How many exercises are there in NCERT Class 8 Maths Chapter 7?

Chapter 7 (Comparing Quantities) has 2 exercises. Exercise 7.1 covers percentages, discounts, profit-loss, and tax. Exercise 7.2 focuses on compound interest and its applications including population growth and depreciation.

What is the difference between simple interest and compound interest?

Simple interest is calculated only on the original principal: $\text{SI} = \dfrac{P \times R \times T}{100}$. Compound interest is calculated on the principal plus previously accumulated interest, so the amount grows faster. After $2$ years, CI is always greater than SI (for the same rate). The difference for $2$ years is exactly $P(R/100)^2$.

How do I handle half-yearly or quarterly compounding?

For half-yearly compounding, divide the annual rate by $2$ and multiply the number of years by $2$. For quarterly compounding, divide the rate by $4$ and multiply the years by $4$. Then use the standard CI formula with these adjusted values. For example, $10\%$ p.a. compounded half-yearly for $1$ year means rate $= 5\%$, periods $= 2$.

Are successive discounts the same as adding the percentages?

No. Two successive discounts of $20\%$ and $10\%$ are not equal to a single $30\%$ discount. First apply $20\%$ to get a reduced price, then apply $10\%$ to that reduced price. The effective single discount is $28\%$, not $30\%$. The formula for effective single discount is $(a + b - ab/100)\%$.

How do I solve depreciation problems?

Depreciation (decrease in value) uses the same CI formula but with a minus sign: $A = P(1 - R/100)^n$. Here $P$ is the original value, $R$ is the rate of depreciation per year, and $n$ is the number of years. This applies to machines, vehicles, electronics, and any asset that loses value over time.

What is the formula for the difference between CI and SI for 2 years?

For $2$ years at rate $R\%$ on principal $P$: $\text{CI} - \text{SI} = P \times \left(\dfrac{R}{100}\right)^2$. This shortcut is extremely useful in competitive exams. For example, if $P = 10000$ and $R = 10\%$, the difference $= 10000 \times 0.01 = 100$.

Is this chapter important for competitive exams?

Very important. Compound interest, percentages, and profit-loss questions appear in almost every competitive exam including NTSE, Olympiad, and later in banking and SSC exams. The CI-SI difference shortcut, successive discount formula, and population growth problems are especially popular in competitive papers.

How many exercises are there in NCERT Class 8 Maths Chapter 7?

Chapter 7 (Comparing Quantities) has 2 exercises. Exercise 7.1 covers percentages, discounts, profit-loss, and tax. Exercise 7.2 focuses on compound interest and its applications including population growth and depreciation.

What is the difference between simple interest and compound interest?

Simple interest is calculated only on the original principal: $\text{SI} = \dfrac{P \times R \times T}{100}$. Compound interest is calculated on the principal plus previously accumulated interest, so the amount grows faster. After $2$ years, CI is always greater than SI (for the same rate). The difference for $2$ years is exactly $P(R/100)^2$.

How do I handle half-yearly or quarterly compounding?

For half-yearly compounding, divide the annual rate by $2$ and multiply the number of years by $2$. For quarterly compounding, divide the rate by $4$ and multiply the years by $4$. Then use the standard CI formula with these adjusted values. For example, $10\%$ p.a. compounded half-yearly for $1$ year means rate $= 5\%$, periods $= 2$.

Are successive discounts the same as adding the percentages?

No. Two successive discounts of $20\%$ and $10\%$ are not equal to a single $30\%$ discount. First apply $20\%$ to get a reduced price, then apply $10\%$ to that reduced price. The effective single discount is $28\%$, not $30\%$. The formula for effective single discount is $(a + b - ab/100)\%$.

How do I solve depreciation problems?

Depreciation (decrease in value) uses the same CI formula but with a minus sign: $A = P(1 - R/100)^n$. Here $P$ is the original value, $R$ is the rate of depreciation per year, and $n$ is the number of years. This applies to machines, vehicles, electronics, and any asset that loses value over time.

What is the formula for the difference between CI and SI for 2 years?

For $2$ years at rate $R\%$ on principal $P$: $\text{CI} - \text{SI} = P \times \left(\dfrac{R}{100}\right)^2$. This shortcut is extremely useful in competitive exams. For example, if $P = 10000$ and $R = 10\%$, the difference $= 10000 \times 0.01 = 100$.

Is this chapter important for competitive exams?

Very important. Compound interest, percentages, and profit-loss questions appear in almost every competitive exam including NTSE, Olympiad, and later in banking and SSC exams. The CI-SI difference shortcut, successive discount formula, and population growth problems are especially popular in competitive papers.

Solved Examples

NCERT Solutions for Class 8 Maths Chapter 7: Comparing Quantities — Free PDF

Master compound interest, discounts, tax, and profit-loss with complete step-by-step solutions.

CBSEClass 8

The SparkEd Authors (IITian & Googler)15 March 202635 min read

Chapter 7 Overview: Comparing Quantities

$Chapter 7 builds on the percentage, profit-loss, and simple interest concepts you learnt in Class 7, and introduces the powerful concept of compound interest (CI) . You will also learn to calculate discounts, sales tax/VAT/GST, and solve real-world problems involving successive percentage changes .$

$These concepts are among the most practically useful in all of school mathematics — from understanding bank deposits and loans to comparing shopping deals. Whether you are calculating EMIs on a future purchase or figuring out the best discount offer during a sale, the formulas in this chapter will serve you for life.$

$The chapter has 2 exercises . Exercise 7.1 covers percentages, profit and loss, discount, and tax (GST/VAT). Exercise 7.2 is entirely dedicated to compound interest and its applications, including population growth and depreciation. Together, the exercises contain around 20 problems of varying difficulty. Mastering this chapter is essential because the same concepts reappear in Class 9, Class 10, and virtually every competitive exam.$

Key Concepts and Definitions

$Before solving the exercises, let us clearly define every concept tested in this chapter.$

$100$

$Profit and Loss: When the selling price (SP) is greater than the cost price (CP), there is a profit. When SP is less than CP, there is a loss. Profit and loss are always calculated as a percentage of the cost price.$

$Discount: The reduction offered on the marked price (MRP) of an article. Discount is always calculated on the marked price, never on the cost price. The price after discount is the selling price.$

$Sales Tax / GST: A tax levied by the government on the sale of goods. It is calculated on the selling price (after discount) and added to the bill amount.$

$\text{SI} = \dfrac{P \times R \times T}{100}$

$Compound Interest (CI): Interest calculated on the principal plus the interest accumulated in previous periods. The amount grows faster than with simple interest because you earn "interest on interest".$

$Successive Percentage Changes: When two percentage changes are applied one after another (e.g., two successive discounts, or population growth over multiple years), the changes are not simply added — each is applied to the result of the previous one.$

Key Formulas

$1. Percentage Increase/Decrease:$

\text{New value} = \text{Original} \times \left(1 \pm \dfrac{\text{rate}}{100}\right)

$2. Profit and Loss:$

\text{Profit \%} = \dfrac{\text{SP} - \text{CP}}{\text{CP}} \times 100, \quad \text{Loss \%} = \dfrac{\text{CP} - \text{SP}}{\text{CP}} \times 100

\text{SP} = \text{CP} \times \left(1 + \dfrac{\text{Profit \%}}{100}\right), \quad \text{SP} = \text{CP} \times \left(1 - \dfrac{\text{Loss \%}}{100}\right)

$3. Discount:$

\text{Discount} = \text{Marked Price} - \text{Selling Price}

\text{Discount \%} = \dfrac{\text{Discount}}{\text{Marked Price}} \times 100

$4. Sales Tax / GST:$

\text{Bill Amount} = \text{SP} + \text{Tax} = \text{SP} \times \left(1 + \dfrac{\text{tax rate}}{100}\right)

$5. Compound Interest:$

A = P\left(1 + \dfrac{R}{100}\right)^n, \quad \text{CI} = A - P

$P$

$R/2$

$R/4$

$8. CI vs SI shortcut for 2 years:$

\text{CI} - \text{SI} = P\left(\dfrac{R}{100}\right)^2

$9. Depreciation (decrease in value):$

A = P\left(1 - \dfrac{R}{100}\right)^n

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Exercise 7.1 — Solved Examples

$800$

$Solution:$

\text{Discount} = 15\% \text{ of } 800 = \dfrac{15}{100} \times 800 = 120

\text{SP} = 800 - 120 = \text{Rs } 680

$---$

$12{,}000$

$Solution:$

\text{Profit} = 13500 - 12000 = \text{Rs } 1500

\text{Profit \%} = \dfrac{1500}{12000} \times 100 = 12.5\%

$---$

$5\%$

$Solution:$

$P$

P \times \dfrac{105}{100} = 315

P = \dfrac{315 \times 100}{105} = \text{Rs } 300

$---$

$1{,}500$

$Solution:$

$= 1500$

\text{Discount} = 10\% \text{ of } 2000 = 200

\text{SP} = 2000 - 200 = 1800

\text{Profit} = 1800 - 1500 = 300

\text{Profit \%} = \dfrac{300}{1500} \times 100 = 20\%

$Note: The discount is on the marked price, but profit percentage is on the cost price.$

$---$

$40{,}000$

$Solution:$

$10\%$

\text{Price}_1 = 40000 \times \dfrac{90}{100} = 36000

$5\%$

\text{SP} = 36000 \times \dfrac{95}{100} = 34200

$10\%$

$---$

$18{,}000$

$Solution:$

$= 18000 + 2000 = 20000$

\text{Profit} = 22000 - 20000 = 2000

\text{Profit \%} = \dfrac{2000}{20000} \times 100 = 10\%

Exercise 7.2 — Solved Examples (Compound Interest)

$10{,}000$

$Solution:$

A = P\left(1 + \dfrac{R}{100}\right)^n = 10000 \times \left(\dfrac{110}{100}\right)^2

= 10000 \times \dfrac{121}{100} = \text{Rs } 12{,}100

\text{CI} = 12100 - 10000 = \text{Rs } 2{,}100

$= \dfrac{10000 \times 10 \times 2}{100} = 2000$

$-$

$---$

$8{,}000$

$Solution:$

$= 5/2 = 2.5\%$

A = 8000 \times \left(1 + \dfrac{2.5}{100}\right)^2 = 8000 \times (1.025)^2

= 8000 \times 1.050625 = \text{Rs } 8{,}405

\text{CI} = 8405 - 8000 = \text{Rs } 405

$---$

$50{,}000$

$Solution:$

$Population growth uses the same formula as CI:$

\text{Population} = 50000 \times \left(1 + \dfrac{4}{100}\right)^2 = 50000 \times (1.04)^2

= 50000 \times 1.0816 = 54{,}080

$---$

$1{,}20{,}000$

$Solution:$

$For depreciation, use the decrease formula:$

A = P\left(1 - \dfrac{R}{100}\right)^n = 120000 \times \left(\dfrac{90}{100}\right)^3

= 120000 \times 0.729 = \text{Rs } 87{,}480

$32{,}520$

$---$

$5{,}000$

$Solution:$

A = P\left(1 + \dfrac{R}{100}\right)^n

5832 = 5000\left(1 + \dfrac{R}{100}\right)^2

\left(1 + \dfrac{R}{100}\right)^2 = \dfrac{5832}{5000} = \dfrac{729}{625}

1 + \dfrac{R}{100} = \sqrt{\dfrac{729}{625}} = \dfrac{27}{25}

\dfrac{R}{100} = \dfrac{27}{25} - 1 = \dfrac{2}{25}

R = \dfrac{2 \times 100}{25} = 8\%

$---$

$6{,}250$

$Solution:$

\left(1 + \dfrac{8}{100}\right)^n = \dfrac{7290}{6250} = \dfrac{729}{625}

\left(\dfrac{27}{25}\right)^n = \left(\dfrac{27}{25}\right)^2

$\dfrac{729}{625} = \left(\dfrac{27}{25}\right)^2$

n = 2 \text{ years}

Additional Worked Examples

$Here are more problems covering common exam and competitive-exam patterns.$

$10\%$

$Solution:$

$-$

50 = P \times \left(\dfrac{10}{100}\right)^2 = P \times \dfrac{1}{100}

P = 50 \times 100 = \text{Rs } 5{,}000

$---$

$4{,}840$

$Solution:$

$= 5324 - 4840 = 484$

$2$

R = \dfrac{484}{4840} \times 100 = 10\%

$Now find the principal:$

4840 = P\left(1 + \dfrac{10}{100}\right)^2 = P \times (1.1)^2 = 1.21P

P = \dfrac{4840}{1.21} = \text{Rs } 4{,}000

$---$

$25\%$

$Solution:$

$= 100$

$= 100 + 25 = 125$

$= 10\%$

$= 125 - 12.5 = 112.5$

$= 112.5 - 100 = 12.5$

$\% = \dfrac{12.5}{100} \times 100 = 12.5\%$

$---$

$5\%$

$Solution:$

$5\%$

P_1 = 200000 \times \dfrac{95}{100} = 190000

$10\%$

P_2 = 190000 \times \dfrac{110}{100} = 209000

$= 2{,}09{,}000$

$---$

$8\%$

$Solution:$

$= \dfrac{8}{4} = 2\%$

A = 10000 \times \left(1 + \dfrac{2}{100}\right)^4 = 10000 \times (1.02)^4

$(1.02)^2 = 1.0404$

A = 10000 \times 1.08243216 \approx \text{Rs } 10{,}824.32

\text{CI} \approx \text{Rs } 824.32

$= 10000 \times 1.08 - 10000 = 800$

Common Mistakes to Avoid

$Students frequently lose marks in this chapter due to these errors:$

$Mistake 1: Calculating discount on cost price instead of marked price. Discount is always a percentage of the marked price (MRP). The cost price is what the shopkeeper paid, and the customer does not know it.$

$20\%$

$= \dfrac{PRT}{100}$

$R/2$

$A = P(1 - R/100)^n$

Exam Tips for Comparing Quantities

$2$

Practice Questions with Answers

$Test your understanding with these additional questions.$

$2{,}400$

$= 2400 \times \dfrac{85}{100} = \text{Rs } 2{,}040$

$---$

$15{,}000$

$A = 15000 \times (1.12)^2 = 15000 \times 1.2544 = 18816$

$---$

$80{,}000$

$P = 80000 \times (0.95)^3 = 80000 \times 0.857375 = 68{,}590$

$---$

$25{,}000$

$= 25000 \times 0.80 = 20000$

$---$

$5\%$

$20 = P \times (5/100)^2 = P/400$

Key Takeaways

$A = P(1 + R/100)^n$

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