Compound Interest Questions with Answers (Class 8 - 10)

The topic of compound interest in mathematics and daily life is very important. Compound interest means that you earn interest not only on the money you first invested (called the principal), but also on the interest already added. In simple words, in compound interest, your money increases rapidly because you gain "interest on interest".

For example, if you deposit ₹100 in the bank with a 10% interest rate, after one year, you earn ₹10 interest and have ₹110. In the second year, the interest is calculated on ₹110 (not just 100), so you get ₹11. In this way, the amount increases.

This page covers compound interest questions with step-by-step solutions for Class 8, 9 and 10 students. You will find solved examples level-wise (easy, medium and advanced), a formula chart and practice questions. By the end of this page, you will be able to solve any compound interest problem confidently.

Table of Contents

• Compound Interest Definition
• Compound Interest Formula Chart
• Solved Compound Interest Questions
• Practice Questions on Compound Interest
• Conclusion
• Frequently Asked Questions on Compound Interest Questions
  
  

Compound Interest Definition

Compound interest is the interest calculated not just on the original principal but also on the interest that has already been added. This is why it is often called "interest on interest." Because of this, money grows much faster under compound interest compared to simple interest.

For example, if you deposit ₹1,000 in a bank at 10% interest per year:

• After Year 1: Interest = ₹100, Total = ₹1,100
• After Year 2: Interest is calculated on ₹1,100 (not ₹1,000), so Interest = ₹110, Total = ₹1,210

As you can see, the interest amount increases every year because the previous interest is added to the principal.

The formula for Compound Interest is:

A = P × (1 + R/100)ⁿ

CI = A − P

Where:

• A = Final Amount
• P = Principal (original money)
• R = Rate of Interest (% per annum)
• n = Time (in years)
• CI = Compound Interest

Compound Interest Formula Chart

Type	Formula	Trick
Annual Compounding	A = P (1 + R/100)¹	Rate and time as-is
Half-Yearly Compounding	A = P (1 + R/200)²¹	Rate ÷ 2, Time × 2
Quarterly Compounding	A = P (1 + R/400)&sup4;¹	Rate ÷ 4, Time × 4
Monthly Compounding	A = P (1 + R/1200)¹²¹	Rate ÷ 12, Time × 12
Depreciation	A = P (1 − R/100)¹	Minus sign instead of plus
Population Growth	A = P (1 + R/100)¹	Same as annual, P = population
Different Rates Each Year	A = P × (1 + R&sub1;/100) × (1 + R&sub2;/100) × (1 + R&sub3;/100)	Multiply each year separately
Finding Principal	P = A / (1 + R/100)¹	Rearrange A formula
Finding Rate	R = [(A/P)^1/n − 1] × 100	Rearrange A formula
Finding Time	(1 + R/100)¹ = A/P	Solve by trial method or logarithm
CI from Amount	CI = A − P	Always subtract principal

Where:

• A = Final Amount
• P = Principal
• R = Rate of Interest (% per annum)
• n = Time (in years)
• CI = Compound Interest
  
  

Example Questions for Each Condition

1. Annual Compounding

Formula: A = P (1 + R/100)ⁿ, CI = A − P

Example: Find CI on ₹10,000 at 10% p.a. for 2 years.

A = 10000 × (1.10)² = 10000 × 1.21 = ₹11,000

CI = 11000 − 10000 = ₹1,000

2. Half-Yearly Compounding

Formula: A = P (1 + R/200)²ⁿ

Example: Find CI on ₹8,000 at 10% p.a. for 1 year, compounded half-yearly.

R = 10/2 = 5%, n = 1 × 2 = 2
A = 8000 × (1.05)² = 8000 × 1.1025 = ₹8,820
CI = 8820 − 8000 = ₹820

3. Quarterly Compounding

Formula: A = P (1 + R/400)⁴ⁿ

Example: Find CI on ₹12,000 at 8% p.a. for 1 year, compounded quarterly.

R = 8/4 = 2%, n = 1 × 4 = 4
A = 12000 × (1.02)⁴ = 12000 × 1.08243 = ₹12,989.16
CI = 12989.16 − 12000 = ₹989.16

4. Monthly Compounding

Formula: A = P (1 + R/1200)¹²ⁿ

Example: Find CI on ₹5,000 at 12% p.a. for 1 year, compounded monthly.

R = 12/12 = 1%, n = 1 × 12 = 12
A = 5000 × (1.01)¹² = 5000 × 1.12683 = ₹5,634.15
CI = 5634.15 − 5000 = ₹634.15

5. Depreciation

Formula: A = P (1 − R/100)ⁿ

Example: A bike costs ₹80,000 and depreciates at 10% per year. Find its value after 2 years.

A = 80000 × (0.90)² = 80000 × 0.81 = ₹64,800
Loss = 80000 − 64800 = ₹15,200

6. Population Growth

Formula: A = P (1 + R/100)ⁿ

Example: A city has 2,00,000 people and grows at 3% per year. Find the population after 2 years.

A = 200000 × (1.03)² = 200000 × 1.0609 = 2,12,180

7. Different Rates in Different Years

Formula: A = P × (1 + R₁/100) × (1 + R₂/100) × (1 + R₃/100)

Example: ₹10,000 is invested at 8% in year 1, 10% in year 2 and 12% in year 3. Find the amount.

Year 1: 10000 × 1.08 = ₹10,800
Year 2: 10800 × 1.10 = ₹11,880
Year 3: 11880 × 1.12 = ₹13,305.60
CI = 13305.60 − 10000 = ₹3,305.60

8. Finding Principal (P)

Formula: P = A / (1 + R/100)ⁿ

Example: The amount after 2 years at 10% p.a. CI is ₹12,100. Find the principal.

P = 12100 / (1.10)² = 12100 / 1.21 = ₹10,000

9. Finding Rate (R)

Formula: R = [(A/P)^(1/n) − 1] × 100

Example: ₹5,000 becomes ₹5,832 in 2 years. Find the rate.

(1 + R/100)² = 5832/5000 = 1.1664
1 + R/100 = √1.1664 = 1.08
R = 0.08 × 100 = 8% p.a.

10. Finding Time (n)

Formula: Solve (1 + R/100)ⁿ = A/P by trial method or logarithm

Example: In how many years will ₹6,000 become ₹7,986 at 10% p.a. CI?

(1.10)ⁿ = 7986/6000 = 1.331
(1.10)³ = 1.331
n = 3 years

Read more:

• Simple Interest Questions
• BODMAS Rule Questions
• Profit and Loss Questions
• Percentage Questions

Solved Compound Interest Questions

Easy Level

Question 1: Find the compound interest on ₹5,000 at 10% p.a. for 2 years, compounded annually.

Solution: P = ₹5,000, R = 10%, T = 2 years

A = 5000 × (1.10)² = 5000 × 1.21 = ₹6,050

CI = 6050 − 5000 = ₹1,050

Question 2: What is the amount on ₹8,000 at 5% p.a. for 2 years, compounded annually?

Solution: P = ₹8,000, R = 5%, T = 2 years

A = 8000 × (1.05)² = 8000 × 1.1025 = ₹8,820

CI = 8820 − 8000 = ₹820

Question 3: Find CI on ₹12,000 at 8% p.a. for 1 year, compounded annually.

Solution: P = ₹12,000, R = 8%, T = 1 year

A = 12000 × 1.08 = ₹12,960

CI = 12960 − 12000 = ₹960

Question 4: A principal of ₹6,000 earns CI at 10% p.a. for 3 years. Find the amount.

Solution: P = ₹6,000, R = 10%, T = 3 years

A = 6000 × (1.10)³ = 6000 × 1.331 = ₹7,986

CI = 7986 − 6000 = ₹1,986

Question 5: Find the difference between SI and CI on ₹4,000 at 10% p.a. for 2 years.

Solution: SI = (4000 × 10 × 2) / 100 = ₹800

A = 4000 × (1.10)² = 4000 × 1.21 = ₹4,840
CI = 4840 − 4000 = ₹840

Difference = 840 − 800 = ₹40

Medium Level

Question 6: Find the CI on ₹16,000 at 10% p.a. for 1.5 years, compounded half-yearly.

Solution: R = 10/2 = 5%, n = 1.5 × 2 = 3

A = 16000 × (1.05)³ = 16000 × 1.157625 = ₹18,522

CI = 18522 − 16000 = ₹2,522

Question 7: Find the amount on ₹20,000 at 8% p.a. for 1 year, compounded quarterly.

Solution: R = 8/4 = 2%, n = 1 × 4 = 4

A = 20000 × (1.02)⁴ = 20000 × 1.08243 = ₹21,648.60

CI = 21648.60 − 20000 = ₹1,648.60

Question 8: A sum becomes ₹9,261 in 3 years at 5% p.a. CI. Find the principal.

Solution: 9261 = P × (1.05)³ = P × 1.157625

P = 9261 / 1.157625 = ₹8,000

Question 9: At what rate will ₹5,000 amount to ₹5,832 in 2 years, compounded annually?

Solution: (1 + R/100)² = 5832/5000 = 1.1664
1 + R/100 = 1.08
R = 8% p.a.

Question 10: In how many years will ₹3,000 become ₹3,993 at 10% p.a. CI?

Solution: (1.10)ⁿ = 3993/3000 = 1.331 = (1.10)³
n = 3 years

Advanced Level

Question 11: A car worth ₹4,00,000 depreciates at 10% per year. Find its value after 3 years.

Solution: A = 400000 × (0.90)³ = 400000 × 0.729 = ₹2,91,600

Loss = 400000 − 291600 = ₹1,08,400

Question 12: A town's population is 80,000 and grows at 5% per year. What will it be after 2 years?

Solution: A = 80000 × (1.05)² = 80000 × 1.1025 = 88,200

Question 13: ₹15,000 is invested for 3 years at 10% in year 1, 12% in year 2 and 15% in year 3. Find the amount and CI.

Solution:
Year 1: 15000 × 1.10 = ₹16,500
Year 2: 16500 × 1.12 = ₹18,480
Year 3: 18480 × 1.15 = ₹21,252
CI = 21252 − 15000 = ₹6,252

Question 14: The CI on a sum for 2 years is ₹832 and SI is ₹800. Find the rate and principal.

Solution: SI for 1 year = ₹400
Difference = 832 − 800 = ₹32
32 = 400 × R/100 → R = 8%
800 = P × 8 × 2 / 100 → P = ₹5,000

Question 15: A machine costs ₹50,000 and depreciates at 8% p.a. After how many years will its value fall below ₹40,000?

Solution:
After 1 year: 50000 × 0.92 = ₹46,000
After 2 years: 46000 × 0.92 = ₹42,320
After 3 years: 42320 × 0.92 = ₹38,934.40
After 3 years the value falls below ₹40,000.

Practice Questions on Compound Interest

1. Find the compound interest on ₹6,000 at 10% p.a. for 2 years, compounded annually.

2. What is the amount on ₹15,000 at 8% p.a. for 3 years, compounded annually?

3. Find the CI on ₹10,000 at 12% p.a. for 1 year, compounded half-yearly.

4. A sum of ₹25,000 is deposited at 6% p.a. CI for 2 years. Find the amount and CI.

5. The population of a town is 1,20,000. It grows at 5% per year. What will the population be after 2 years?

6. A laptop costs ₹60,000 and depreciates at 10% per year. Find its value after 2 years.

7. At what rate will ₹8,000 amount to ₹9,261 in 3 years, compounded annually?

8. In how many years will ₹5,000 become ₹6,655 at 10% p.a. CI?

9. Find the difference between SI and CI on ₹10,000 at 10% p.a. for 3 years.

10. ₹40,000 is invested at 5% p.a. CI for 2 years, compounded half-yearly. Find the amount.

Answers:

1. CI = ₹1,260
2. A = ₹18,895.68
3. CI = ₹1,236
4. A = ₹28,090, CI = ₹3,090
5. 1,32,300
6. ₹48,600
7. R = 5%
8. 3 years
9. Difference = ₹310
10. A = ₹44,152.51
    
    

Conclusion

Compound interest means earning interest not only on the original money but also on the interest already added. This is why money grows faster in compound interest. We learnt the formulas, solved examples, and saw how it is used in banks, loans, savings, and investments. By practising the sum, the topic becomes simple and very useful in real life.

Frequently Asked Questions on Compound Interest Questions

Q1. What is the difference between simple interest and compound interest?

Answer: In simple interest, interest is calculated only on the principal every year. In compound interest, interest is calculated on the principal plus the interest already added. This makes compound interest grow faster than simple interest over time.

Q2. What is the compound interest on ₹8,000 at 5% per annum for 2 years?

Answer: A = 8000 × (1.05)² = 8000 × 1.1025 = ₹8,820
CI = 8820 − 8000 = ₹820

Q3. What is the compound interest on ₹10,000 at 8% per annum for 2 years?

Answer: A = 10000 × (1.08)² = 10000 × 1.1664 = ₹11,664
CI = 11664 − 10000 = ₹1,664

Q4. What will be the compound interest on ₹25,000 after 3 years at 12% per annum?

Answer: A = 25000 × (1.12)³ = 25000 × 1.404928 = ₹35,123.20
CI = 35123.20 − 25000 = ₹10,123.20

Q5. Is compound interest always more than simple interest?

Answer: Yes, for the same principal, rate and time period of more than 1 year, compound interest is always greater than simple interest. The difference increases as the time period gets longer.

Q6. What happens when compounding is done half-yearly instead of annually?

Answer: When compounding is half-yearly, the rate is divided by 2 and the time is multiplied by 2. This results in slightly more interest compared to annual compounding because interest is calculated more frequently.

Q7. Where is compound interest used in real life?

Answer: Compound interest is used in bank fixed deposits, recurring deposits, home loans, car loans, credit card interest, mutual fund returns and provident fund calculations.