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NCERT Solutions for Class 8 Maths Chapter 7 Comparing Quantities

NCERT Solutions for Class 8 Maths Chapter 7 Comparing Quantities

Edited By Komal Miglani | Updated on Jul 12, 2025 09:28 AM IST

From understanding discounts in the market to seeing how your savings grow with compound interest, Comparing Quantities shows how ratios, percentages, taxes, and interest formulas connect maths to everyday life. When you buy a ₹1,000 shirt at a 20% discount, calculate GST on your bill, or check how your money grows in a bank using compound interest, you are using Comparing Quantities in your daily life. Mathematicians often use ratios, percentages and proportions to compare costs or values. For example, if an item is highly priced and increases in price by a percentage, then the total cost of all related items will also increase. Likewise, calculating interest helps us define how money can grow over time, or what extra we may pay over time. To make these concepts easier, the NCERT Solutions for Class 8 by Careers360 are aligned with the CBSE 2025-26 syllabus and pattern, ensuring that students learn and apply Comparing Quantities confidently.

This Story also Contains
  1. Comparing Quantities Class 8 Questions And Answers PDF Free Download
  2. Comparing Quantities Class 8 Maths NCERT Solutions (Exercise)
  3. Comparing Quantities Class 8 Solutions - Important Formulae
  4. NCERT Solutions for Class 8 Maths Chapter 7 Comparing Quantities - Topics
  5. NCERT Solutions for Class 8 Maths: Chapter Wise
  6. NCERT Solutions for Class 8 - Subject Wise
  7. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 8 Maths Chapter 7 Comparing Quantities
NCERT Solutions for Class 8 Maths Chapter 7 Comparing Quantities

Comparing Quantities teaches how numbers, savings, and spending are connected in reality. In Computing Quantities Class 8, students will learn how to solve practical problems involving percentages, profit and loss, discount, simple interest, and compound interest. The solutions address all the exercises in the NCERT textbook and include worked examples, real-world word problems, and simple steps to solve each problem. These NCERT solutions for Class 8 Maths Chapter 7 Comparing Quantities help the students cover all kinds of problems related to ratios, percentages, discounts, profit-loss, as well as simple compound interest. For the full syllabus mapping, simple notes, and PDFs, please visit: NCERT.

Comparing Quantities Class 8 Questions And Answers PDF Free Download

Students who wish to access the NCERT solutions for class 8, chapter 7, Comparing Quantities, can click on the link below to download the entire solution in PDF.

Comparing Quantities Class 8 Maths NCERT Solutions (Exercise)

Class 8 Maths Chapter 7 Question Answer: 7.1
Total Questions: 6
Page number: 81-82

Question 1(a): Find the ratio of the following.

The speed of a cycle is 15 km per hour to the speed of a scooter is 30 km per hour.

Answer:

The ratio of the speed of the cycle to the speed of the scooter

=15 km/hour30 km/hour=1530=12

= 1:2

Question 1(b): Find the ratio of the following.

5m to 10km

Answer:

To find the ratio, we need to make both quantities of the same unit.

We know, 1 km = 1000 m.

10 km = 10 × 1000 m = 10000 m

Therefore the required ratio =

510×1000=510000=12000

= 1 : 2000

Question 1(c): Find the ratio of the following.

50 paise to Rs. 5

Answer:

To find the ratio, we first make the quantities of the same unit.

We know, that Rs. 1 = 100 paisa.
Rs. 5 = 5x100 = 500 paisa

Therefore, the required ratio =

50 paisaRs. 5=50 paisa5×100 paisa=110

= 1:10

Question 2 (a): Convert the following ratios into percentages.

(a) 3:4

Answer:

To convert a ratio to a percentage, we multiply the ratio by 100%.

The required percentage of the ratio 3:4 =

34×100 %=75%

Question 2(b): Convert the following ratios to percentages.

(b) 2:3

Answer:

To convert a ratio to a percentage, we multiply the ratio by 100%.

The required percentage of ratio 2:3 =

23×100 %=66.67%

Question 3: 72% of 25 students are interested in mathematics. How many are not interested in mathematics?

Answer:

Given;

72% of 25 students are interested in mathematics

Percentage of students not interested in mathematics = (100 - 72)% = 28 %

The number of students not interested in mathematics = 28% of 25

=28100×25=284

=7

Therefore, 7 students (out of 25) are not interested in mathematics.

Question 4: A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all?

Answer:

Given,

Win percentage of the team = 40%

This means that they won 40 matches out of 100 matches played.

They won 10 matches out of 10040×10 = 25 matches played.

Therefore, they played a total of 25 matches in all.

Question 5: If Chameli had Rs 600 left after spending 75% of her money, how much did she have in the beginning?

Answer:

Let the amount of money Chameli had in the beginning = Rs. X

She spends 75% of the money.

Percentage of money left = (100 - 75)% = 25%

Since she has Rs. 600 left.

25% of X = 600

25100×X=600

X4=600

X=600×4

X=2400

Therefore, Chameli had Rs. 2400 with her in the beginning.

Question 6: If 60% of people in the city like cricket, 30% like football, and the remaining like other games, then what percent of the people like other games? If the total number of people is 50 lakh, find the exact number who like each type of game.

Answer:

Given,

Total number of people = 50 lakhs

Percentage of people who like cricket = 60%

Percentage of people who like football = 30%

Since the remaining people like other games,

Percentage of people who like other games
= {100 - (60 + 30) } = (100 - 90) = 10%

10% of the people like other games.

Now,

Number of people who like cricket = 60% of 50 lakhs

=60100×50=30 lakhs

Number of people who like football = 30% of 50 lakhs

=30100×50=15 lakhs

Number of people who like other games = 10% of 50 lakhs

=10100×50=5 lakhs

Class 8 Maths Chapter 7 Question Answer: 7.2
Total Questions: 5
Page number: 85

Question 1: During a sale, a shop offered a discount of 10% on the market prices of all the items. What would a customer have to pay for a pair of jeans marked at Rs 1450 and two shirts marked at Rs 850 each?

Answer:

Since the 10% discount is on all the items, we can calculate the Selling price by totalling the Cost price of all items bought.

Now,

Total Cost price(CP) of the items he bought
= CP of a pair of jeans + CP of two shirts
= Rs.(1450 + 850 + 850) = Rs. 3150

The selling price of these items
= (100- 10)% of Rs. 3150 = 90% x Rs.3150 = Rs. 2835

The customer has to pay Rs. 2835.

Question 2: The price of a TV is Rs 13,000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it.

Answer:

Given,

Cost price of the TV = Rs. 13000

Sales tax at the rate of 12%

Selling price
= CP + Sales Tax
= CP + 12% of CP

= 112% of CP
=112100×13000=Rs.14560

Vinod will have to pay Rs. 14,560.

Question 3: Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is Rs 1,600, find the marked price.

Answer:

The discount given was 20% which means if CP is Rs. 100 then the SP is Rs. 80

If SP is Rs. 80 then CP is Rs. 100

For SP of Rs. 1600,
CP =10080×1600=Rs.2000

The marked price is Rs. 2000.

Question 4: I purchased a hair dryer for Rs 5,400 including 8% VAT. Find the price before VAT was added.

Answer:

Given,

VAT = 8%

Let the original price be Rs. 100

Original price + VAT = Rs. 100 + Rs. (8%of100)

Original price + VAT = Rs. 100 + Rs. 8 = Rs. 108

If the price after VAT is Rs.5400, then the price before VAT is
Rs.100108×5400=Rs.5000

The price before VAT was added is Rs. 5000.

Question 5: An article was purchased for Rs 1,239 including a GST of 18%. Find the price of the article before GST was added.

Answer:

Given,

GST = 18%

Cost with GST included = Rs. 1239

Cost without GST = x Rs.

x+(18100×x)=1239

Cost before GST+ GST = cost with GST

x+(9x50)=1239

x=1050

Price before GST = 1050 rupees

Class 8 Maths Chapter 7 Question Answer: 7.3
Total Questions: 3
Page number: 89

Question 1(i): The population of a place increased to 54,000 in 2003 at a rate of 5% per annum

(i) Find the population in 2001.

Answer:

Let the population in 2001 be P

Compound rate of increase = 5% p.a.

The population in 2003 will be more than in 2001

Time period, n = 2 years (2001 to 2003)

54000=P(1+5100)2
P=54000×100×100105×10548980

Therefore, the population in 2001 was 48980 (approx)

Question 1(ii): The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.

(ii) what would be its population in 2005?

Answer:

Let the population in 2001 be P'

Compound rate of increase = 5% p.a.

The population in 2005 will be more than in 2003

Time period, n = 2 years (2003 to 2005)

P=54000(1+5100)2
P=54000×105×105100×10059535

Therefore, the population in 2005 will be 59535 (approx)

Question 2: In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5, 06,000.

Answer:

Given,

The initial count of bacteria, P = 5, 06,000 (Principal Amount)

Rate of increase, R = 2.5% per hour.

Time period, n = 2 hours

(This question is done in a similar manner as compound interest)

Number of bacteria after 2 hours = P(1+R100)n

=506000(1+2.5100)2531616

Therefore, the number of bacteria at the end of 2 hours will be 531616 (approx)

Question 3: A scooter was bought at Rs 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Answer:

Given,

Principal = Rs 42,000

Rate of depreciation = 8% p.a

Reduction = 8% of Rs 42000 per year

=42000×8×1100=Rs 3360

Value at the end of 1 year = Rs (42000 – 3360) = Rs 38,640

Comparing Quantities Class 8 Solutions - Important Formulae

  • Profit: Profit = Selling price - Cost price
  • Loss: Loss = Cost price - Selling price

  • Profit or Loss Determination:

  • If SP > CP, then it is a profit.

  • If SP = CP, then it is neither a profit nor a loss.

  • If CP > SP, then it is a loss.

  • Discount: Discount = Marked Price - Sale Price

  • Discount Percentage: Discount % = (Discount × 100) / Marked Price

  • Profit Percentage: Profit Percentage = (Profit / Cost Price) × 100

  • Loss Percentage: Loss Percentage = (Loss / Cost Price) × 100

  • Percentage Increase: Percentage Increase = (Change in Value / Original Value) × 100

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

  • Compound Interest Formula: Compound Interest = Amount - Principal

  • Sales Tax or VAT: Sales tax or VAT = (Cost Price × Rate of Sales Tax) / 100

  • Billing Amount: Billing Amount = Selling price + VAT

NCERT Solutions for Class 8 Maths Chapter 7 Comparing Quantities - Topics

The topics discussed in the NCERT Solutions for class 8, chapter 7, Comparing Quantities, are:

  • Recalling Ratios and Percentages

  • Finding Discounts

  • Sales Tax/Value Added Tax/Goods and Services Tax

  • Compound Interest

  • Deducing a Formula for Compound Interest

  • Applications of the Compound Interest Formula

NCERT Solutions for Class 8 Maths: Chapter Wise

Access all NCERT Class 8 Maths solutions from one place using the links below.

NCERT Solutions for Class 8 - Subject Wise

Students can find subject-wise solutions for various subjects from the links below.

NCERT Books and NCERT Syllabus

The following links will take students to the latest CBSE syllabus and some reference books.

Articles

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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