Class – 7 CH-8 COMPARING QUANTITIES

MATHS NCERT SOLUTIONS

Exercise 8.1

Question 1:

Find the ratio of:

(a) ₹5 to 50 paise (b) 15 kg to 210 g

SOLUTION 1:

To find ratios, both quantities should be in same unit.

(a) ₹5 to 50 paise

 5 x 100 paise to 50 paise [ ₹ 1 = 100 paise]

 500 paise to 50 paise

500 10

Thus, the ratio is =  = 10 : 1

50 1

(b) 15 kg to 210 g

 15 x 1000 g to 210 g [ 1 kg = 1000 g]

 15000 g to 210 g

15000 500

Thus, the ratio is =  = 500 : 7

210 7

 9 x 100 cm to 27 cm [ 1 m = 100 cm]

 900 cm to 27 cm

900 100

Thus, the ratio is =  = 100 : 3

27 3

(d) 30 days to 36 hours

 30 x 24 hours to 36 hours [ 1 day = 24 hours]

 720 hours to 36 hours

720 20

Thus, the ratio is = = 20 : 1

36 1

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

SOLUTION 2:

6 students need = 3 computers

 1 student needs = computers

 24 students need = 24 = 12 computers

Thus, 12 computers will be needed for 24 students.

Question 3:

Population of Rajasthan = 570 lakhs and population of U.P. = 1660 lakhs. Area of Rajasthan = 3 lakh km2 and area of U.P. = 2 lakh km2.

(i) How many people are there per km2 in both states? (ii) Which state is less populated?

SOLUTION 3:

2 = Population (i) People present per km

Area

570 lakhs 2

In Rajasthan = 3 lakhs per km2 = 190 people km

1660 lakhs 2

In U.P. = 2 lakh per km2 = 830 people per km

(ii) Rajasthan is less populated.

Exercise 8.2

Question 1:

Convert the given fractional numbers to percent:

(a) (b) (c)

SOLUTION 1:

1 25

(a) = 100% % = 12.5%

8 2

(b) = 100% 5 x 25% = 125%

3 3 15

40 2 2

2 200 4

(d) = 100% % 28 %

7 7 7

Question 2:

Convert the given decimal fractions to per cents:

(a) 0.65 (b) 2.1 (c) 0.02

SOLUTION 2:

(a) 0.65 = 100% = 65%

(b) 2.1 = × 100% = 210%

(b) 12.35 = × 100% = 1235%

(d)

(d) 12.35

Estimate what part of the figures is coloured and hence find the percent which is coloured.

SOLUTION 3:

(i) Coloured part =

 Percent of coloured part = 100% = 25% (ii) Coloured part =

 Percent of coloured part = 100% = 60%

(iii) Coloured part =

 Percent of coloured part = 100% = 25%

= 37.5%

Question 4:

Find:

(a) 15% of 250 (b) 1% of 1 hour (c) 20% of ₹2500 (d) 75% of 1 kg

SOLUTION 4:

(a) 15% of 250 = 250 = 15 x 2.5 = 37.5

(b) 1% of 1 hours = 1% of 60 minutes = 1% of (60 x 60) seconds

=  60 60 = 6 x 6 = 36 seconds

(d) 75% of 1 kg = 75% of 1000 g = 1000 = 750 g = 0.750 kg

Question 5:

Find the whole quantity if:

(a) 5% of it is 600 (b) 12% of it is ₹1080

(e) 8% of it is 40 litres

SOLUTION 5:

Let the whole quantity be x in given questions:

(a) 5% of x = 600

  x 600

 x = 12,000

(b) 12% of x = ₹1080

  x 1080

 x = ₹ 9,000

  x 500

 x = 1,250 km

(d) 70% of x = 14 minutes

  x 14

 x = 20 minutes

(e) 8% of x = 40 litres

  x 40

 x = 500 litres

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

(a) 25% (b) 150% (c) 20% (d) 5% SOLUTION 6:

S. No. Per cents Fractions Simplest form Decimal form

(a) 25% 0.25

(b) 150% 1.5

(d) 5% 0.05

Question 7:

In a city, 30% are females, 40% are males and remaining are children. What percent are children?

SOLUTION 7:

Given: Percentage of females = 30% Percentage of males = 40%

Total percentage of females and males = 30 + 40 = 70%

Percentage of children = Total percentage – Percentage of males and females

= 100% – 70%

= 30%

Hence, 30% are children.

Question 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?

SOLUTION 8:

Total voters = 15,000

Percentage of voted candidates = 60%

Percentage of not voted candidates = 100 – 60 = 40% Actual candidates, who did not vote = 40% of 15000

= 15000 = 6,000

Hence, 6,000 candidates did not vote.

Question 9: Meeta saves ₹ 400 from her salary. If this is 10% of her salary. What is her salary?

SOLUTION 9:

Let Meera’s salary be ₹x.

Now, 10% of salary = ₹ 400

 10% of x = ₹ 400

  x 400

 x

 x 4,000

Hence, Meera’s salary is ₹ 4,000.

Question 10:

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

SOLUTION 10:

Number of matches played by cricket team = 20

Percentage of won matches = 25%

Total matches won by them = 25% of 20

= 20

= 5

Hence, they won 5 matches.

Exercise 8.3

Question 1:

Tell what is the profit or loss in the following transactions. Also find profit percent or loss percent in each case.

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

(b) A refrigerator bought ₹12,000 and sold at ₹ 13,500.

(a) Cost price of gardening shears = ₹ 250

Selling price of gardening shears = ₹ 325

Since, S.P. > C.P., therefore here is profit.

 Profit = S.P. – C.P. = ₹325 – ₹250 = ₹ 75

Profit

Now Profit% = 100

C.P.

= 100 = 30%

Therefore, Profit = ₹75 and Profit% = 30%

(b) Cost price of refrigerator = ₹ 12,000

Selling price of refrigerator = ₹13,500

Since, S.P. > C.P., therefore here is profit.  Profit = S.P. – C.P. = ₹13500 – ₹12000 = ₹1,500

Profit

Now Profit% = 100

C.P.

= 100 = 12.5%

Therefore, Profit = ₹1,500 and Profit% = 12.5%

Selling price of cupboard = ₹ 3,000

Since, S.P. > C.P., therefore here is profit.

 Profit = S.P. – C.P. = ₹3,000 – ₹2,500 = ₹ 500

Profit

Now Profit% = 100

C.P.

= 100 = 20%

Therefore, Profit = ₹ 500 and Profit% = 20%

(d) Cost price of skirt = ₹ 250

Selling price of skirt = ₹ 150

Since, C.P. > S.P., therefore here is loss.

 Loss = C.P. – S.P. =₹250 – ₹150 = ₹100

Loss Now Loss% = 100

C.P.

= 100 = 40%

Therefore, Profit = ₹ 100 and Profit% = 40%

Question 2:

Convert each part of the ratio to percentage:

(a) 3 : 1 (b) 2 : 3 : 5 (c) 1 : 4

SOLUTION 2:

(a) 3 : 1

Total part = 3 + 1 = 4

3 1

Therefore, Fractional part = :

4 4

3 1

 Percentage of parts = 100: 100

4 4

 Percentage of parts = 75% : 25%

(b) 2 : 3 : 5

Total part = 2 + 3 + 5 = 10

2 3 5

Therefore, Fractional part = : :

10 10 10

2 3 5

 Percentage of parts = 100: 100: 100

10 10 10

 Percentage of parts = 20% : 30% : 50%

Total part = 1 + 4 = 5

1 4

Therefore, Fractional part = :

5 5

1 4

 Percentage of parts = 100: 100

5 5

 Percentage of parts = 20% : 80%

(d) 1 : 2 : 5

Total part = 1 + 2 + 5 = 8

1 2 5

Therefore, Fractional part = : :

8 8 8

1 2 5

 Percentage of parts = 100: 100: 100

8 8 8

 Percentage of parts = 12.5% : 25% : 62.5%

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

SOLUTION 3:

The decreased population of a city from 25,000 to 24,500.

Population decreased = 25,000 – 24,500 = 500

Population decreased

Decreased Percentage = 100

Original population

= 100 = 2%

Hence, the percentage decreased is 2%.

Question 4:

Arun bought a car for ₹3,50,000. The next year, the price went up to ₹3,70,000. What was the percentage of price increase?

SOLUTION 4:

Increased in price of a car from ₹ 3,50,000 to ₹ 3,70,000.

Amount change = ₹ 3,70,000 – ₹ 3,50,000 = ₹ 20,000.

Amount of change

Therefore, Increased percentage = 100

Original amount

= 100 = 5 %

Hence, the percentage of price increased is 5 %.

Question 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?

SOLUTION 5:

The cost price of T.V. = ₹ 10,000

Profit percent = 20%

Now, Profit = Profit% of C.P.

= 10000

= ₹ 2,000

Selling price = C.P. + Profit = ₹10,000 + ₹2,000 = ₹ 12,000 Hence, he gets ₹12,000 on selling his T.V.

Question 6:

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

SOLUTION 6:

Selling price of washing machine = ₹13,500

Loss percent = 20%

Let the cost price of washing machine be ₹x. Since, Loss = Loss% of C.P.

20 x

 Loss = 20% of ₹ x =  x

100 5

Therefore, S.P. = C.P. – Loss x

 13500 = x

 13500 =

 x = ₹16,875

Hence, the cost price of washing machine is ₹16,875.

(i) Chalk contains Calcium, Carbon and Oxygen in the ratio 10:3:12. Find the percentage of Carbon in chalk.

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

SOLUTION 7:

(i) Given ratio = 10 : 3 : 12

Total part = 10 + 3 + 12 = 25

Part of Carbon =

Percentage of Carbon part in chalk = 100 = 12%

(ii) Quantity of Carbon in chalk stick = 3 g Let the weight of chalk be x g. Then, 12% of x = 3

  x 3

 x = 25 g

Hence, the weight of chalk stick is 25 g.

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

SOLUTION 8:

The cost of a book = ₹275

Loss percent = 15%

Loss = Loss% of C.P. = 15% of ₹275

= 275 = ₹ 41.25

Therefore, S.P. = C.P. – Loss = ₹275 – ₹41.25 = ₹233.75 Hence, Amina sells a book for ₹233.75.

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

(a) Principal = ₹1,200 at 12% p.a. (b) Principal = ₹ 7,500 at 5% p.a.

SOLUTION 9:

(a) Here, Principal (P) = ₹1,200, Rate (R) = 12% p.a., Time (T) = 3 years

P   R T 1200 12 3 

Simple Interest = =

100 100

= ₹ 432

Now, Amount = Principal + Simple Interest

= ₹1200 + ₹432

= ₹1,632

(b) Here, Principal (P) = ₹7,500, Rate (R) = 5% p.a., Time (T) = 3 years

P   R T 7500 5 3 

Simple Interest = =

100 100

= ₹1,125

Now, Amount = Principal + Simple Interest

= ₹7,500 + ₹1,125

= ₹ 8,625

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

SOLUTION 10:

Here, Principal (P) = ₹56,000, Simple Interest (S.I.) = ₹280, Time (T) = 2 years

P   R T

Simple Interest =

100

56000 R 2

 280 =

 R =

 R = 0.25%

Hence, the rate of interest on sum is 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?

SOLUTION 11:

Simple Interest = ₹45, Rate (R) = 9% p.a., Time (T) = 1 years

P   R T

Simple Interest =

100 P 9 1 

 45 =

100

 P =

 P = ₹ 500

Hence, she borrowed ₹ 500.

🌈⚡Key To Enjoy Learning Maths☘🌈

Pages