  # NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities download the pdf given below. If students aim to score higher in Maths, it requires the right amount of practice for every topic. More marks can be achieved by referring to NCERT Solutions for Class 7 Maths. Problems are solved step by step by CoolGyan’S experts with neat explanations.

Chapter 8 – Comparing Quantities contains 3 exercises, and the NCERT Solutions for Class 7 Maths available on this page provide solutions to the questions present in the exercises. Following are the concepts covered in this chapter of Class 7 Maths NCERT Solutions.

• Equivalent Ratios
• Percentage-Another way of comparing quantities
• Converting fraction numbers to percentage
• Converting Decimals to Percentage
• Converting percentage to Fraction or Decimals
• Use of Percentage
• Interpreting Percentage
• Converting Percentage to “How Many”
• Ratios to Percent
• Increase or Decrease as Percent
• Price Related to an Item or Buying and Selling
• Profit or Loss as a Percentage
• Charge Given on Borrowed Money or Simple Interest
• Interest for Multiple Years

## Download the PDF of NCERT Solutions For Class 7 Maths Chapter 8 Comparing Quantities                  ### Access Exercises of NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities

Exercise 8.1 Solutions

Exercise 8.2 Solutions

Exercise 8.3 Solutions

### Access answers to Maths NCERT Solutions For Class 7 Chapter 8 – Comparing Quantities

Exercise 8.1 Page: 157

1. Find the ratio of:

(a) ₹ 5 to 50 paise

Solution:-

We know that,

₹ 1 = 100 paise

Then,

₹ 5 = 5 × 100 = 500 paise

Now we have to find the ratio,

= 500/50

= 10/1

So, the required ratio is 10: 1.

(b) 15 kg to 210 g

Solution:-

We know that,

1 kg = 1000 g

Then,

15 kg = 15 × 1000 = 15000 g

Now we have to find the ratio,

= 15000/210

= 1500/21

= 500/7 … [∵divide both by 3]

So, the required ratio is 500: 7.

(c) 9 m to 27 cm

Solution:-

We know that,

1 m = 100 cm

Then,

9 m = 9 × 100 = 900 cm

Now we have to find the ratio,

= 900/27

= 100/3 … [∵divide both by 9]

So, the required ratio is 100: 3.

(d) 30 days to 36 hours

Solution:-

We know that,

1 day = 24 hours

Then,

30 days = 30 × 24 = 720 hours

Now we have to find the ratio,

= 720/36

= 20/1 … [∵divide both by 36]

So, 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?

Solution:-

From the question it is given that,

Number of computer required for 6 students = 3

So, number of computer required for 1 student = (3/6)

= ½

So, number of computer required for 24 students = 24 × ½

= 24/2

= 12

∴Number of computer required for 24 students is 12 computers.

3. Population of Rajasthan = 570 lakhs and population of UP = 1660 lakhs.

Area of Rajasthan = 3 lakh km2 and area of UP = 2 lakh km2.

(i) How many people are there per km2 in both these States?

(ii) Which State is less populated?

Solution:-

(i) From the question, it is given that,

Population of Rajasthan = 570 lakh

Area of Rajasthan = 3 lakh Km2

Then, population of Rajasthan in 1 km2 area = (570 lakh)/ (3 lakh km2)

= 190 people per km2

Population of UP = 1660 Lakh

Area of UP = 2 Lakh km2

Then, population of UP in 1 lakh km2 area = (1660 lakh)/ (2 lakh km2)

= 830 people per km2

(ii) By comparing the two states Rajasthan is the less populated state.

Exercise 8.2 Page: 164

1. Convert the given fractional numbers to percent.

(a) 1/8

Solution:-

In order to convert a fraction into a percentage multiply the fraction by 100 and put the percent sign %.

= (1/8) × 100 %

= 100/8 %

= 12.5%

(b) 5/4

Solution:-

In order to convert a fraction into a percentage multiply the fraction by 100 and put the percent sign %.

= (5/4) × 100 %

= 500/4 %

= 125%

(c) 3/40

Solution:-

In order to convert a fraction into a percentage multiply the fraction by 100 and put the percent sign %.

= (3/40) × 100 %

= 300/40 %

= 30/4 %

= 7.5%

(d) 2/7

Solution:-

In order to convert a fraction into a percentage multiply the fraction by 100 and put the percent sign %.

= (2/7) × 100 %

= 200/7 %

= %

2. Convert the given decimal fraction to percent.

(a) 0.65

Solution:-

First we have to remove the decimal point,

= 65/100

Now,

Multiply by 100 and put the percent sign %.

We have,

= (65/100) × 100

= 65%

(b) 2.1

Solution:-

First we have to remove the decimal point,

= 21/10

Now,

Multiply by 100 and put the percent sign %.

We have,

= (21/10) × 100

=210%

(c) 0.02

Solution:-

First we have to remove the decimal point,

= 2/100

Now,

Multiply 100 and put the percent sign %.

We have,

= (2/100) × 100

= 2%

(d) 12.35

Solution:-

First we have to remove the decimal point,

= 1235/100

Now,

Multiply by 100 and put the percent sign %.

We have,

= (1235/100) × 100)

= 1235%

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

(i) Solution:-

By observing the given figure,

We can able to identify that 1 part is shaded out of 4 equal parts.

It is represented by a fraction = ¼

Then,

= ¼ × 100

= 100/4

= 25%

Hence, 25% of figure is coloured.

(ii) Solution:-

By observing the given figure,

We can able to identify that 3 part is shaded out of 5 equal parts.

It is represented by a fraction = 3/5

Then,

= (3/5) × 100

= 300/5

= 60%

Hence, 60% of figure is coloured.

(iii) Solution:-

By observing the given figure,

We can able to identify that 3 part is shaded out of 8 equal parts.

It is represented by a fraction = 3/8

Then,

= (3/8) × 100

= 300/8

= 37.5%

Hence, 37.5% of figure is coloured.

4. Find:

(a) 15% of 250

Solution:-

We have,

= (15/100) × 250

= (15/10) × 25

= (15/2) × 5

= (75/2)

= 37.5

(b) 1% of 1 hour

Solution:-

We know that, 1 hour = 60 minutes

Then,

1% of 60 minutes

1 minute = 60 seconds

60 minutes = 60 × 60 = 3600 seconds

Now,

1% of 3600 seconds

= (1/100) × 3600

= 1 × 36

= 36 seconds

(c) 20% of ₹ 2500

Solution:-

We have,

= (20/100) × 2500

= 20 × 25

= ₹ 500

(d) 75% of 1 kg

Solution:-

We know that, 1 kg = 1000 g

Then,

75% of 1000 g

= (75/100) × 1000

= 75 × 10

= 750 g

5. Find the whole quantity if

(a) 5% of it is 600

Solution:-

Let us assume the whole quantity be x,

Then,

(5/100) × (x) = 600

X = 600 × (100/5)

X = 60000/5

X = 12000

(b) 12% of it is ₹ 1080.

Solution:-

Let us assume the whole quantity be x,

Then,

(12/100) × (x) = 1080

X = 1080 × (100/12)

X = 540 × (100/6)

X = 90 × 100

X = ₹ 9000

(c) 40% of it is 500k km

Solution:-

Let us assume the whole quantity be x,

Then,

(40/100) × (x) = 500

X = 500 × (100/40)

X = 500 × (10/4)

X = 500 × 2.5

X = 1250 km

(d) 70% of it is 14 minutes

Solution:-

Let us assume the whole quantity be x,

Then,

(70/100) × (x) = 14

X = 14 × (100/70)

X = 14 × (10/7)

X = 20 minutes

(e) 8% of it is 40 liters

Solution:-

Let us assume the whole quantity be x,

Then,

(8/100) × (x) = 40

X = 40 × (100/8)

X = 40 × (100/8)

X = 40 × 12.5

X = 500 liters

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

(a) 25%

Solution:-

First convert the given percentage into fraction and then put the fraction into decimal form.

= (25/100)

= ¼

= 0.25

(b) 150%

Solution:-

First convert the given percentage into fraction and then put the fraction into decimal form.

= (150/100)

= 3/2

= 1.5

(c) 20%

Solution:-

First convert the given percentage into fraction and then put the fraction into decimal form.

= (20/100)

= 1/5

= 0.2

(d) 5%

Solution:-

First convert the given percentage into fraction and then put the fraction into decimal form.

= (5/100)

= 1/20

= 0.05

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

Solution:-

From the question, it is given that

Percentage of female in a city =30%

Percentage of male in a city = 40%

Total percentage of male and female both = 40% + 30%

= 70%

Now we have to find the percentage of children = 100 – 70

= 30%

So, 30% are children.

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:-

From the question, it is given that

Total number of voters in the constituency = 15000

Percentage of people who voted in the election = 60%

Percentage of people who did not voted in the election = 100 – 60

= 40%

Total number of voters who did not voted in the election = 40% of 15000

= (40/100) × 15000

= 0.4 × 15000

= 6000 voters

∴ 6000 voters did not vote.

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

Solution:-

Let us assume Meeta’s salary be ₹ x,

Then,

10% of ₹ x = ₹ 4000

(10/100) × (x) = 4000

X = 4000 × (100/10)

X = 4000 × 10

X = ₹ 40000

∴ 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?

Solution:-

From the question, it is given that

Total matches played by a local team = 20

Percentage of matches won by the local team = 25%

Then,

Number of matches won by the team = 25% of 20

= (25/100) × 20

= 25/5

= 5 matches.

∴The local team won 5 matches out of 20 matches.

Exercise 8.3 Page: 171

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.

Solution:-

From the question, it is given that

Cost price of gardening shears = ₹ 250

Selling price of gardening shears = ₹ 325

Since (SP) > (CP), so there is a profit

Profit = (SP) – (CP)

= ₹ (325 – 250)

= ₹ 75

Profit % = {(Profit/CP) × 100}

= {(75/250) × 100}

= {7500/250}

= 750/25

= 30%

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

Solution:-

From the question, it is given that

Cost price of refrigerator = ₹ 12000

Selling price of refrigerator = ₹ 13500

Since (SP) > (CP), so there is a profit

Profit = (SP) – (CP)

= ₹ (13500 – 12000)

= ₹ 1500

Profit % = {(Profit/CP) × 100}

= {(1500/12000) × 100}

= {150000/12000}

= 150/12

= 12.5%

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

Solution:-

From the question, it is given that

Cost price of cupboard = ₹ 2500

Selling price of cupboard = ₹ 3000

Since (SP) > (CP), so there is a profit

Profit = (SP) – (CP)

= ₹ (3000 – 2500)

= ₹ 500

Profit % = {(Profit/CP) × 100}

= {(500/2500) × 100}

= {50000/2500}

= 500/25

= 20%

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

Solution:-

Since (SP) < (CP), so there is a loss

Loss = (CP) – (SP)

= ₹ (250 – 150)

= ₹ 100

Loss % = {(Loss/CP) × 100}

= {(100/250) × 100}

= {10000/250}

= 40%

2. Convert each part of the ratio to percentage:

(a) 3 : 1

Solution:-

We have to find total parts by adding the given ratio = 3 + 1 = 4

1st part = ¾ = (¾) × 100 %

= 3 × 25%

= 75%

2nd part = ¼ = (¼) × 100%

= 1 × 25

= 25%

(b) 2: 3: 5

Solution:-

We have to find total parts by adding the given ratio = 2 + 3 + 5 = 10

1st part = 2/10 = (2/10) × 100 %

= 2 × 10%

= 20%

2nd part = 3/10 = (3/10) × 100%

= 3 × 10

= 30%

3rd part = 5/10 = (5/10) × 100%

= 5 × 10

= 50%

(c) 1:4

Solution:-

We have to find total parts by adding the given ratio = 1 + 4 = 5

1st part = (1/5) = (1/5) × 100 %

= 1 × 20%

= 20%

2nd part = (4/5) = (4/5) × 100%

= 4 × 20

= 80%

(d) 1: 2: 5

Solution:-

We have to find total parts by adding the given ratio = 1 + 2 + 5 = 8

1st part = 1/8 = (1/8) × 100 %

= (100/8) %

= 12.5%

2nd part = 2/8 = (2/8) × 100%

= (200/8)

= 25%

3rd part = 5/8 = (5/8) × 100%

= (500/8)

= 62.5%

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

Solution:-

From the question, it is given that

Initial population of the city = 25000

Final population of the city = 24500

Population decrease = Initial population – Final population

= 25000 – 24500

= 500

Then,

Percentage decrease in population = (population decrease/Initial population) × 100

= (500/25000) × 100

= (50000/25000)

= 50/25

= 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?

Solution:-

From the question, it is given that

Arun bought a car for = ₹ 350000

The price of the car in the next year, went up to = ₹ 370000

Then increase in price of car = ₹ 370000 – ₹ 350000

= ₹ 20000

The percentage of price increase = (₹ 20000/ ₹ 350000) × 100

= (2/35) × 100

= 200/35

= 40/7

= 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?

Solution:-

From the question, it is given that

Cost price of the T.V. = ₹ 10000

Percentage of profit = 20%

Profit = (20/100) × 10000

= ₹ 2000

Then,

Selling price of the T.V. = cost price + profit

= 10000 + 2000

= ₹ 12000

∴ I will get it for ₹ 12000.

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:-

From the question, it is given that

Selling price of washing machine = ₹ 13500

Percentage of loss = 20%

Now, we have to find the cost price washing machine

By using the formula, we have:

CP = ₹ {(100/ (100 – loss %)) × SP}

= {(100/ (100 – 20)) × 13500}

= {(100/ 80) × 13500}

= {1350000/80}

= {135000/8}

= ₹ 16875

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

Solution:-

From the question it is given that,

The ratio of calcium, carbon and oxygen in chalk = 10: 3: 12

So, total part = 10 + 3 + 12 = 25

In that total part amount of carbon = 3/25

Then,

Percentage of carbon = (3/25) × 100

= 3 × 4

= 12 %

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

Solution:-

From the question it is given that,

Weight of carbon in the chalk = 3g

Let us assume the weight of the stick be x

Then,

12% of x = 3

(12/100) × (x) = 3

X = 3 × (100/12)

X = 1 × (100/4)

X = 25g

∴The weight of the stick is 25g.

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

Solution:-

From the question, it is given that

Cost price of book = ₹ 275

Percentage of loss = 15%

Now, we have to find the selling price book,

By using the formula, we have:

SP = {((100 – loss %) /100) × CP)}

= {((100 – 15) /100) × 275)}

= {(85 /100) × 275}

= 23375/100

= ₹ 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.

Solution:-

Given: – Principal (P) = ₹ 1200, Rate (R) = 12% p.a. and Time (T) = 3years.

If interest is calculated uniformly on the original principal throughout the loan period, it is called Simple interest (SI).

SI = (P × R × T)/100

= (1200 × 12 × 3)/ 100

= (12 × 12 × 3)/ 1

= ₹432

Amount = (principal + SI)

= (1200 + 432)

= ₹ 1632

(b) Principal = ₹ 7,500 at 5% p.a.

Solution:-

Given: – Principal (P) = ₹ 7500, Rate (R) = 5% p.a. and Time (T) = 3years.

If interest is calculated uniformly on the original principal throughout the loan period, it is called Simple interest (SI).

SI = (P × R × T)/100

= (7500 × 5 × 3)/ 100

= (75 × 5 × 3)/ 1

= ₹ 1125

Amount = (principal + SI)

= (7500 + 1125)

= ₹ 8625

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

Solution:-

Given: – P = ₹ 56000, SI = ₹ 280, t = 2 years.

We know that,

R = (100 × SI) / (P × T)

= (100 × 280)/ (56000 × 2)

= (1 × 28) / (56 × 2)

= (1 × 14) / (56 × 1)

= (1 × 1) / (4 × 1)

= (1/ 4)

= 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:-

From the question it is given that, SI = ₹ 45, R = 9%, T = 1 year, P =?

SI = (P × R × T)/100

45 = (P × 9 × 1)/ 100

P = (45 ×100)/ 9

= 5 × 100

= ₹ 500

Hence, she borrowed ₹ 500.

## Frequently Asked Questions on NCERT Solutions for Class 7 Maths Chapter 8

### Does NCERT Solutions for Class 7 Maths Chapter 8 help you to clear board exams?

Yes, NCERT Solutions for Class 7 Maths Chapter 8 is one of the important chapters of Class 7 Maths of NCERT Solutions. These solutions are focused on learning various Mathematics tricks and shortcuts for quick and easy calculations. This makes them learn and clear the Maths subject in board exams.

### How NCERT Solutions for Class 7 Maths Chapter 8 helps in learning areas and volumes of geometrical shapes?

NCERT Solutions for Class 7 Maths Chapter 8 is created by our faculty in an interactive manner to make it easy for the students to understand the concepts. Students can refer to the PDF of solutions as a major study material to improve their speed in solving problems accurately.

### Give a short summary on NCERT Solutions for Class 7 Maths Chapter 8 ?

8th Chapter of NCERT Solutions for Class 7 Maths has 3 exercises. The topics covered in NCERT Solutions for Class 7 Maths Chapter 8 are
1. Equivalent Ratios
2. Percentage-Another way of comparing quantities
3. Converting fraction numbers to percentage
4. Converting Decimals to Percentage
5. Converting percentage to Fraction or Decimals
6. Use of Percentage
7. Interpreting Percentage
8. Converting Percentage to “How Many”
9. Ratios to Percent
10. Increase or Decrease as Percent
11. Price Related to an Item or Buying and Selling
12. Profit or Loss as a Percentage
13. Charge Given on Borrowed Money or Simple Interest
14. Interest for Multiple Years.