Gujarat BoardĀ GSEB Textbook Solutions Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

Question 1.

Convert the given fractional numbers to per cents.

(a) \(\frac { 1 }{ 8 }\)

(b) \(\frac { 5 }{ 4 }\)

(c) \(\frac { 3 }{ 40 }\)

(d) \(\frac { 2 }{ 7 }\)

Solution:

(a) We have \(\frac { 1 }{ 8 }\) = \(\frac { 1 }{ 8 }\) x \(\frac { 100 }{ 100 }\)

= \(\frac { 100 }{ 8 }\)%

= \(\frac { 1 }{ 8 }\)% or 12.5%

= \(\frac { 1 }{ 8 }\) = \(\frac { 25 }{ 2 }\)% or 12.5%

(b) We have \(\frac { 5 }{ 4 }\) = \(\frac { 5 }{ 4 }\) x \(\frac { 100 }{ 100 }\)

= \(\frac { 5Ć100 }{ 4 }\)%

= (5 Ć 25)% = 125%

Thus, = \(\frac { 5 }{ 4 }\) = 12.5%

(c) We have \(\frac { 3 }{ 40 }\) = \(\frac { 3 }{ 40 }\) x \(\frac { 100 }{ 100 }\)

= \(\frac { 3Ć100 }{ 40 }\)%

= \(\frac { 3Ć25 }{ 10 }\)%

= \(\frac { 75 }{ 10 }\)%

or 7.5%

Thus, \(\frac { 2 }{ 7 }\) = 28\(\frac { 4 }{ 7 }\)%

(d) We have \(\frac { 2 }{ 7 }\) = \(\frac { 2 }{ 7 }\) x \(\frac { 100 }{ 100 }\)

= \(\frac { 200 }{ 7 }\)%

= 28\(\frac { 4 }{ 7 }\)%

Thus, \(\frac { 2 }{ 7 }\) = 28\(\frac { 4 }{ 7 }\)%

Question 2.

Convert the given decimal fractions to percents.

(a) 0.65

(b) 2.1

(c) 0.02

(d) 12.35

Solution:

Question 3.

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

Solution:

(i) āµ \(\frac { 1 }{ 4 }\) Part is shaded.

and \(\frac { 1 }{ 4 }\) = \(\frac { 1 }{ 4 }\) = \(\frac { 100 }{ 4 }\) = 25%

Thus, coloured part is 25%.

(ii) āµ 3 parts out of 5 parts are shaded.

ā“ \(\frac { 3 }{ 5 }\) part is shaded

and \(\frac { 3 }{ 5 }\) x 100% = 3 x 20% = 60%

Thus, 60% part is shaded

(iii) Here, 3 parts out of 8 parts are shaded.

ā“ \(\frac { 3 }{ 5 }\) part is shaded

and \(\frac { 3 }{ 8 }\) x 100% = \(\frac { 3 }{ 2 }\) x 25% = \(\frac { 75 }{ 2 }\)% = 37.5%

Thus, 37.5% part is shaded.

Question 4.

Find:

(a) 15% of 250

(b) 1% of 1 hour

(c) 20% of ā¹ 2500

(d) 75% of 1 kg

Solution:

(a) 15% of 250 = \(\frac { 15 }{ 100 }\) of 250

= \(\frac { 15Ć250 }{ 100 }\) = \(\frac { 75 }{ 2 }\)

= 37.5

(b) 1% of 1 hour = \(\frac { 100 }{ 60 }\) x 60 minutes

(āµ 1 hour = 60 minutes)

= \(\frac { 3 }{ 5 }\) minutes = \(\frac { 3 }{ 5 }\) x 60 seconds

= 36 seconds

Thus, 1% of 1 hour = 36 seconds

(c) 20% of ā¹ 2500 = \(\frac { 20 }{ 100 }\) of 2500

= \(\frac { 20Ć2500 }{ 100 }\)

= 20 x 25 = 500

Thus, 20% of ā¹ 2500 = ā¹ 500

(d) 75% of 1 kg = 75% of 1000 g

(āµ 1 kg = 100g)

= \(\frac { 75}{ 100 }\) x 1000g

= \(\frac { 75Ć1000 }{ 100 }\)

= 750 g

Thus, 75% of 1 kg = 750 g

Question 5.

Find the whole quantity if

(a) 5% of it is 600.

(b) 12% of it is ā¹ 1080.

(c) 40% of it is 500 km.

(d) 70% of it is 14 minutes.

(e) 8% of it is 40 litres.

Solution:

(a) 5% of a quantity is 600.

Let the quantity be x.

ā“ 5% of x = 600

\(\frac { 5 }{ 100 }\) Ć x = 600 ā x = \(\frac { 600Ć100 }{ 5 }\)

or x = 120 x 100 = 12000 .

Thus, the required quantity is 12000.

(b) 12% of a quantity is ā¹ 1080.

Suppose the required quantity = x

ā“ 12% of x = ā¹ 1080

or \(\frac { 12 }{ 100 }\) Ć x = ā¹ 1080 100

or x = ā¹ \(\frac { 1080Ć100 }{ 12 }\)

= ā¹ 9000

Thus, the required amount = ā¹ 9000.

(c) 40% of a quantity is 500 km.

Let the quantity be x.

ā“ 40% of x = 500 km

or \(\frac { 40 }{ 100 }\) Ć x = 500 km

or x = \(\frac { 500Ć100 }{ 40 }\) = 125 x 10 km

or x = 1250 km

Thus, the required quantity = 1250 km.

(d) 70% of a quantity is 14 minutes.

Let the required quantity be x.

ā“ 70% of x = 14 minutes

or \(\frac { 70 }{ 100 }\) Ć x = 14 minutes

or x = \(\frac { 14Ć100 }{ 70 }\)

= 20 minutes

Thus, the required quantity = 20 minutes.

(e) 8% of a quantity is 40 litres.

Let the quantity be x.

ā“ 8% of x = 40 litres

or x = \(\frac { 40Ć100 }{ 8 }\)

x = 500 litres.

Thus, the required quantity is 500 litres.

Question 6.

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

(a) 25%

(b) 150%

(c) 20%

(d) 5%

Solution:

(a) We have 25% = \(\frac { 25 }{ 100 }\) = \(\frac { 1 }{ 4 }\)

Thus, 25% = \(\frac { 1 }{ 4 }\)

(b) We have 150% = \(\frac { 150 }{ 100 }\) = \(\frac { 3 }{ 2 }\)

Thus, 150% = \(\frac { 3 }{ 2 }\)

(c) We have 20% = \(\frac { 20 }{ 100 }\) = \(\frac { 1 }{ 5 }\)

Thus, 20% = \(\frac { 1 }{ 5 }\)

(d) We have 5% = \(\frac { 5 }{ 100 }\) = \(\frac { 1 }{ 20 }\)

Thus, 5% = \(\frac { 1 }{ 20 }\)

Question 7.

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

Solution:

āµ Females are 30% and males are 40%.

ā“ Remaining part of the population

= 100% – (30 + 40)%

= 100% – 70% = 30%

Thus, children are 30% of the population.

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:

Total voters = 15000

Part of voters who voted = 60%

ā“ Part of voters who did not vote

= 100% – 60%

= (100 – 60)% = 40%

Now, 40% of 15000 = \(\frac { 40 }{ 100 }\) x 15000

= 40 x 150 = 6000

Thus, 6000 voters did not vote.

Question 9.

Meeta saves ā¹ 400 from her salary. If this is 10% of her salary, what is her salary?

Solution:

Saving = 10% of the salary

Consider the salary = ā¹ x

Saving = 10% of x

So, we have 10% of x = ā¹ 400 10

Thus, the salary = ā¹ 4000.

Question 10.

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

Solution:

Total number of matches played = 20

Part of matches won = 25%

āµ 25% of 20 = \(\frac { 25 }{ 100 }\) x 20 = 5 100

ā“ The team won 5 matches.