NCERT Solutions For Class 8 Maths Chapter 7 Comparing Quantities - 2025-26

Exercise 7.1:

 Introduction to Ratios

• Definition and concept of ratios.

• Solving problems involving the comparison of quantities using ratios.

Introduction to Percentages:

• Definition and conversion between fractions, decimals, and percentages.

• Solving problems related to finding percentages of given quantities.

Increase and Decrease Percentages:

• Understanding how to calculate the increase or decrease in percentage.

• Solving problems involving percentage increase and decrease.

Exercise 7.2:

Finding the Cost Price and Selling Price:

• Concept of cost price (CP) and selling price (SP).

• Solving problems to find CP and SP using given information.

Profit and Loss Calculations:

• Understanding profit and loss.

• Calculating profit and loss percentages based on CP and SP.

Discount:

• Definition and calculation of discount.

• Solving problems involving discount percentages and final prices.

Exercise 7.3:

Simple Interest:

• Definition and formula for calculating simple interest.

• Solving problems to find simple interest, principal, rate, and time.

Compound Interest:

• Basic understanding of compound interest.

• Difference between simple and compound interest.

Word Problems on Comparing Quantities:

• Solving various word problems that involve the application of ratios, percentages, profit and loss, and simple interest.

Access NCERT Solutions for Class 8 Maths Chapter 7 – Comparing Quantities

Exercise: 7.1

1. Find the ratio of the following:

(a) Speed of a cycle $15$km per hour to the speed of scooter $30$km per hour.

Ans: Speed of a cycle $ = $ $15$km

Speed of scooter $ = $$30$km

Ratio of the speed of a cycle to the speed of scooter 

$ = \dfrac{{15}}{{30}}$

$ = \dfrac{1}{2}$ 

The required ratio is $1:2$.

(b) $5$m to $10$km.

Ans:  $5$m to $10$km.

Since 1km $ = $$1000$ m

$ \Rightarrow $$\dfrac{{5\;{\text{m}}}}{{10\;{\text{km}}}}\; = \,\dfrac{5}{{10}} \times \dfrac{1}{{1000}}$

$ = \dfrac{1}{{2000}}$

$ \Rightarrow 1:2000$

The required ratio is $1:2000$.

(c) $50$ paise to Rs $5$.

Ans:  $50$ paise to Rs $5$

Since Rs 1 $ = $ $100$ paise      

$ \Rightarrow \dfrac{{50paise}}{{Rs5}}$                     

$ = \dfrac{{50}}{5} \times \dfrac{1}{{100}}$               

$ = \dfrac{1}{{10}}$                    

$ \Rightarrow 1:10$ 

The required ratio is $1:10$.

2. Convert the following ratios to percentages.

(a) $3:4$

Ans: $3:4$ $ = $ $\dfrac{3}{4}$

$ \Rightarrow $$\dfrac{3}{4} \times \dfrac{{100}}{{100}}$                        

$ = $$0.75 \times 100\% $ 

$ = 75\% $

The required ratio to percentage is $75\% $

(b) $2:3$

Ans: $2:3$ $ = $ $\dfrac{2}{3}$

$ = \dfrac{2}{3} \times \dfrac{{100}}{{100}}$

$ = \dfrac{{200}}{3}\% $ 

$ = 66\dfrac{2}{3}\% $

The required ratio to percentage is $66\dfrac{2}{3}\% $

3. $72\% $ of  $25$ students are good in mathematics. How many are not good in mathematics?

Ans: Total number of students $ = $ 25 .

Percentage of students are good in mathematics  $ = $$72\% $

Percentage of students who are not good in mathematics  $ = $ $(100 - 72)\% $

$ \Rightarrow 28\% $

$\therefore $ Number of students who are not good in mathematics $ = $$28\%  \times 25$

$ \Rightarrow \dfrac{{28}}{{100}} \times 25$

$ \Rightarrow \dfrac{{28}}{4}$

$ \Rightarrow 7$

Students are not good in mathematics $ = 7$

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 ?

Ans: The total number of matches won by the football team $ = 10$.

Percentage of team $ = 40\% $

The total number of matches played by the team $ = ?$

The total number of matches played by the team 

$ \Rightarrow 40\%  \times x = 10$

$ \Rightarrow \dfrac{{40}}{{100}} \times x = 10$

$ \Rightarrow x = 10 \times \dfrac{{100}}{{40}}$

$ \Rightarrow x = \dfrac{{100}}{4}$

$ \Rightarrow x = 25$

The total number of matches played by the team $ = 25$.

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

Ans:Chameli’s money after spend $ = 600$

Percentage of money after spend $ = 75\% $

Beginning amount of chameli $ = ?$

Percentage of beginning amount 

$ = \left( {100{\text{ }} - {\text{ }}75} \right)\% \qquad $

$ = 25\% $

Beginning amount of chameli

$ \Rightarrow 25\%  \times x = 600$

$ \Rightarrow \left( {\dfrac{{25}}{{100}}} \right) \times x = 600$

$ \Rightarrow \left( {\dfrac{1}{4}} \right) \times x = 600$

$ \Rightarrow x = 600 \times 4$

$ \Rightarrow x = 2400$

Beginning amount of chameli $ = 2400$.

6.  If $60\% $ people in city like cricket,$30\% $like football and the remaining like other games, then what per cent of the people like other games? If the total number of people are $50$ lakh, find the exact number who like each type of game.

Ans: Total number of people in city $ = 50$lakh

Percentage of people like cricket $ = 60\% $

Percentage of people like football $ = 30\% $

Percentage of people like other games $ = ?$

Number of people like each type of game $ = ?$

Percentage of people like other games 

$ = \left( {100 - 60 - 30} \right)\% $ 

$ = \left( {100 - 90} \right)\% $

$ = 10\% $

Percentage of people like other games$ = 10\% $

Number of people like cricket 

$ = \left( {60\%  \times 50} \right)$

$ = \left( {\dfrac{{60}}{{100}} \times 50} \right)$

$ = \dfrac{{60}}{2}$

$ = 30$lakh

Number of people like cricket$ = 30$lakh

Number of people like football

$ = \left( {30\%  \times 50} \right)$

$ = \left( {\dfrac{{30}}{{100}} \times 50} \right)$

$ = \dfrac{{30}}{2}$

$ = 15$lakh

Number of people like football$ = 15$lakh

Number of people like other games

$ = \left( {10\%  \times 50} \right)$

$ = \left( {\dfrac{{10}}{{100}} \times 50} \right)$

$ = \dfrac{{10}}{2}$

$ = 5$lakh.

Number of people like other games$ = 5$lakh.

Exercise 7.2

1. During a sale, a shop offered a discount of $10\% $ on the marked 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?

Ans: Discount percentage $ = 10\% $

Price of pair jeans $ = $ Rs.$1450$

Price of shirt $ = $ Rs. $850$

Formula:

Discount $ = $ Marked price $ - $ Sale price.

One shirt price $ = $ Rs. $850$

Two shirt price = $2 \times $ Rs. $850$

$ \Rightarrow $ Rs. $1,700$

Total marked price = Rs. $\left( {1,450 + 1,700} \right)$

= Rs. $3,150$

Discount = Rs. $\left( {10\%  \times 3150} \right)$

= Rs. $\left( {\dfrac{{10}}{{100}} \times 3150} \right)$

Discount = Rs. $315$.

Discount $ = $ Total Marked price $ - $ Sale price

$ \Rightarrow $ Rs $315$ = Rs $3150$ − Sale price

$ \Rightarrow $Sale price $ = $ Rs $\left( {3150 - 315} \right)$

$ \Rightarrow $Sale price $ = $ Rs $2835$

Customer paid $ = $ Rs $2835$.

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.

Ans: Price of TV $ = $ Rs. $13,000$.

Sales tax percentage $ = 12\% $

Vinod have to pay $ = ?$

if  Rs.$100$, then Tax to be pay is Rs. $12$.

when  Rs. $13,000$

Tax to be pay $ = \left( {\dfrac{{12}}{{100}} \times 13,000} \right)$

Tax to be pay $ = 12 \times 130$

Tax to be pay $ = $Rs. $1,560$.

Vinod have to pay $ = $ price of TV $ + $ Tax to be pay

$ = $ Rs. $13,000$$ + $ Rs. $1560$

$ = $ Rs. $14560$

Vinod have to pay$ = $ Rs. $14560$.

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.

Ans: Discount in skates $ = 20\% $

Total amount $ = 1,600$

Marked price $ = x$

Formula:

Discount percent\[\]$ = \left( {\dfrac{{Discount}}{{Marked{\text{ }}price}}} \right) \times 100$ 

Discount percent $ = \left( {\dfrac{{Discount}}{{Marked{\text{ }}price}}} \right) \times 100$

$ \Rightarrow 20 = \left( {\dfrac{{Discount}}{x}} \right) \times 100$

Discount $ = \dfrac{{20 \times x}}{{100}}$

Discount $ = \dfrac{{1 \times x}}{5}$

Discount $ = $ Marked price $ - $ Total amount

$ \Rightarrow \dfrac{{1 \times x}}{5}$$ = x - 1600$

$ \Rightarrow 1600 = x - \dfrac{1}{5}x$

$ \Rightarrow 1600 = \dfrac{{5x - x}}{5}$

$ \Rightarrow 1600 = \dfrac{{4x}}{5}$

$ \Rightarrow \dfrac{{1600 \times 5}}{4} = x$

$ \Rightarrow x = 400 \times 5$

$ \Rightarrow x = 2000$

Marked price $ = 2000$.

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

Ans: Hair-dryer rate include VAT $ = 5,400$

Tax percentage $ = 8\% $

Rate before VAT $ = ?$

VAT$ = 8\% $

If VAT without Rs.$100$, then price is Rs. $108$

when  Rs. $5400$

Rate before VAT $ = \left( {\dfrac{{100}}{{108}} \times 5400} \right)$

Rate before VAT $ = 100 \times 50$

Rate before VAT $ = $ Rs. $5000$.

5. An article was purchased for ₹1239 including GST of 18%. Find the price of the article before GST was added?

Ans: Let’s assume the price of the article before GST as ₹x.

Given that the article was purchased for 1239 rupees, including GST of 18%, so we can write the equation as

x + GST = 1239    ….(1)

We know that GST is 18% of the price before GST, so:

GST = 0.18x

Substituting this into (1)

x + 0.18x = 1239

1.18x = 1239

x = $\frac{{1239}}{{1.18}}$

x = 1050

So, the price of the article before GST was added is approximately 1050 rupees.

Exercise 7.3 

1. The population of a place increased to $54000$ in $2003$ at a rate of $5\% $ per annum

(i) find the population in $2001$

(ii) what would be its population in $2005$?

Ans:  Population in the year $2003$$ = 54,000$

(i) $54000$ $ = $population in 2001 $ \times $ ${\left( {1 + \dfrac{5}{{100}}} \right)^2}$

$54000$$ = $population in 2001 $ \times $${\left( {\dfrac{{100 + 5}}{{100}}} \right)^2}$

$54000$$ = $population in 2001 $ \times $${\left( {\dfrac{{105}}{{100}}} \right)^2}$

$54000$$ = $population in 2001 $ \times $${\left( {1.05} \right)^2}$

$54000$$ = $population in 2001 $ \times $$1.1025$

$\dfrac{{54000}}{{1.1025}}$$ = $population in 2001 

population in 2001 $ = 48979.591$

(ii) population in 2005

population in 2001 $ = $$54000{\left( {1 + \dfrac{5}{{100}}} \right)^2}$

population in 2001 $ = 54000$${\left( {\dfrac{{100 + 5}}{{100}}} \right)^2}$

population in 2001  $ = 54000{\left( {\dfrac{{105}}{{100}}} \right)^2}$

population in 2001 $ = 5400{\left( {1.05} \right)^2}$

population in 2001 $ = 54000 \times 1.1025$

population in 2001 $ = 59.535$

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 $5,06,000$.

Ans:  Initial count of bacteria $ = 5,06,000$

Bacteria at the end of $2$ hours

$ = 506000{\left( {1 + \dfrac{{2.5}}{{100}}} \right)^2}$

$ = 506000{\left( {1 + 0.025} \right)^2}$

$ = 506000{\left( {1.025} \right)^2}$

$ = 506000\left( {1.050625} \right)$

$ = 531616.25$(approximately)

The count of bacteria at the end of 2 hours$ = 531616.25$.

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.

Ans:  Principal (P) $ = $ Rs $42,000$

Rate(R)             $ = $ $8\% $ per annum .

Number(n)      $ = $ $1$ year.

Formula:

S.I $ = \dfrac{{P \times R \times T}}{{100}}$

S.I \[ = \] Rs \[\left( {\dfrac{{42000 \times 8 \times 1}}{{100}}} \right)\]

S.I \[ = \] Rs \[8 \times 420\]

S.I \[ = \] Rs \[3360\].

Value after $1$ year \[ = \] Rs. \[\left( {42000 - 3360} \right)\]

\[ = \] Rs. \[38,640\].

Overview of Deleted Syllabus for CBSE Class 8 Comparing Quantities Maths Chapter 7

Class 8 Maths Chapter 7: Exercises Breakdown

Conclusion

The NCERT Class 8th Maths Chapter 7, "Comparing Quantities," teaches fundamental mathematical concepts like ratios, percentages, and profit and loss. It emphasizes practical skills like discount calculation, sales price estimation, and understanding simple and compound interest. Students should master percentage calculations and conversion between percentages, fractions, and decimals for exams and real-life financial literacy. The chapter introduces concepts like compound interest through clear examples and practice problems, ensuring students understand compounded growth.

Other Study Material for CBSE Class 8 Maths Chapter 7

Chapter-Specific NCERT Solutions for Class 8 Maths

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