# Important Questions Class 7 Maths Chapter 8

## Important Questions Class 7 Mathematics Chapter 8 – Comparing Quantities

Mathematics is an important subject that students study in school. It is not a subject that students study in school, but rather one that we require in everyday life.It helps us solve daily life’s problems with more complicated issues in technology, construction, or the economy.

Chapter 8 is about comparing quantities. In previous classes, students have learned about ratios and proportions, which help us to compare quantities of different units. In this chapter, they will learn about percentages and simple interests. These are important concepts; students must practice questions to build the ideas.

Extramarks is a leading company that provides all the important study materials related to CBSE and NCERT. Our experts have made the Important Questions Class 7 Mathematics Chapter 8 to help students in practice. They have collected different types of questions from CBSE sample papers, textbook exercises, important reference books, NCERT Exemplars, and CBSE past years’ question papers. They have also solved the questions so that students can follow the answers. Thus, it will help students  score better in exams.

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## Important Questions Class 7 Mathematics Chapter 8 – With Solutions

Extramarks is a leading company that provides all the important study materials related to CBSE and NCERT. Our experts have collected the important questions from sources such as textbook exercises, CBSE sample papers, CBSE past years’ question papers, NCERT Exemplars, and important reference books. They have also solved the questions, and experienced professionals have further checked the answers to ensure the best quality of the content. Thus, Chapter 8 Class 7 Mathematics Important Questions will help students score better in exams. The important questions are-

Question 1. Find the ratio of

(a) ₹ 5 to 50 paise

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

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

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

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.

Question 2. In a computer lab, there are three computers for every six students. How many computers will be needed if there are 24 students?

From the above question, it is given that,

Then, Number of computers required for six students = is 3

So, the number of computers required for one student = (3/6)

= ½

Hence, the Number of computers required for the 24 students = 24 × ½

= 24/2

= 12

Therefore, the Number of computers required for 24 students is 12 computers.

Question 3. It is given that the Population of Rajasthan = is 570 lakhs, and the population of UP = is 1660 lakhs.

The area of Rajasthan is = 3 lakh km sq, and area of UP = 2 lakh km sq.

(a) How many people are present per km sq in both these States?

(b) Which of the above State is less populated?

(a) From the above question, it is given that,

The population of the state Rajasthan = 570 lakh

Area of the state Rajasthan = 3 lakh Km2

Then, the final population of the state Rajasthan in 1 km sq area = (570 lakh)/ (3 lakh km2)

= 190 people are there per km2

The population of UP is = 1660 Lakh

Area of UP is = 2 Lakh km2

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

= 830 people per km sq

(ii) By comparing the two states, Rajasthan was found to be the less populated state.

Question 4. Convert the given fractional numbers given below to percent.

(i) 1/8

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

= (1/8) × 100 %

= 100/8 %

= 12.5%

(ii) 5/4

In order to convert a fraction into percentage, multiply the fraction by 100 and then put the per cent sign %.

= (5/4) × 100 %

= 500/4 %

= 125%

(iii) 3/40

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

= (3/40) × 100 %

= 300/40 %

= 30/4 %

= 7.5%

Question 5. Convert the following given decimal fraction to percent.

(i) 0.65

Firstly we have to remove the decimal point,

= 65/100

Now,

Multiply it by 100 and then put the percent sign %.

We have,

= (65/100) × 100

= 65%

(ii) 2.1

Firstly we have to remove the decimal point,

= 21/10

Now,

Multiply it by 100 and then put the percent sign %.

We have,

= (21/10) × 100

=210%

(iii) 0.02

Firstly we have to remove the decimal point,

= 2/100

Now,

Multiply it by 100 and put the percent sign %.

We have,

= (2/100) × 100

= 2%

(iv) 12.35

Firstly we have to remove the decimal point,

= 1235/100

Now,

Multiply it by 100 and put the percent sign %.

We have,

= (1235/100) × 100)

= 1235%

Question 6. Find:

(a) 15% of 250

We have,

= (15/100) × 250

= (15/10) × 25

= (15/2) × 5

= (75/2)

= 37.5

(b) 1% of 1 hour

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

We have,

= (20/100) × 2500

= 20 × 25

= ₹ 500

(d) 75% of 1 kg

We know that 1 kg = 1000 g

Then,

75% of 1000 g

= (75/100) × 1000

= 75 × 10

= 750 g

Question 7. Find the whole quantity if

(i) 5% of it is 600

Let us assume that the whole quantity is x,

Then,

(5/100) × (x) = 600

X = 600 × (100/5)

X = 60000/5

X = 12000

(ii) 12% of it is ₹ 1080.

Let us assume that the whole quantity is x,

Now,

(12/100) × (x) = 1080

X = 1080 × (100/12)

X = 540 × (100/6)

X = 90 × 100

X = ₹ 9000

(iii) 40% of it is 500 km

Let us assume that the whole quantity is x,

Now,

(40/100) × (x) = 500

X = 500 × (100/40)

X = 500 × (10/4)

X = 500 × 2.5

X = 1250 km

(iv) 70% of it is 14 minutes

Let us assume the whole quantity is x,

Now,

(70/100) × (x) = 14

X = 14 × (100/70)

X = 14 × (10/7)

X = 20 minutes

(v) 8% of it is 40 litres

Let us assume the whole quantity is x,

Now,

(8/100) × (x) = 40

X = 40 × (100/8)

X = 40 × (100/8)

X = 40 × 12.5

X = 500 litres

Question 8. Convert the given percent to decimal fractions and also convert these fractions in simplest forms:

(i) 25%

Firstly convert the given above percentage into a fraction and then convert the fraction into decimal form.

= (25/100)

= ¼

= 0.25

(ii) 150%

Firstly convert the given above percentage into a fraction and then convert the fraction into decimal form.

= (150/100)

= 3/2

= 1.5

(iii) 20%

Firstly convert the given above percentage into a fraction and then convert the fraction into decimal form.

= (20/100)

= 1/5

= 0.2

(iv) 5%

First, convert the above-given percentage into a fraction and then put these fractions into decimal form.

= (5/100)

= 1/20

= 0.05

Question 9. In a city, 30% of the public are females, 40% of them are males, and the remaining are children. What percent of them are children?

From the above question, it is given that

Percentage of females in the city =30%

Percentage of males in the city = 40%

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

= 70%

Now we will have to find the percentage of children present in the city = 100 – 70

= 30%

So, 30% are children.

Question 10. Out of 15,000 voters in a constituency, 60% of people voted. Find the percentage of voters who didn’t vote. Now find out how many of them actually did not vote.

From the above question, it is given that

Total Number of voters present in the constituency = 15000

The percentage of people that voted in the election is = 60%

The percentage of people who did not vote in the election is = 100 – 60

= 40%

The total number of voters who did not vote in the election is = 40% of 15000

= (40/100) × 15000

= 0.4 × 15000

= 6000 voters

∴ Six thousand voters did not vote.

Question 11. Meeta saves ₹ 4000 from her salary. Suppose this is 10% of her salary. Then What is her total salary?

Let us assume that

Meeta’s salary be ₹ x,

Now,

10% of ₹ x = ₹ 4000

(10/100) × (x) = 4000

X = 4000 × (100/10)

X = 4000 × 10

X = ₹ 40000

Hence Meeta’s salary is ₹ 40000.

Question 12. A local cricket team played around 20 matches in one season. They won 25% of the matches. Hence how many matches did they win?

From the above question, it is given that the

Total matches played by the local team = 20

Percentage of matches won by these local teams = 25%

Now,

The number of matches won by the team is = 25% of the 20

= (25/100) × 20

= 25/5

= 5 matches.

Hence, The local team won 5 matches out of 20 matches.

Question 13. Tell the profit or loss in the following transactions. Also, find the profit per cent or the loss per cent in each of the below cases.

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

From the above question, it is given that

The cost price of the gardening shears = ₹ 250

The selling price of the gardening shears = ₹ 325

Since (SP) > (CP), hence there is a profit.

Profit is = (SP) – (CP)

= ₹ (325 – 250)

= ₹ 75

Profit % = {(Profit/CP) × 100}

= {(75/250) × 100}

= {7500/250}

= 750/25

= 30%

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

From the above question, it is given that the

The cost price of the refrigerator is = ₹ 12000

The selling price of the refrigerator is = ₹ 13500

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

Profit = (SP) – (CP)

= ₹ (13500 – 12000)

= ₹ 1500

Profit % = {(Profit/CP) × 100}

= {(1500/12000) × 100}

= {150000/12000}

= 150/12

= 12.5%

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

From the above question, it is given that

The cost price of the cupboard is = ₹ 2500

The selling price of the cupboard is = ₹ 3000

Since (SP) > (CP), hence there is a profit.

Profit = (SP) – (CP)

= ₹ (3000 – 2500)

= ₹ 500

Profit % = {(Profit/CP) × 100}

= {(500/2500) × 100}

= {50000/2500}

= 500/25

= 20%

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

Since (SP) < (CP), hence there is a loss.

Loss is = (CP) – (SP)

= ₹ (250 – 150)

= ₹ 100

Loss % = {(Loss/CP) × 100}

= {(100/250) × 100}

= {10000/250}

= 40%

Question 14. Convert each part of the ratio given below to a percentage:

(i) 3: 1

We have to find the 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%

(ii) 2: 3: 5

We have to find the total parts by adding the given ratio as = 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%

(iii) 1:4

We have to find the 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%

(iv) 1: 2: 5

We have to find the total parts by adding the given ratio as = 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%

Question 15. The population of a city gets decreases from 25,000 to 24,500. Find the percentage decrease.

From the above question, it is given that.

The Initial population of the city is = 25000

The final population of the city is = 24500

Population decrease is equal to = Initial population – Final population

= 25000 – 24500

= 500

Now,

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

= (500/25000) × 100

= (50000/25000)

= 50/25

= 2%

Question 16. Ishan buys a T.V. for ₹ 10,000 and then sells it at a profit of 20%. How much money did Ishan get for it?

From the above question, it is given that.

The cost price of the T.V. is = ₹ 10000

Percentage of profit is = 20%

Profit = (20/100) × 10000

= ₹ 2000

Then,

The selling price of the T.V. is = cost price + profit

= 10000 + 2000

= ₹ 12000

Hence Ishan will get it for ₹ 12000.

Question 17. Juhi sells a washing machine for ₹ 13,500. She loses 20% in the bargain. What was the initial price at which she bought it?

From the above question, it is given that.

The selling price of the washing machine is = ₹ 13500

Percentage of loss is = 20%

Now, we have to find the initial cost of price washing machine.

By using the given formula, we have:

CP = ₹ {(100/ (100 – loss %)) × SP}

= {(100/ (100 – 20)) × 13500}

= {(100/ 80) × 13500}

= {1350000/80}

= {135000/8}

= ₹ 16875

Question 18. (a) Chalk contains calcium, carbon, and oxygen in the ratio of 10:3:12. Find the total percentage of carbon in chalk.

From the above question, it is given that,

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

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

In that, the total part amount of carbon = 3/25

Then,

Percentage of carbon is = (3/25) × 100

= 3 × 4

= 12 %

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

From the above question, it is given that,

The Weight of carbon in the chalk is = 3g

Let us assume that the weight of the stick is 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.

Question 19. Amina buys a book for ₹ 275 and then sells it at a loss of 15%. How much does she initially sell it for?

From the above question, it is given that

The cost price of the book = ₹ 275

Percentage of loss of the book = 15%

Now, we have to find the selling price of a book,

By using the below formula, we have:

SP = {((100 – loss %) /100) × CP)}

= {((100 – 15) /100) × 275)}

= {(85 /100) × 275}

= 23375/100

= ₹ 233.75

Question 20. Find the final amount to be paid at the end of three years in each case:

(i) Principal = ₹ 1,200 at 12% p.a.

Given in the above question.

P (Principal) = ₹ 1200, Rate (R) = 12% p.a. and T (time) = 3years.

If the interest is calculated uniformly on an original principal throughout the loan period, then 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

(ii) Principal = ₹ 7,500 at 5% p.a.

Given in the question: –

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

If the interest is calculated uniformly on an original principal throughout the loan period, then 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

Question 21. What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in two years?

Given:

P(principal) = ₹ 56000, SI = ₹ 280, t = 2 years.

We know the formula,

R (rate) = (100 × SI) / (P × T)

= (100 × 280)/ (56000 × 2)

= (1 × 28) / (56 × 2)

= (1 × 14) / (56 × 1)

= (1 × 1) / (4 × 1)

= (1/ 4)

= 0.25%

Question 22. If Meena gives an interest of ₹ 45 for one year at a 9% rate per annum. Then What is the sum she has borrowed?

From the above question, it is given that the 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.

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Extramarks is a reputed company in India that provides all the important study materials related to CBSE and NCERT. You will find CBSE past years’ question papers, CBSE sample papers, NCERT exemplars, NCERT important questions, NCERT solutions, CBSE revision notes, CBSE extra questions, important formulas, and many more. The links to the study materials are given below-

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## FAQs (Frequently Asked Questions)

### 1. How can the Important Questions Class 7 Mathematics Chapter 8 help students?

Practice is very important to score better in exams. Especially in Mathematics, students must practice questions as much as possible to score better in exams. The experts atExtramarks have collected different questions from various sources and included them in this article. They have taken help from the textbook exercises, CBSE sample papers, CBSE past years’ question papers, NCERT exemplars and important reference books. They have also provided the answers to the questions. Experienced professionals have further checked the answers to ensure the best quality of the content. Thus, the Important Questions Class 7 Mathematics Chapter 8 will help students generate knowledge in the subject matter, clarify their doubts, and score better in the exams.