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

### 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, = = **10 : 1**

**b) 15 kg to 210 g**

1 kg = 1000 g

15 kg = 1000 g x 15 = 15000 g

The required ratio is, = = **500 : 7**

**c) 9 m to 27 cm**

1 m = 100 cm

9 m = 100 cm x 9 = 900cm

The required ratio, = = **100 : 3**

**d) 30 days to 36 hours**

1 day = 24 hours

30 days = 24 hours x 30 = 720 hours

The required ratio is, = = **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 = computers

Therefore 24 students need 24 x = **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} =

In Rajasthan = = **190 lakhs people per km ^{2}**

In UP = = **830 lakhs people per km ^{2}**

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

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

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

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

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

**a) **

= x 100%

= %

= %

= %

= 12.5%

**b) **

= x 100%

= 125%

**c) **

= x 100%

= x 5%

= %

= %

= 7.5%

**d) **

= x 100%

= %

= %

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

a) 0.65

x 100%

=**65%**

b) 2.1

x 100%

=**210%**

**c) 0.02**

x 100%

= 2%

**d) 12.35**

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 =

Therefore, per cent of coloured part = 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 =

Therefore, per cent of coloured part = 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 =

Therefore, per cent of coloured part = x 100%

**= 37.5%**

**4. Find :**

**a) 15% of 250**

= 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

x 36000

=**36 seconds**

**c) 20% of ₹2500**

x 2500

=20 x 25

= **₹ 500**

**d) 75% of 1 kg**

1 kg = 1000 g

Therefore 75% of 1000g is

x 1000

= 750 g

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

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

Let the whole value = a

x a = 600

a =

**a = 12000**

**b) 12% of it is ₹1080**

Let the whole value = ₹a

x ₹a = ₹1080

a =

**a = ₹ 9000**

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

Let the whole value = a km

x a km = 500 km

a =

**a = 1250 km**

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

Let the whole value = a minutes

x a minutes = 14 minutes

a =

**a = 20 minutes**

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

Let the whole value = a litres

x a litres = 40 litres

a =

**a = 500 litres**

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

**a) 25%**

Fractions :

Simplest form :

Decimal form : **0.25**

**b) 150%**

Fractions :

Simplest form :

Decimal form : **1.5**

**c) 20%**

Fractions:

Simplest form :

Decimal form : **0.2**

**d) 5%**

Fractions:

Simplest form :

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 = 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: x a = ₹ 4000

a =

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 = x 20

= 5

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

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

### 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 % = x 100

= 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% = 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 % = x 100

= 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 % = x 100

= x 100

**Loss % = 40%**

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

**a) 3 : 1**

Total part = 3 + 1 = 4

I^{st} Part =

II^{nd} Part =

**Percentage of I ^{st}part = **

**x 100 = 75%**

**Percentage ofII ^{nd} part= x 100**

**= 25%**

**b) 2 : 3 : 5**

Total part = 2 + 3 + 5 =10

1^{st} part =

2^{nd} part =

3^{rd} part =

**Percentage of 1 ^{st} part = x 100 = 20%**

**Percentage of 2 ^{nd} part = x 100 = 30%**

**Percentage of 3 ^{rd} part = x 100 = 50%**

** **

**c) 1 : 4**

Total part = 1 + 4 = 5

1^{st} part =

2^{nd} part =

**Percentage of 1 ^{st} part = x 100 = 20%**

**Percentage of 2 ^{nd} part = x 100 = 80%**

**d) 1 : 2 : 5**

Total part = 1 + 2 + 5 = 8

1^{st} part =

2^{nd} part =

3^{rd} part =

**Percentage of 1 ^{st} part = x 100 = 12.5%**

**Percentage of 2 ^{nd} part = x 100 = 25%**

**Percentage of 3 ^{rd} part = 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 = x 100

=

**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 = x 100

=

**Percentage of price increase = ****%**

**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% = x 100

Therefore, profit = Profit% x CP/ 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 % = x 100

Therefore Loss = Loss% x CP/ 100

=

= ₹

We know that Loss = CP – SP

₹ = X – ₹13500

X – = 13500

= 13500

X =

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 =

Percentage of Carbon = 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

x X = 3

X =

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

=

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 =

=

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 =

=

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 =

Therefore R =

=

**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 =

Therefore, P =

=

**P = ₹ 500**

**Hence, she borrowed ₹ 500**

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

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