f 15,000 is * available for f14, 400. Find the discount given £ and the discount percent.
Solution:

Marked price ofthe table =₹ 15,000

Sale price ofthe table =  ₹ 14,400

∴  Discount given

= Marked price – Sale price

= ₹ 15,000 – ₹14,400

= ₹ 600

∴  Discount per cent

= \(\left(\frac{\text { Discount }}{\text { Marked price }} \times 100\right) \% \)

= \(\left(\frac{600}{15000} \times 100\right) \%=4 \%\)

Question 3. An almirah is sold at ₹5,225 after allowing a discount of 5%. Find its marked price.
Solution:

Let the marked price bet x.

Discount rate = 5%

Discount = 5% of ₹ x

=\(₹ \frac{5}{100} \times x=₹ \frac{x}{20}\)

∴ Sale price

= Marked price -Discount

= \(₹ x-₹ \frac{x}{20}=₹ \frac{19 x}{20}\)

According to the question,

∴ \(\frac{19 x}{20}=5225\)

x = \(\frac{5225 \times 20}{19}\)

x = ₹ 5500

Hence, the marked price of the almirah is ₹5500

Question 4. Try estimating 20% of the same bill amount.

1. Round of the bill to the nearest tens.
2. Find the amount of discount.
3. Reduce the bill amount by the discount amount

Solution:

Your bill in a shop is ₹  577.80.

1. Round off the bill to the nearest tens of ₹ 577.80, i.e., to ₹ 580.

2. Find 20% of this,

⇒  \(₹ \frac{20}{100} \times 580\)

= ₹  116

3. Therefore estimated 20% of the same bill amount is ₹116 or ₹120 (rounded off to the nearest tens).

Question 5. Try finding 15% of  ₹ 375.
Solution:

Find 10% of ₹ 375, i.e.

⇒ ₹\(₹ \frac{10}{100} \times 375\)

= ₹ 37.5

= ₹  40

(rounded off to the nearest tens)

2. Take half of this, i.e \(\frac{1}{2} \times ₹ 40\)

= ₹ 20

(3) Add (2) and (1) to get ₹ 60.

Therefore, an estimated 15% of 7 375 is

= 60

Sales Tax/Value Added Tax/Goods And Services Tax

Sales tax was charged at a specified rate on the sale price of an item by the state government and was added to the bill amount. It is different for different items and also for different states.

• Amount of Sales Tax = Tax% of bill amount
• These days, the prices include the tax known as Value Added Tax (VAT).
• From July 1, 2017, the Government of India introduced GST which stands for Goods and Services Tax which is levied on the supply of goods or services or both

Question 1. Find the buying price of each of the following when 5%STis added to the purchase

1. A towel at ₹ 50
2. Two bars of soap at ₹ 35 each
3. 5 kg of flour at ₹15 per kg.

Solution:

1. Cost of the towel

Rate of ST = 5%

∴ ST =5% of 7 50

=₹ \(\frac{5}{100} \times 50\)

= ₹ 2.50

∴ Buying price of the towel

= Cost of the towel + ST

= ₹ 50 + ₹ 2.50 =₹ 52.50.

2. The cost of two bars of soap

= ₹35 × 2 = ₹ 70

Rate of  ST= 5%

ST = 5% of ₹ 70

= \(₹ \frac{5}{100} \times 70\)

= ₹ 3.50

∴ Buying price of two bars of soap

= Cost of two bars of soap + ST

= ₹ 70 +₹ 3.50 = ₹ 73.50

3. Cost of 5 kg of flour

= ₹ 15 × 5 = ₹ 75

Rate of ST = 5%

ST = 5% of ₹ 75

= \(₹ \frac{5}{100} \times 75\)

= ₹ 3.75

∴ The buying price of 5 kg of flour + ST

= Cost of 5 kg of flour + ST

= ₹ 75 +₹ 3.75

=₹ 78.75.

Question 2. If 8% is included in the prices, find the original price of

1. A TV bought for ₹13,500
2. A shampoo bottle was bought for ₹ 180.
   

Solution:

1. When the price of a TV including

VAT is ₹108,

original price = ₹ 100

When the price of the TV including VAT is

₹ 1, original price = ₹ \( \frac{100}{108}\)

The price of a TV including VAT is ₹ 13500,

Original price = \(₹ \frac{100}{108} .\)

= ₹ 12500

Hence, the original price of the TV Is ₹ 12500.

Aliter:

Price of TV including VAT

= ₹ 13500

Rate of VAT = 8%

Let, the original price of the TV be ₹x

Then,

x+8 %  of x =₹ 13500

⇒ \(x+\frac{8}{100} x =₹ 13500 \)

⇒ 1+ \(\frac{8}{100}\)x =₹ 13500

⇒  \(\frac{108}{100} x =₹ 13500\)

= ₹ \(\frac{13500 \times 100}{108}\)

Hence, the original price of the T.V. is ₹12500.

2. When the price of a shampoo bottle including VAT is ₹ 108,

Original price = ₹ 100

When the price of a shampoo bottle including VAT is ₹1, the original price

= \(₹ \frac{100}{108} .\)

∴ The price of a shampoo bottle including VAT is? 180,

Original price = \(=₹ \frac{100}{108} \times 180 \)

⇒ \(₹ \frac{100 \times 5}{3}=₹ \frac{500}{3}\)

=\(₹ 166 \frac{2}{3}\)

Hence, the original price ofthe shampoo Hence, the original price ofthe shampoo bottle is \(166 \frac{2}{3}\)

Alitor: Price of shampoo bot t lo including

VAT = ₹ 180

Rate of VAT = 8%

Let the original price of shampoo bottle be ₹x

Then,

x+ 8%of x = ₹180

⇒ \(x+\frac{8}{100} x=₹ 180\)

⇒  \(\left(1+\frac{8}{100}\right) x=₹ 180\)

⇒  \(\frac{108}{100} x=₹ 180\)

x =\(₹ \frac{180 \times 100}{108}\)

x = \(₹ \frac{500}{3}=₹ 166 \frac{2}{3}\)

Hence, the original price ofthe shampoo bottle is \(166 \frac{2}{3}\)

NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities Exercise 7.2

Question 1. During a sale, a shop offered a discount of 10% on the marked prices of all the items. W7iat would a customer have to pay for a pair of jeans marked at ₹ 1450 and two shirts marked at f850 each
Solution:

For a pair of jeans

Marked price = ₹ 1450

Discount rate = 10%

Discount = 10% of marked price

= 10% of  ₹ 1450

= \(₹ \frac{10}{100} \times 1450\)

= ₹ 145

Sale price – Marked price – Discount

= ₹1450-₹ 145

= ₹ 1305

For two shirts

Marked price =₹ 850 × 2 = ₹ 1700

Discount rate = 10%

Discount = 10% of marked price

= 10% of  ₹ 1700

=₹ \(\frac{10}{100} \times 1700=₹ 170\)

Sale price

= Marked price – Discount

= ₹1700 – ₹70 =₹1530

Total payment made by the customer

= ₹ 1305 + ₹ 1530 =₹ 2835

Hence, the customer will have to pay. 2835 for a pair of jeans and two shirts.

Question 2. The price ofa W is f13,000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it
Solution: 

Price of TV = ₹ 13,000

Sales tax charged on it

= 12% of ₹ 13,000

= \(₹ \frac{12}{100} \times 13,000\)

= ₹ 1560

Amount to be paid

= Price + Sales Tax

= ₹ 13,000 + ₹ 1,560

= ₹14,560.

Hence, the amount that Vinod will haveto pay for the TV if he buys it is ₹14,560

Question 3. Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is  ₹ 1600, find the marked price
Solution:

Let the marked price of the pair of 6  shoes be ₹ 100.

Kate of discount = 20

Discount = 20% of ₹ 100

= ₹ \(\frac{12}{100} \times 100\)

=₹ 20

Amount paid = Marked price -Discount

= ₹ 100 – ₹20 – = ₹ 80

If the amount paid is ₹80,

Then the marked price – \(₹ \frac{100}{80}\)

If the amount paid is ₹ 1, then the marked price = \(₹ \frac{100}{80} \times 1600\)

= ₹ 2000

Hence, the marked price of the pair of shoes is ₹ 2000.
.
Question 4. I purchased a hair dryer for f 5,400 including 8% VAT. Find the price before it was added,
Solution:

Price of hair-dryer including VAT

=  ₹ 5400

VAT rate = 8%

Let the original price be ₹ 100

∴ VAT = ₹ 8

Price including VAT = Original price + VAT

= ₹100 + ₹ 8

= ₹ 108

If the price including VAT is ₹108, then

Original price =  ₹ 100

If the price including VAT is ₹ 1, then

Orginalprice = \(₹ \frac{100}{108}\)

If the price including VAT is  ₹5,400, then

Original price = ₹ \(\frac{100}{108} \times 5,400\)

= ₹ 5000

Hence, ‘the price before VAT was added is ₹5,000

Question 5. An article was purchased for ₹1239 including a GST of 18%. Find the price of the article before GST was added,
Solution:

Let the original price of the article be 100. GST Rate = 18% ,

Price after GST is included

= ₹(100 + 18) = ₹118

If the selling price is? 118 then original price = ₹100

If the selling price is ₹ 1, then

Original Price = \(₹ \frac{100}{108} \times 5,400\)

If the selling price is ₹ 1239, then

original price = \(₹ \frac{100}{118} \times 1239\)

= 100 × 10.5

= ₹ 1050

Hence, the price ofthe article before GST was added is? 1050

Compound Interest

1. Interest:

Interest is the extra money paid by institutions like banks or post offices on money that is deposited (kept) with them. Interest is also paid by people when they borrow money. The money deposited or borrowed is called the principal. Interest is generally given in percent for one year.

2. Simple interest (SI):

The interest is called simple when the principal does not change.

3. The formula for simple interest:

Simple interest on a principal of 1 P at R% rate of interest per year for T years is given by

Simple Interest = \(\frac{\text { Principal } \times \text { Rate } \times \text { Time }}{100}\)

SI =  \(\frac{\text { P R T }}{100}\)

4. Amount :

Amount (A)= Principal (P) + Simple Interest(SI)

Question 1.   Find interest and amount to be paid on 115,000 at 5% per annum after 2 years
Solution:

Interest on₹ 100 for 1 year is = 5

Interest on ₹ for 1 year is

= \(₹ \frac{5}{100}\)

Interest On ₹ for 1 year is

= \(₹ \frac{5}{100} \times 15,000\)

Interest on ₹ 15000 for 2 years is

= \(₹ \frac{5 \times 15,000 \times 2}{100}\)

= ₹  1,500

∴ Amount = Principal + Interest

=  ₹ 15,000 + ₹ 1,500

= ₹ 16,500

Aliter:

Here, P = ₹ 15,000

R = 5% per annum

T = 2 years

∴ \(\text { S.I. }=\frac{\text { PRT }}{1000}\)

= \(₹ \frac{15000 \times 5 \times 2}{100}\)

₹ 1500

Amount = Principal (P) + SimpleInterest (SI)

= ₹ 15000 + ₹ 1500

= ₹ 1,500 = ₹ 16500

Deducing A Formula For Compound Interest

A = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n\)

where P = Principal

R = Rate of interest per annum compounded annually

n = Number of years

A = Amount

CI = A-P

Question 1. Find Cl on a sum of ₹ 8000 for 2 years at 5% per annum compounded annually.
Solution:

Here,

P =₹ 8000

R = 5%p.a.

n = 2 years

A = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n\)

= \(8000\left(1+\frac{5}{100}\right)^2\)

= \(8000\left(1+\frac{1}{20}\right)^2\)

= \(8000\left(\frac{21}{20}\right)^2 \)

= \(8000 \times \frac{21}{20} \times \frac{21}{20}\)

= 20 × 21 × 21

=₹ 8820

Cl = A – P

= ₹ 8820 – ₹ 8000

= ₹ 820.

∴ Hence, the compound interest is ₹ 820

Applications Of Compound Interest Formula

We use the compound interest formula to find

1. Increase (or decrease) in population.
2. The growth of a bacteria if the rate of growth is known.
3. The value of an item, if its price increases or decreases in the intermediate years.

Note: For increase, R is positive and for decrease, R is negative.

Question 1. A machinery worth f 10,500 depreciated by 5%. Find its value after one year.
Solution:

P = ₹ 10,500

R = -5% per annum

There is depreciation

n = year

A = \(P\left(1-\frac{R}{100}\right)^n \)

= \(10,500\left(1-\frac{5}{100}\right)\)

= \(10,500\left(1-\frac{1}{20}\right) \)

= \(10,500 \times \frac{19}{20}\)

= 9,975

Hence value after 1 year is ₹ 9,975.

Question 2. Find the population ofa city after 2 years, which is at present 12 lacks, if the rate of increase is 4%.
Solution:

P = 12,00,000

R = 4% per annum

n = 2years

A = \(P\left(1+\frac{R}{100}\right)^n\)

= \(12,00,000\left(1+\frac{4}{100}\right)^2 \)

= \(12,00,000\left(1+\frac{1}{25}\right)^2\)

= \(12,00,000\left(\frac{26}{25}\right)^2 \)

= \(12,00,000 \times \frac{26}{25} \times \frac{26}{25} \)

= 12,97,920

Hence, the population of the city after 2 years is 12,97,920

Question 3. Calculate the amount and compound interest on

1. 10,800 for 3 years at \(2 \frac{1}{2} \)% perannum compounded annually.
2. ₹ 18,000 for  \(2 \frac{1}{2} \)% years at  perannum compounded annually.
3. ₹  62,500 for \(1 \frac{1}{2} \)% years at  perannum compounded half-yearly.
4. ₹  8,000 for 1 year at   9% per annum compounded half yearly.You could use the year-by-year calculation to verify).
5. ₹  10,000 for a year at 8% per annum compounded halfyearly

Solution:

1. By using year-by-year calculation

Here, P =₹ 10800

R = ₹ 12% per annum

T = ₹ year

SI on ₹ 10,800 at \(12 \frac{1}{2}\) per annum for

= \(10,800 \times \frac{25}{2} \times \frac{1}{100}\)

= ₹ 1,350

∴ Amount at the end of 1st year

= ₹ 10,800 + ₹ 1,350 (A = P + SI)

= ₹ 12,150

= Principal for 2nd year.

SI on ₹ 12,150 at \(12 \frac{1}{2} \% \)per annum

For 1 year = \(\frac{\text { PRT }}{100}\)

=’\(12,150 \times \frac{25}{2} \times \frac{1}{100}\)

= ₹  1,518.75

∴Amount at the end of the 2nd year

= ₹ 12,150+ ₹ 1,518.75

= ₹ 13,608.75

= Principal for 3rd year

SI on  ₹ 13,668.75 at 12\(\frac{1}{2}\)% per annum

For 1year =\(\frac{P R T}{100}\)

=\(13,668.75 \times \frac{25}{2} \times \frac{1}{100} \)

=\(13,668.75 \times \frac{1}{8}\)

= ₹1,708.59

Amount at the end of the year

= ₹13,668.75 + ₹ 1,708.59

=₹ 15,377.34

This is the required amount. Now, CI = ₹15,377.34 – ₹ 10,800

= ₹ 4,577.34

By using the compound interest formula

P =  ₹ 10,800

R = 12\(\frac{1}{2}\) per annum

= \(\frac{25}{2}\)% per annum

n = 3 years

A = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n \)

= \(10,800\left(1+\frac{25}{2 \times 100}\right)^3 \)

= \(10,800\left(1+\frac{1}{8}\right)^3\)

= \(10,800\left(\frac{9}{8}\right)^3\)

= \(10,800 \times \frac{9}{8} \times \frac{9}{8} \times \frac{9}{8}\)

= ₹  15,377.34

2. By using year-by-year calculation

Here

P = ₹ 18000

R = ₹ 10% per annum

T = ₹ year

SI on ₹ 18,000 at 10% p.a. for 1 year

= \(\frac{\mathrm{PRT}}{100}\)

= \(\frac{18,000 \times 10 \times 1}{100}\)

= ₹ 1,800

∴ Amount at the end of the year

= ₹ 18,000 + ₹ 1,800

= ₹ 19,800

= Principal for 2nd year

SI on ₹ 19,800 at 10% p.a. for 1 year

= \(\frac{19,800 \times 10 \times 1}{100}\)

= ₹ 1,980

Amount at the end of 2nd year

= ₹ 19,800 + ₹1,980

= ₹ 21,780

= Principal for 3rd year

SI on ₹ 21,780 at 10% p.a. for \(\frac{1}{2}\)year

⇒ \(\frac{\mathrm{PRT}}{100}=\frac{21,780 \times 10 \times 1}{2 \times 100}\)

= ₹1089

∴ Amount at the end of \(2 \frac{1}{2}\) year

= ₹ 21,780 +₹1,089

= ₹ 22,869

This is the required amount.

Now, CI = ₹ 22,869 – ₹ 18,000

= ₹ 4,869

By using the compound interest formula

P = ₹ 18,000

R = 10% p.a.

n = 2 years

A = \(P\left(1+\frac{R}{100}\right)^n \)

= \(18,000\left(1+\frac{10}{100}\right)^2 \)

= \(18,000\left(1+\frac{1}{10}\right)^2 \)

= \(18,000\left(\frac{11}{10}\right)^2\)

=18,000  × \(\frac{11}{10} \times \frac{11}{10}\)

=₹ 21,780

SI on? 21,780 at 10% p.a. for \(\frac{1}{2}\) year

=\(\frac{21,780 \times 10 \times 1}{2 \times 100}\)

= ₹ 1089

Amount at the end of \(\frac{1}{2}\) years

= ₹21,780 + ₹ 1,089

= ₹ 22,869

Cl = A- P

= ₹22,869- ₹ 18,000

= ₹ 4,869

3. By using half-year by half-year calculation

Here, P = ₹ 62500

R = 8% per annum

= \(\frac{8}{2}\) per half-year

T = 1 year

∴ SI on ₹ 62,500 at 8% p.a. for 1st half year

= \(\frac{P R T}{100} \)

= \(\frac{62,500 \times 8 \times 1}{2 \times 100}=\frac{62500}{25}\)

= ₹ 2,500

∴  Amount at the end of 1st half-year

= ₹ 62,500 +₹ 2,500

= ₹65,000

= Principal for 2nd half year ₹ 65,000 at 8% p.a. for 2nd half-year

= \(\frac{P R T}{100}\)

= \(\frac{65,000 \times 8 \times 1}{2 \times 100}\)

= ₹2,600

Amount at the end of 2nd halfyear

= ₹ 65,000 + ₹ 2,600

= ₹ 67,600

= Principal for 3rd halfyear

SI on ₹ 67,600 at 8% p.a. for 3rd half-year

= \(\frac{P R T}{100} \)

⇒ \(\frac{67,600 \times 8 \times 1}{2 \times 100}\)

= ₹ 24704

∴ Amount at the end of 3rd half-year

= ₹ 67,600 + ₹ 2,704

= ₹ 70,304

This is the required amount.

Now, Cl = Amount – Principal

= ₹ 70,304 – ₹ 62,500

= ₹7,804

Cl = ₹ 2,500 + ₹ 2,600 +₹ 2,704

= ₹ 7,804

By using the compound interest 2 formula

P =₹ 62,500

= \(\frac{1}{2} \times 8 \% \text { per half year } \)

= 4% per half-year

n = \(1 \frac{1}{2} \text { year } \)

= \(1 \frac{1}{2} \times 2 \text { half years }\)

3 half years

A =\(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n \)

= \(62,500\left(1+\frac{4}{100}\right)^3 \)

= \(62,500\left(1+\frac{1}{25}\right)^3 \)

= \(62,500\left(\frac{26}{25}\right)^3 \)

= \(62,500 \times \frac{26}{25} \times \frac{26}{25} \times \frac{26}{25}\)

= \(4\times 26 \times 676\)

= 4 × 26 × 676

= ₹ 70,304

Cl = A- P = ₹ 70,304- ₹ 62,500

=₹ 7804.

By using the compound interest formula

P =₹  62,500

R = 8%p.a

= \(\frac{1}{2} \times 8 \% \text { per half year }\)

1% per half-year

n =\(\frac{1}{2}\)

= 1 \(\frac{1}{2} \times 2\)

= 3 half years

A = \(P\left(1+\frac{R}{100}\right)^n\)

= \(62,500\left(1+\frac{4}{100}\right)^3 \)

= \(62,500\left(1+\frac{1}{25}\right)^3\)

= \(62,500\left(\frac{26}{25}\right)^3\)

=\(62,500 \times \frac{26}{25} \times \frac{26}{25} \times \frac{26}{25}\)

4. By using the compound interest formula

P = ₹ 8,000

R = 9% p.a.

= \(\frac{9}{2}\) % per half-year

= \(8,000\left(1+\frac{9}{2 \times 100}\right)^2\)

= \(8,000\left(1+\frac{9}{200}\right)^2\)

=\(8,000\left(\frac{209}{200}\right)^2 \)

= \(8,000 \times \frac{209}{200} \times \frac{209}{200} \)

\(=\frac{209 \times 209}{5}\)

= 4 ×  26× 676

= ₹  70,304

Cl = A- P = ₹ 70,304 – ₹ 62,500

= ₹ 7804

By using compound interest formula

P=8000

R=9%p.a

=\(\frac{9}{2} \%\)per halfyear

n=1 year = 1 × 2 half-year

= 2 half years

∴ A =\(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n\)

= \(8,000\left(1+\frac{9}{2 \times 100}\right)^2\)

= \(8,000\left(1+\frac{9}{200}\right)^2 \)

= \(8,000\left(\frac{209}{200}\right)^2\)

= \(8,000 \times \frac{209}{200} \times \frac{209}{200} \)

=\(\frac{209 \times 209}{5}\)

= ₹8,736.20

Cl = A-P

= ₹ 8,736.20 – ₹8,000

=₹ 736.20

5. By using the compound interest formula

P = ₹ 10,000

= 8% per annum

= \(\frac{8}{2}\) per half-year

= 4 % per half-year

n = year = 1 × 2 half years

= 2 half years

= \(P\left(1+\frac{R}{100}\right)^n \)

= \(10,000\left(1+\frac{4}{100}\right)^2 \)

=\(10,000\left(1+\frac{1}{25}\right)^2 \)

= \(10,000 \times \frac{26}{25} \times \frac{26}{25}\)

16 × 26 × 26

= ₹ 10,816

∴ Cl = A-P

= ₹ 10,816 – ₹ 10,000

= ₹ 816

Question 4.  Kamala borrowed ₹26,400from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of the year and 4 months to clear the loan 

(Hint: Find A for 2 years, interest is compounded yearly and then find SI on the 2nd year amount for \(\frac{4}{12}\)
Solution:

P = ₹ 26,400

R = 15% p.a.

n = 2 years

A = \(P\left(1+\frac{R}{100}\right)^n \)

= \(26,400\left(1+\frac{15}{100}\right)^2 \)

= \(26,400\left(1+\frac{3}{20}\right)^2 \)

= \(26,400\left(\frac{23}{20}\right)^2\)

= \(26,400 \times \frac{23}{20} \times \frac{23}{20}\)

= \(66 \times 23 \times 23 \)

= ₹ 34,914

S.I. on X 34,914 at 15% p.a. for 4 months

=\(\text { (i.e., } \left.\frac{4}{12} \text { year, i.e., } \frac{1}{3} \text { year }\right) \)

= \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100} \)

= \(34,914 \times \frac{15}{100} \times \frac{1}{3} \)

= \(\frac{34,914 \times 15 \times 1}{3 \times 100}\)

= 1,745.70

Required amount

= ₹ 34,914 +₹ 1,745.70

= ₹ 36,659.70

Hence, the amount that Kavita will pay is ₹ 36,659.70.

NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities Exercise 7.3

Question 1. The population of the place increased to 54,000 in 2003 at a rate of 5% per annum. Find the population in 2001. What, would be its population in 2005?
Solution:

1. Let the population in 2001 be P.

R = 5% p.a.

n = 2 years

A = \(P\left(1+\frac{R}{100}\right)^n=P\left(1+\frac{5}{100}\right)^2 \)

= \(P\left(1+\frac{1}{20}\right)^2=P\left(\frac{21}{20}\right)^2\)

According to the question

⇒ \(\mathrm{P}\left(\frac{21}{20}\right)^2=54,000\)

P = \(54,000\left(\frac{20}{21}\right)^2\)

= \(54,000 \times \frac{20}{21} \times \frac{20}{21} \)

= 48,980(approx.)

Hence, the population in 2001 was 48,980 (approx.)

Initial population in 2003

(P) = 54,000

R = 5% p.a.

n = 2 years (2005 – 2003 = 2)

A=\(\left(1+\frac{R}{100}\right)^n\)

= \(54,000 \times \frac{20}{21} \times \frac{20}{21}\)

= \(54,000\left(1+\frac{1}{20}\right)^2\)

= \(54,000\left(\frac{21}{20}\right)^2\)

= \(54,000 \times \frac{21}{20} \times \frac{21}{20}\)

=\(135 \times 21 \times 21=59,535\)

Hence, the population in 2005 would be 59,535. .1

Question 2. In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of2 hours, if the count was initially 5,06,000
Solution:

The initial count of bacteria

(P) = 5,06,000

R = 2.5% per hour

n = 2 hours

=\(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n \)

=\(5,06,000\left(1+\frac{2.5}{100}\right)^2 \)

= \(5,06,000\left(1+\frac{1}{40}\right)^2 \)

= \(5,06,000\left(\frac{41}{40}\right)^2 \)

= \(5,06,000 \times \frac{41}{40} \times \frac{41}{40} \)

= \(\frac{1265}{4} \times 41 \times 41\)

= 531616.25

= 5,31,616 (approx.)

Hence, the bacteria count at the end UJ of 2 hours is 5,31,616 (approx.).

Question 3. A scooter was bought at ₹42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Solution:

Initial value of the scooter

P =₹ 42,000

R = -8% per annum

I v There is depreciation

A = \(P\left(1-\frac{R}{100}\right)^n \)

=\(42,000\left(1-\frac{8}{100}\right)^1\)

= \(42,000\left(1-\frac{2}{25}\right)\)

= \(42,000 \times \frac{23}{25}\)

= \(1680 \times 23=₹ 38,640\)

= 1680 × 23 = ₹ 38,640

Hence, its value after 1 year is ₹ 38,640.

NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities Multiple Choice Questions

Question 1. The ratio of 50 paise to  ₹ 1 is

1. 1:2
2. 2:1
3. 1:1
4. 1:5.

Solution: 1. 1:2

1 = 100 paise

50paise : 100 paise or 50 : 100 or 1:2

Question 2. The ratio of 10 m to 1 km is

1. 1:10
2. 10:1
3. 1:100
4. 100 :1.

Solution: 3.1:100

1 km = 1000 m

10 m: 1000 m or 10: 1000 or 1: 100

Question 3. The ratio of 10 km per hour to 30 km per hour is

1. 3:1
2. 1: 2
3. 1: 3
4. 2:1.

Solution: 3. 1: 3

10 : 30 = 1:3

Question 4. The ratio 1: 4 converted to percentage is

1. 50%
2. 25%
3. 75%
4. 4%.

Solution: 2. 25%

1: 4 = \(\frac{1}{4} \times 100 \%=25 \% \text {. }\)

Question 5. The ratio 4:25 converted to percentage is

1. 8%
2. 4%
3. 16%
4. 25%.

Solution: 3. 16%

4: 25 = \(\frac{4}{25} \times 100 \%=16 \%\)

Question 6. The fraction ^ converted to a percentage is

1. 20%
2. 30%
3. 40%
4. 50%.

Solution: 3.40%

⇒ \(\frac{2}{5}\) = \(\frac{2}{5} \times 100 \%=40 \% \text {. }\)

Question 7. The fraction\(\frac{1}{8}\) converted to percentage

1. 8
2. 12%
3. 25%
4. 8%
5. 16%.

Solution: 1. 8

⇒ \(\frac{1}{8}=\frac{1}{8} \times 100 \%=12 \frac{1}{2} \%\)

Question 8. Out of 40 students in a class, 25% passed. How many students passed?

1. 20
2. 10
3. 30
4. 40.

Solution: 2.10

Out of 100, passed = 25

Out of40, passed =\(\frac{25}{100} \times 40\)

⇒ \(25 \% \text { of } 40=\frac{25}{100} \times 40\)

= 10

Question 9. Out of 100 students in a class, 30% like to watch T.V. How many students like to watch T.V.?

1. 70
2. 50
3. 60
4. 30.

Solution: 4.30

Number of students who like to watch

TV =30% of\(100=\frac{30}{100} \times 100\)

= 30

Question 10. There are 50 students in a class of which 40 are boys and the rest are girls. The ratio of the number of boys and the number of girls is

1. 2: 3
2. 1:5
3. 4:1
4. 2:5.

Solution: 3. 4:1

Number of Girls = 50 – 40 – 10

Required ratio = 40 : 10 = 4 : 1

Question 11. 40% of 50 students in a class are good at Science. How many students are not good at Science?

1. 20
2. 30
3. 10
4. 40.

Solution: 2. 30

Out of 100, good at Science = 40

Out of 50, good at Science

Number of students good at science

= \(40 \% \text { of } 50=\frac{40}{100} \times 50=20\)

Or

Number of students good at science

= 40% of 50

= \(40 \% \text { of } 50=\frac{40}{100} \times 50=20\)

Number of students not good at science

= 50-20 = 30

Question 12. Apala has ₹ 200 with her. She spent 80% amount she had. How much money is left with her?

1. 10
2. 20
3. 30
4. 40.

Solution: 4.40

Out of  100, money spent =? 80

Out of? 200, money spent

= \(\frac{80}{100} \times 200\)

Money spent = 80% of₹ 200

= \(200 \times \frac{80}{100}\)

= ₹ 160

Money left = ₹ 200 – ₹  160

= ₹ 40.

Question 13. A toy marked at t 40 is available for 132. What percent discount is given on the marked price?

1. 10%
2. 20%
3. 25%
4. 40%.

Solution: 2.20%

Discount = ₹  40 – ₹ 32 = ?8

Discount on ₹  40 = ? 8

Discount on ₹  100

⇒\(\frac{80}{40}\) x100

= ₹ 20

Question 14. Find the simple interest on? 1000 for 2 years at 8% per annum.

1. ₹ 80
2. ₹ 40
3. ₹ 120
4. ₹ 60.

Solution: 4. ₹ 60

S.I = \(\frac{1000 \times 2 \times 8}{100}\)

= ₹ 160

Question 15. A sofa-set was bought for l 10000. Its value depreciated at the rate of 10% per annum. Find its value after one year.

1. ₹11000
2. ₹19000
3. ₹10000
4. ₹1000.

Solution: 2. ₹19000

Depreciation in 1 year =\(10000 \times \frac{10}{100}\)

= 1000

Value after 1 year

= 10000 – 1000 =₹ 9000.

Question 16. There are 1275 trees in Chaudhary Farm. Out of these 36% trees are of fruits. How many trees of fruits are there in Chaudhary Farm?

1. 459
2. 549
3. 945
4. 954.

Solution: 1.459

Required number = \(\frac{1275 \times 36}{100}\)

= 459

Question 17. The quantity of protein in a particular variety of pulse is 25%. Find the amount of protein in 4 kg of pulse.

1. 1 kg
2. 2 kg
3. 3 kg
4. kg.

Solution: 1. 1 kg

Amount of protein =\(4 \times \frac{25}{100}\)

= 1kg

Question 18. In a mixture, the amount of zinc is 45%. Find the amount of zinc in the 400 g mixture.

1. 60 g
2. 120 g
3. 180 g
4. 200 g.

Solution: 3. 180 g

Amount of zinc =\(400\times \frac{45}{100}\)

= 80kg

Question 19. In a school out of 340 students, 55% of students are of Science. The remaining students are of Commerce. Find the number of students of Commerce.

1. 135
2. 153
3. 315
4. 140.

Solution: 2. 153

Number of Science students

= \(340 \times \frac{55}{100}\) = 187

Number of students of Commerce

= 340 – 187 = 153

Question 20. The salary of Manish is ₹ 10000. His salary gets increased by 10%. Find his increased salary.

1. ₹ 9000
2. ₹ 11000
3. ₹ 8000
4. ₹ 12000.

Solution: 2.₹ 11000

Increase in salary

= ₹  \(10000 \times \frac{10}{100}=₹ 1000\)

Question 21. A shopkeeper purchased 2 refrigerators for ₹ 9800 and  ₹ 8200 respectively. He sold them for ₹ 16920. Find loss%.

1. 2%
2. 4%
3. 5%
4. 6%.

Solution: 4.6%

Total C.P. =₹ 9800 + ₹ 8200 = ₹18000

S.P. = ₹ 16920

Loss = ₹18000 – ₹16920 = ₹1080

=\(\frac{1080}{18000} \times 100 \%=6 \%\)

Question 22. In selling a plot of land for ₹ 61200, a profit of 20% is made. The cost price of the plot is

1. ₹ 51000
2. ₹ 50000
3. ₹  49000
4. ₹  52000.

Solution: 1. ₹ 51000

If S.P. is ₹120(100 + 20),

C.P. =₹ 100

If S.P. is ₹ 61200,

C.P =\(\frac{100}{120} \times 61200\)

=  ₹ 51000

Question 23. The simple interest on X 2000 for 4 years is ? 400. The rate percent of interest is

1. \(\frac{2000 \times 100}{400 \times 4}\)

2. \(\frac{400 \times 4}{2000 \times 10} \)

3. \(\frac{400 \times 100}{2000 \times 4}\)

4. \( \frac{2000 \times 4}{400 \times 100} \)

Solution: 3.\(\frac{400 \times 100}{2000 \times 4}\)

⇒ \(\frac{2000 \times 4 \times \mathrm{R}}{100}=400\)

R= \(\frac{400 \times 100}{2000 \times 4}\)

Question 24. The simple interest of ₹ 500 at the rate of 5% is ₹ 100. This interest is of the time

1. 1 year
2. 4 years
3. 10 years
4. 20 years.

Solution: 2. 4 years

⇒ \(\frac{500 \times 5 \times \mathrm{T}}{100}=100\)

⇒ \(\mathrm{T}=\frac{100 \times 100}{500 \times 5}=4\)

Question 25. The S.I. of ₹ 100 for 1 year at the rate of 3 paise per rupee per month is

1. ₹ 30
2. ₹ 36
3. ₹ 24
4. ₹ 8.

Solution: 2.₹ 36

S.I. = 100 × 1 × (3 ×  12)

= 3600 paise

= ₹ 36

NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities True-False

1. VAT is always calculated on the selling price – True

2. A machinery worth .₹ P is depreciated by R% per annum. Its value after 1 year will be \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)\)– False

3. Comparison of two quantities by division is called ratio-True

NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities Fill In the Blanks

1. The overhead changes are added to → CP

2. If the conversion period is not specified, then it is taken as → Oneyear

3. If a discount of ₹ 4y is available on the market price of ₹ 8y, then the discount percent – 50%

4. If — % of a number is 154, then the number is → 4900

5. Find 10% of [100- 20% of 300] → 4

6. Explain 0.1234 as a percentage → 12.34

7 A jacket work for? 4000 is offered for sale at? 3600. What per cent discount is offered during the sale →  10%

8. Apala got 100 marks out of 200 and Meenu got 120 works out of 300. Whose performance is better → Apalas.