**Class 7 Maths NCERT Solutions Chapter 8 – Comparing Quantities comprises of the 3 Exercises**

**This Chapter contains the Exercises relating to the following topics , which are discussed in Chapter 8 – Comparing Quantities Class 7 NCERT book : –**

- 8.1 INTRODUCTION
- 8.2 EQUIVALENT RATIOS
- 8.3 PERCENTAGE – ANOTHER WAY OF COMPARING QUANTITIES
- 8.3.1 Meaning of Percentage
- 8.3.2 Converting Fractional Numbers to Percentage
- 8.3.3 Converting Decimals to Percentage
- 8.3.4 Converting Percentages to Fractions or Decimals
- 8.3.5 Fun with Estimation
- 8.4 USE OF PERCENTAGES
- 8.4.1 Interpreting Percentages
- 8.4.2 Converting Percentages to “How Many”
- 8.4.3 Ratios to Percents
- 8.4.4 Increase or Decrease as Per Cent
- 8.5 PRICES RELATED TO AN ITEM OR BUYING AND SELLING
- 8.5.1 Profit or Loss as a Percentage
- 8.6 CHARGE GIVEN ON BORROWED MONEY OR SIMPLE INTEREST
- 8.6.1 Interest for Multiple Years

### NCERT Solutions for Class 7 Maths Chapter 8 Exercise 8.1

### NCERT Solutions for Class 7 Maths Chapter 8 Exercise 8.2

### NCERT Solutions for Class 7 Maths Chapter 8 Exercise 8.3

### NCERT Solutions for Class 7 Maths Chapter 8 Exercise 8.1

**1. Find the ratio of :**

**a) ₹5 to 50 paise**

₹1 = 100 paise

₹5 = 100 x 5 = 500 paise

The required ratio is, 500/50 = 10/1 = **10 : 1**

**b) 15 kg to 210 g**

1 kg = 1000 g

15 kg = 1000 g x 15 = 15000 g

The required ratio is, 15000/210 = 500/7 = **500 : 7**

**c) 9 m to 27 cm**

1 m = 100 cm

9 m = 100 cm x 9 = 900cm

The required ratio, 900/27 = 100/3 = **100 : 3**

**d) 30 days to 36 hours**

1 day = 24 hours

30 days = 24 hours x 30 = 720 hours

The required ratio is, 720/36 = 20/1 = **20 : 1**

**2. In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students?**

Given : 6 students need = 3 computers

1 student need = 3 x 1/6 = 1/2 computers

Therefore 24 students need 24 x 1/2 = **12 computers**

**3. Population of Rajasthan = 570 lakhs and population of UP = 1660 lakhs. Area of Rajasthan = 3 lakh km ^{2} and area of UP = 2 lakh km^{2}**

**i) How many people are there per km ^{2} in both these States?**

People present per km^{2} = Population/Area

In Rajasthan = (570 lakhs)/(3 lakhs per km²) = **190 lakhs people per km ^{2}**

In UP = (1660 lakhs)/(2 lakhs per km²) = **830 lakhs people per km ^{2}**

**ii) Which State is less populated?**

From the above data**, Rajasthan is less populated** compared to UP.

### NCERT Solutions for Class 7 Maths Chapter 8 Exercise 8.2

**1. Convert the given fractional numbers to per cents.**

**a) 1/8**

= 1/8 x 100%

= 100/8 %

= 25/2 %

= 12 1⁄2 %

= 12.5%

**b) 5/4**

= 5/4 x 100%

= 125%

**c) 3/40**

= 3/40 x 100%

= 3/2 x 5%

= 15/2%

= 7 1⁄2 %

= 7.5%

**d) 2/7**

= 2/7 x 100%

= 200/7%

= 28 4⁄7 %%

**2. Convert the given decimal fractions to per cents**

a) 0.65

(65/100) x 100%

=**65%**

b) 2.1

(21/10) x 100%

=**210%**

**c) 0.02**

(2/100) x 100%

= 2%

**d) 12.35**

(1235/100) x 100%

= **1235%**

**3. Estimate what part of the figures is coloured and hence find the per cent which is coloured**

**i) **

The circle can be divided into 4 equal quadrants. In the given figure one quadrant is coloured.

i.e coloured part = 1/4

Therefore, per cent of coloured part = (1/4) x 100%

=**25%**

**ii) **

In the given figure the circle is divided into 5 equal parts.

3 out of 5 parts are coloured.

i.e coloured part =3/5

Therefore, per cent of coloured part = (3/5) x 100%

= **60%**

**iii)**

In the given figure the circle is divided into 8 equal parts.

3 out of 8 parts are coloured.

i.e coloured part = 3/8

Therefore, per cent of coloured part = (3/8) x 100%

**= 37.5%**

**4. Find :**

**a) 15% of 250**

= (15/100) x 250

= 15 x 2.5

= **37.5%**

**b) 1% of 1 hour**

1 hour = 60 minutes

= 60 x 60 seconds

1 hour = 3600 seconds

Therefore, 1% of 3600 seconds is

1/100 x 36000

=**36 seconds**

**c) 20% of ₹2500**

20/100 x 2500

=20 x 25

= **₹ 500**

**d) 75% of 1 kg**

1 kg = 1000 g

Therefore 75% of 1000g is

75/100 x 1000

= 750 g

**5. Find the whole quantity if:**

**a) 5% of it is 600**

Let the whole value = a

(5/100) x a = 600

a = (600 x 100)/5

**a = 12000**

**b) 12% of it is ₹1080**

Let the whole value = ₹a

(12/100) x ₹a = ₹1080

a = (1080 x 100)/12

**a = ₹ 9000**

**c) 40% of it is 500 km**

Let the whole value = a km

(40/100) x a km = 500 km

a = (500 x 100)/40

**a = 1250 km**

**d) 70% of it is 14 minutes**

Let the whole value = a minutes

(7/100) x a minutes = 14 minutes

a = (40 x 100)/70

**a = 20 minutes**

**e) 8% of it is 40 litres**

Let the whole value = a litres

(8/100) x a litres = 40 litres

a = (40 x 100)/8

**a = 500 litres**

**6. Convert given percents to decimal fractions and also to fractions in simplest forms:**

**a) 25%**

Fractions : 25/100

Simplest form : 1/4

Decimal form : **0.25**

**b) 150%**

Fractions : 150/100

Simplest form : 3/2

Decimal form : **1.5**

**c) 20%**

Fractions: 20/100

Simplest form : 1/5

Decimal form : **0.2**

**d) 5%**

Fractions: 5/100

Simplest form : 1/20

Decimal form : **0.05**

**7. In a city, 30% are females, 40% are males and remaining are children. What per cent are children?**

Given: Percentage of females = 30%

Percentage of males = 40%

Since remaining are children,

Percentage of children = 100% – (Percentage of females + males)

= 100% – (30 + 40)%

= 100% – 70%

**Percentage of children = 30%**

**8. Out of 15,000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?**

Given: Total number of voters = 15000

Percentage of people who voted = 60%

Percentage of people who did not vote = 100% – 60%

= 40%

Therefore the number of people who did not vote = (40 x 100) x 15000

= 6000

**Thus, 6000 people did not vote.**

**9. Meeta saves ₹ 4000 from her salary. If this is 10% of her salary. What is her salary?**

Let Meeta’s salary = ₹ a

Given: (10/100) x a = ₹ 4000

a = (4000 x 100)/10

a = ₹ 40000

**Thus, Meeta’s salary is ₹ 40000**

**10. A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?**

Given: Total number of matches played = 20

Percentage of matches won = 25%

Number of matches won = (25/100) x 20

= 5

**Thus, the number of matches won by the team is 5**

### NCERT Solutions for Class 7 Maths Chapter 8 Exercise 8.3

**1. Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case**

**a) Gardening shears bought for ₹ 250 and sold for ₹325.**

CP of gardening shears = ₹ 250

SP of gardening shears = ₹ 325

SP > CP , therefore profit

Profit = SP – CP

= 325 – 250

Profit = ₹ 75

Profit % = (Profit/CP) x 100

= (75/250) x 100

**Profit % = 30%**

**b) A refrigerator bought for ₹ 12,000 and sold at ₹ 13,500**

CP of refrigerator = ₹12000

SP of refrigerator = ₹13500

SP > CP, therefore profit

Profit = SP – CP

= 13500 – 12000

Profit = ₹ 1500

Profit % = Profit/CP

= 1500/1200

**Profit% = 12.5 %**

**c) A cupboard bought for ₹ 2,500 and sold at ₹ 3,000.**

CP of cupboard = ₹2500

SP of cupboard = ₹3000

SP > CP, therefore profit

Profit = SP – CP

= 3000 – 2500

Profit = ₹ 500

Profit % = (Profit/CP) x 100

= (500/2500) x 100

**Profit % = 20 %**

**d) A skirt bought for ₹ 250 and sold at ₹ 150.**

SP of skirt = ₹250

CP of skirt = ₹150

SP < CP, therefore loss

Loss = CP – SP

= 250 – 150

Loss = ₹ 100

Loss % = (Loss/CP) x 100

= (100/250) x 100

**Loss % = 40%**

**2. Convert each part of the ratio to percentage:**

**a) 3 : 1**

Total part = 3 + 1 = 4

I^{st} Part = 3/4

II^{nd} Part = 1/4

**Percentage of I ^{st}part = (3/4)**

**x 100 = 75%**

**Percentage of II ^{nd} part= (1/4) x 100**

**= 25%**

**b) 2 : 3 : 5**

Total part = 2 + 3 + 5 =10

1^{st} part = 2/10

2^{nd} part = 3/10

3^{rd} part = 5/10

**Percentage of 1 ^{st} part = (2/10) x 100 = 20%**

**Percentage of 2 ^{nd} part = (3/10) x 100 = 30%**

**Percentage of 3 ^{rd} part = (5/10) x 100 = 50%**

**c) 1 : 4**

Total part = 1 + 4 = 5

1^{st} part = 1/5

2^{nd} part = 4/5

**Percentage of 1 ^{st} part = (1/5) x 100 = 20%**

**Percentage of 2 ^{nd} part = (4/5) x 100 = 80%**

**d) 1 : 2 : 5**

Total part = 1 + 2 + 5 = 8

1^{st} part = 1/8

2^{nd} part = 2/8

3^{rd} part = 5/8

**Percentage of 1 ^{st} part = (1/8) x 100 = 12.5%**

**Percentage of 2 ^{nd} part = (2/8) x 100 = 25%**

**Percentage of 3 ^{rd} part = (5/8) x 100 = 62.5%**

**3. The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.**

Given : Initial population = 25000

Final population = 24500

Decrease in population = 25000 – 24500

= 500

Percentage decrease = (Population decrease/Initial population) x 100

= 500/25000

**Percentage decrease = 2%**

**4. Arun bought a car for ₹ 3,50,000. The next year, the price went upto ₹ 3,70,000. What was the Percentage of price increase?**

Given : Initial price of car = ₹ 350000

Final price of car = ₹ 370000

Increase in price = ₹ 370000 – ₹ 350000

= ₹ 20000

Percentage of price increase = (Increase in price/Initial price) x 100

= 20000/370000

**Percentage of price increase = 5 5⁄7 %**

**5. I buy a T.V. for ₹ 10,000 and sell it at a profit of 20%. How much money do I get for it?**

Given : CP of TV = ₹ 10000

Profit% = 20%

We know that Profit% = (Profit/CP) x 100

Therefore, profit = Profit% x CP/100

= (20 x 10000)/100

Profit = ₹ 2000

We know that Profit = SP – CP

Therefore, SP = Profit + CP

= ₹2000 + ₹10000

SP = ₹12000

**I get ₹12000 on selling the TV at 20% profit.**

**6. Juhi sells a washing machine for ₹ 13,500. She loses 20% in the bargain. What was the price at which she bought it?**

Given : SP of washing machine = ₹13500

Loss% = 20%

Let CP of the washing machine = ₹ X

We know that Loss % = (Loss/CP) x 100

Therefore Loss = Loss% x CP/ 100

= (20 x X)/100

= ₹(X/5)

We know that Loss = CP – SP

₹ (X/5) = X – ₹13500

X – (X/5) = 13500

(4X/5) = 13500

X = (13500 x X)/4

Therefore X = CP = ₹16875

**The price at which she bought the washing machine is ₹16875**

**7. (i) Chalk contains calcium, carbon and oxygen in the ratio 10:3:12. Find the percentage of carbon in chalk.**

Given ratio : 10 : 3 : 12

Total part = 10 + 3 + 12 = 25

Part of Carbon =3/25

Percentage of Carbon = (3/25) x 100

**Percentage of Carbon = 12%**

**(ii) If in a stick of chalk, carbon is 3g, what is the weight of the chalk stick?**

Given : Quantity of carbon in chalk stick = 3 g

Previously we have found that the percentage of Carbon = 12%

Let the weight of chalk = X g

Then, 12 % of X = 3 g

12/100 x X = 3

X = (3 x 100)/12

X = 25 g

**Thus the weight of chalk stick is 25 g**

**8. Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for**

Given: CP of book = ₹ 275

Sold at loss% = 15%

We know that Loss = Loss% x CP/ 100

= (15 x 275)/100

Loss = ₹ 41.25

We know that Loss = CP – SP

Therefore SP = CP – loss

= 275 – 41.25

SP = ₹233.75

**Hence, Amina sells the book for ₹233.75**

**9. Find the amount to be paid at the end of 3 years in each case:**

**a) Principal = ₹ 1,200 at 12% p.a**.

Given: P = ₹ 1200 , T = 3 years, R = 12 %

We know that SI = (P x T x R)/100

= (1200 x 3 x 12)/100

SI = ₹ 432

Amount = P + SI

= 1200 + 432

**Amount = ₹ 1632 **

**b) Principal = ₹ 7,500 at 5% p.a.**

Given: P = ₹ 7500 , T = 3 years , R = 5 % ,

We know that SI = (P x T x R)/100

= (7500 x 3 x 5)/100

SI = ₹ 1125

Amount = P + SI

= 7500 + 1125

**Amount = ₹ 8625**

**10. What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in 2 years? **

Given : P = ₹ 56000 , T = 2 years , SI = ₹280

We know that SI = (P x T x R)/100

Therefore R = (SI x 100)/(P x T)

= (280 x 100)/(56000 x 2)

**Hence, R = 0.25%**

**11. If Meena gives an interest of ₹ 45 for one year at 9% rate p.a.. What is the sum she has borrowed?**

Given : SI = ₹ 45 , T = 1 year , Rate = 9%

We know that,

SI = (SI x 100)/(P x T)

Therefore, P = (SI x 100)/(R x T)

= (45 x 100)/(9 x 1)

**P = ₹ 500**

**Hence, she borrowed ₹ 500**

**With this we come to the end of NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities. We hope these helped you study your subject.**

**Download NCERT Solutions for Class 7 Maths Chapter 8 – Comparing Quantities**

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