# NCERT Solutions for Class 7 Mathematics: Chapter 8 - Comparing Quantities

Check NCERT Solutions for Class 7 Mathematics (Chapter 8 - Comparing Quantities). With this article, you will get solutions to all the questions of Class 7 Maths NCERT Textbook, Chapter 8 - Comparing Quantities.

*CBSE 2020*

Check NCERT Solutions for Class 7 Mathematics, Chapter 7 - Congruence of Triangles. This chapter is one of the most important chapters of CBSE Class 7 Mathematics subject. Students of Class 7 are advised to prepare this chapter well, as concepts given in this chapter are required to solve questions of other chapters.

**NCERT Solutions for Class 7 Mathematics, Chapter 8 - Comparing Quantities:**

**1. **Find the ratio of:

(a) Rs. 5 to 50 paise

(b) 15 kg to 210 g

(c) 9 m to 27 cm

(d) 30 days to 36 hours

**Solution:**

**(a)** Rs. 5 to 50 paise

As, 1 rupee = 100 paise

So, 5 rupee = 500 paise

∴ Ration of Rs. 5 to 50 paise = Rs. 5/50 Paise = 500 paise /50 paise = 10/1.

**(b)** 15 kg to 210 g

As, 1 kg = 1000 g

So, 15 kg = 15000 g

∴Ratio of 15 kg to 210 g = 15kg/210 g = 15000g/210g = 500/7.

**(c)** 9 m to 27 cm

As, 1 m = 100 cm

So, 9 m = 900 cm

∴ Ratio of 9 m to 27 cm = 900cm/27cm = 900/27 = 100/3.

**(d)** 30 days to 36 hours

As, 1 days = 24 hrs

So, 30 days = 24 × 30 = 720 hrs

∴ Ratio of 30 days to 36 hours = 30days/36hrs = 720 hrs/36 hrs = 20/1.

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

**Solutions:**

Given, number of computers required for 6 students = 3

So, the number of computers required for 1 student = 3/6 = 1/2

Now, for 24 students, the number of computers required = 24 × (½) = 12

**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?

(ii) Which State is less populated?

**Solutions:**

(i) Given, population of Rajasthan in 3 km^{2} area = 570 lakh

So the population of Rajasthan in 1 km^{2} area = 570/3 = 190 lakh

Now, Population of U.P in 2 km^{2} area = 1660 lakh

So, Population of U.P in 1 km^{2} area = 1660/2 = 830 lakh

(ii) On comparing the numbers we can easily observe that Rajasthan is less populated.

**Exercise 8.2**

**1. Convert the given fractional numbers to percents.**

(a) 1/8

(b) 5/4

(c) 3/40

(d) 2/7

**Solution:**

**(a)** 1/8 = (1/8) × 100/100

= 1/8 × 100 % = 12.5%

**(b)** 5/4 = (5/4) × 100/100

= 500/4 % = 125 %

**(c) **3/40 = (3/40) × 100/100

= 300/40 % = 7.5 %

**(d)** 2/7 = (2/7) × 100/100

= 200/7 % = 28.57 %

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

(a) 0.65

(b) 2.1

(c) 0.02

(d) 12.35

**Solutions:**

(a) 0.65 = 0.65 × 100%

= (65/100) × 100%

= 65%

(b) 2.1 = 2.1 × 100%

= (21/10) × 100%

= 210%

(c) 0.02 = 0.02 × 100%

= (2/100) × 100%

= 2%

(d) 12.35 = 12.35 × 100%

= (1235/100) × 100%

= 1235%

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

**Solutions:**

(i) 1 out of 4 equal parts is shaded so it represents the fraction 1/4.

Now, 1/4 = (1/4) × 100% = 25%.

(ii) 3 out of 5 equal parts are shaded so it represents the fraction 3/5.

Now, 3/5 = (3/5) × 100% = 60%.

(iii) 3 out of 8 equal parts are shaded so it represents the fraction 3/8.

Now, 3/8 = (3/8) × 100% = 300/8 % = 37.5%.

**4. **Find:

(a) 15% of 250

(b) 1% of 1 hour

(c) 20% of Rs. 2500

(d) 75% of 1 kg

**Solutions:**

(a) 15% of 250 = (15/100) × 250 = 75/2 = 37.5

(b) As, 1 hour = 60 minutes

So,1% of 60 Minutes = (1/100) × 60 Minutes= 3/5 Minutes

(c) 20% of Rs. 2500 = (20/100) × Rs. 2500 = Rs. 500

(d) 75% of 1 Kg = (75/100) × 1 Kg = 0.75 Kg

**5. **Find the whole quantity if

(a) 5% of it is 600.

(b) 12% of it is Rs. 1080.

(c) 40% of it is 500 km.

(d) 70% of it is 14 minutes.

(e) 8% of it is 40 litres.

**Solutions:**

(a) Let the quantity be *x*

According to the question, 5% of *x *= 600

⇒ (5/100) × *x* = 600

⇒ *x* = 600 × (100/5) = 12000

(b) Let the quantity be *x*

According to the question, 12% of *x *= Rs. 1080

⇒ (12 × 100) × *x* = Rs. 1080

*⇒ x *= Rs. 1080 × (100/12) = Rs. 9000

(c) Let the quantity be *x*

According to the question, 40% of *x* = 500 Km

⇒ (40/100) ×* x* = 500 Km

* ⇒ x *= 500 × (100/40) = 1250 Km

(d) Let the quantity be *x*

According to the question, 70% of *x* = 14 min

*⇒ x* × (70/100) = 14 min

*⇒ x* = 14 × (100/70) = 20 min

(e) Let the quantity be *x*

According to the question, 8% of *x* = 40 L

*⇒ x* × (8/100) = 40 L

*⇒ x* = 40 × (100/8) = 500 L

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

(a) 25%

(b) 150%

(c) 20%

(d) 5%

**Solutions:**

(a) 25% = 25/100

Decimal form = 0.25

Fraction form = 1/4

(b) 150%

Decimal form = 150/100 = 1.5

Fraction form = 3/2

(c) 20% = 20/100

Decimal form = 0.2

Fraction form = 1/5

(d) 5% = 5/100

Decimal form = = 0.05

Fraction form = 1/20

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

**Solutions:**

According to the question, 30% are females and 40% are males.

So % of children are = (100 − 30 − 40) % = 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?

**Solutions:**

Percentage of voters who voted = 60%

Percentage of those who did not vote = 100% − 60% = 40%

Number of people who did not vote = 40% of 15000

= (40/100) × 15000 = 6000

Therefore, 6000 people did not vote.

**9. **Meeta saves Rs. 400 from her salary. If this is 10% of her salary. What is her salary?

**Solutions:**

Let Meeta’s salary be Rs* x*.

Given that,

10% of *x *= 400

⇒(10/100) × x = 400

⇒ x/10 = 400

⇒ x = 400 × 10 = Rs. 4000

Therefore, Meeta’s salary is Rs 4000.

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

**Solutions:**

Given, number of games won = 25% of 20 = (25/100) × 20 = 5

Hence, the team won 5 matches.

**Exercise 8.3**

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 Rs. 250 and sold for Rs. 325.

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

(c) A cupboard bought for Rs. 2,500 and sold at Rs. 3,000.

(d) A skirt bought for Rs. 250 and sold at Rs. 150.

**Solutions:**

(a) According to the statement,

Cost price = Rs 250

Selling price = Rs 325

So, Profit = Selling Price - Cost Price 325 − 250 = Rs 75

Now, Profit % = (Profit/Cost Price) × 100 = 75/250 × 100 = 30%

(b) According to the statement,

Cost price = Rs 12000

Selling price = Rs 13,500

So, Profit = 13500 − 12000 = Rs 1500

Now, Profit % = (Profit/Cost Price) × 100 = (500/2500) × 100 = 20%

(d) According to the statement,

Cost price = Rs 250

Selling price = Rs 150

So, loss = 250 − 150 = Rs 100

Now, loss % = (Loss/CP) × 100 = (100/250) × 100 = 40%

2. Convert each part of the ratio to percentage:

(a) 3:1

(b) 2:3:5

(c) 1:4

(d) 1:2:5

**Solutions:**

(a) 3: 1

First part = 3/4 = (3/4) × 100% = 75%

Second part = 1/4 = (1/4) × 100% = 25%

(b) 2: 3: 5

First part = 2/10 = (2/10) × 100% = 20%

Second part = 3/10 = (3/10) × 100% = 30%

Third part = 5/10 = (5/10) × 100% = 50%

(c) 1: 4

First part = 1/5 = (1/5) × 100% = 20%

Second part = 4/5 = (4/5) × 100% = 80%

(d) 1: 2: 5

First part = 1/8 = (1/8) × 100% = 12.5%

Second part = 2/8 = (2/8) × 100% = 25%

Third part = 5/8 = (5/8) × 100% = 62.5%

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

**Solutions:**

According to the question,

Initial population = 25000

Final population = 24500

Net Decrease in population = 500

% Decrease = [(Net Decrease in population)/Initial population] × 100= (500/25000) × 100 = 2%

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

**Solutions:**

According to the question,

Initial price = Rs 350000

Final price = Rs 370000

Increase in price = Rs 20000

% increase = [(Increase in price)/(Initial price)] × 100= (20000/350000) × 100= 5.71 %

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

**Solutions:**

According to the question,

Cost price = Rs 10000

Now, profit = 20% of 10000 = (20/100) × 10000 = Rs 2000

Also, Selling price = Cost price + Profit = 10000 + 2000 = Rs 12,000

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

**Solutions:**

According to the question,

Selling price = Rs 13,500 & Loss % = 20%

Suppose the cost price be *x*.

∴ Loss = 20% of *x*

Also, Cost price − Loss = Selling price

⇒x – (20/100) × *x* = 13500

⇒x – (1/5)*x *= 13500

⇒(4/5)*x *= 13500

⇒x = 13500 × (5/4) = 16875

**7. (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 3g, what is the weight of the chalk stick?

**Solutions:**

(i) According to the question,

Ratio of calcium, carbon, and oxygen = 10: 3: 12

Therefore, percentage of carbon = [3/(10 + 3 + 12)] × 100 = 12 %

(ii) Suppose the weight of the stick be *x** *(in gram)

12 % of *x* = 3

⇒(12/100) × *x *= 3

⇒*x* = 3 × (100/12) = 25 g

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

**Solutions:**

Given, Cost price = Rs 275 & Loss % = 15%

Therefore, loss = 15% of 275

Now, Selling price = Cost price − Loss

⇒Selling Price = 275 – (15/100) × 275

⇒Selling Price = 275 – (4125/100)

⇒Selling Price =275 − 41.25

Therefore, Selling price = Rs 233.75

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

(a) Principal = Rs. 1,200 at 12% p.a.

(b) Principal = Rs. 7,500 at 5% p.a.

**Solutions:**

(a) Given, Principal = Rs 1200, Rate = 12 % p.a., Time = 3 years

Simple Interest = (Principal × Rate × Time)/100 = (1200 × 12 × 3)/100 = Rs 432

Amount = Principal + Simple Interest = 1200 + 432 = Rs 1632

(b) Given, Principal = Rs 7500, Rate = 5% p.a., Time = 3 years

Simple Interest = (Principal × Rate × Time)/100 = (7500 × 5 × 3)/100 = Rs 1125

Amount = Principal + Simple Interest = 7500 + 1125 = Rs 8625

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

**Solutions:**

Simple Interest = (Principal × Rate × Time)/100

⇒280 = (56000 × Rate × 2)/100

Rate = 280/(560 × 2) = 1/4 = 0.25

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

**Solutions:**

Simple Interest = (Principal × Rate × Time)/100

⇒ 45 = (Principal × 9 × 1)/100

Principal = (45 × 100)/9 = Rs 500.

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