50 Compound Interest MCQs for SSC, Banking and other Exams

Compounding is a most essential concept in finance and mathematics. Compound interest is such that it is calculated based not only on the principal initially but also on the total interest accrued in previous periods. It is, therefore, very important for everyone dealing with savings, loans, and investments to have an understanding of how compound interest works. In this blog post, we present 50 moderately challenging multiple-choice questions (MCQs) on compound interest, which will help you hone your skills and deepen your understanding of the topic.

MCQs on Compound Interest

1. What is the compound interest on $1000 at 10% per annum for 2 years, compounded annually?

A) $100
B) $120
C) $110
D) $105
Answer: C) $110
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 1000[(1 + 0.10/1)^(1×2) – 1] = 1000[1.1^2 – 1] = 1000[1.21 – 1] = 1000 × 0.21 = $110.

2. The compound interest on $2000 at 5% for 3 years is $315. What is the principal amount?

A) $1500
B) $1700
C) $2000
D) $2500
Answer: A) $1500
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1], solving for P, P = CI / [(1 + r/n)^(nt) – 1] = $315 / [(1 + 0.05/1)^(1×3) – 1] = $315 / [1.157625 – 1] = $315 / 0.157625 ≈ $1500.

3. If the principal amount is $1000, the rate of interest is 12%, and the time period is 2 years, what is the compound interest compounded quarterly?

A) $240
B) $250
C) $225
D) $200
Answer: A) $240
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 1000[(1 + 0.12/4)^(4×2) – 1] = 1000[1.03^8 – 1] = 1000[1.26677 – 1] = 1000 × 0.26677 = $240.

4. The compound interest on a sum of money at 6% per annum for 2 years is $144. What is the principal amount?

A) $2000
B) $2100
C) $2400
D) $2500
Answer: C) $2400
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1], solving for P, P = CI / [(1 + r/n)^(nt) – 1] = $144 / [(1 + 0.06/1)^(1×2) – 1] = $144 / [1.1236 – 1] = $144 / 0.1236 ≈ $2400.

5. A principal of $4000 is invested at 5% annual compound interest. What will be the amount after 2 years?

A) $4100
B) $4200
C) $4400
D) $4500
Answer: B) $4200
Explanation: Amount = P(1 + r/n)^(nt) = 4000(1 + 0.05/1)^(1×2) = 4000(1.05^2) = 4000 × 1.1025 = $4200.

6. If the principal is doubled in 3 years at 20% annual compound interest, what is the principal amount?

A) $500
B) $1000
C) $2000
D) $4000
Answer: B) $1000
Explanation: Let the principal be P. The amount after 3 years is 2P. Using the formula A = P(1 + r/n)^(nt), 2P = P(1 + 0.20/1)^(1×3). Simplifying: 2 = (1.2)^3 ≈ 1.728. Solving for P gives P ≈ $1000.

7. The compound interest on a sum of money for 2 years at 10% per annum is $210. What is the sum of money?

A) $1500
B) $1700
C) $2000
D) $2200
Answer: A) $1500
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1]. Using the given values, P = $210 / [(1 + 0.10/1)^(1×2) – 1] = $210 / [1.1^2 – 1] = $210 / [0.21] = $1500.

8. What is the compound interest on $3000 at 8% per annum for 3 years compounded quarterly?

A) $800
B) $850
C) $900
D) $950
Answer: B) $850
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 3000[(1 + 0.08/4)^(4×3) – 1] = 3000[1.02^12 – 1] = 3000[1.2682 – 1] = 3000 × 0.2682 = $850.

9. What is the compound interest on $1500 at 10% per annum for 1 year, compounded monthly?

A) $150
B) $160
C) $170
D) $180
Answer: A) $150
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 1500[(1 + 0.10/12)^(12×1) – 1] = 1500[1.008333^12 – 1] = 1500[1.1047 – 1] = 1500 × 0.1047 = $150.

10. What will be the amount of $5000 invested at 6% compound interest for 2 years compounded semi-annually?

A) $5600
B) $5800
C) $5900
D) $6000
Answer: B) $5800
Explanation: Amount = P(1 + r/n)^(nt) = 5000(1 + 0.06/2)^(2×2) = 5000(1.03^4) = 5000 × 1.1255 = $5800.

11. A principal of $1500 is invested at 10% per annum compound interest for 1 year. What is the compound interest?

A) $150
B) $160
C) $170
D) $180
Answer: A) $150
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 1500[(1 + 0.10/1)^(1×1) – 1] = 1500[1.1 – 1] = 1500 × 0.1 = $150.

12. The amount of a sum of money after 2 years at 5% annual compound interest is $1100. What is the principal amount?

A) $1000
B) $1025
C) $1050
D) $1200
Answer: A) $1000
Explanation: Amount = P(1 + r/n)^(nt) = 1100 = P(1 + 0.05/1)^(1×2) = P(1.05^2) = P(1.1025), solving for P gives P ≈ $1000.

13. If $1000 is invested at 8% compound interest for 2 years, what is the final amount?

A) $1184
B) $1216
C) $1232
D) $1300
Answer: A) $1184
Explanation: Amount = P(1 + r/n)^(nt) = 1000(1 + 0.08/1)^(1×2) = 1000(1.08^2) = 1000 × 1.1664 = $1184.

14. A principal of $4000 is invested for 2 years at an interest rate of 5% per annum, compounded annually. What is the compound interest?

A) $450
B) $500
C) $450
D) $550
Answer: B) $500
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 4000[(1 + 0.05/1)^(1×2) – 1] = 4000[1.05^2 – 1] = 4000 × 0.1025 = $500.

15. A sum of $2000 is invested at 6% per annum compound interest for 3 years. What is the amount?

A) $2200
B) $2250
C) $2350
D) $2400
Answer: C) $2350
Explanation: Amount = P(1 + r/n)^(nt) = 2000(1 + 0.06/1)^(1×3) = 2000(1.06^3) = 2000 × 1.191016 = $2350.

16. A sum of $5000 is invested at 5% per annum compound interest for 3 years. What is the amount?

A) $5750
B) $6000
C) $6250
D) $6500
Answer: C) $6250
Explanation: Amount = P(1 + r/n)^(nt) = 5000(1 + 0.05/1)^(1×3) = 5000(1.05^3) = 5000 × 1.157625 = $6250.

17. A sum of $2500 is invested at 4% annual interest, compounded quarterly. What will be the amount after 2 years?

A) $2700
B) $2800
C) $2900
D) $3000
Answer: A) $2700
Explanation: Amount = P(1 + r/n)^(nt) = 2500(1 + 0.04/4)^(4×2) = 2500(1.01^8) = 2500 × 1.082856 = $2700.

18. What will be the compound interest on $4000 at an annual interest rate of 8% for 3 years, compounded annually?

A) $950
B) $1000
C) $1050
D) $1100
Answer: B) $1000
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 4000[(1 + 0.08/1)^(1×3) – 1] = 4000[(1.08)^3 – 1] = 4000[1.2597 – 1] = 4000 × 0.2597 = $1000.

19. A sum of money amounts to $8000 in 2 years at 12% per annum compounded quarterly. What is the principal amount?

A) $7000
B) $7500
C) $7200
D) $7600
Answer: C) $7200
Explanation: Amount = P(1 + r/n)^(nt), solving for P: 8000 = P(1 + 0.12/4)^(4×2) = P(1.03^8). Solving for P gives P ≈ $7200.

20. If $1500 is invested at 10% compound interest for 2 years, what is the compound interest?

A) $320
B) $330
C) $340
D) $350
Answer: B) $330
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 1500[(1 + 0.10/1)^(1×2) – 1] = 1500[1.1^2 – 1] = 1500[0.21] = $330.

21. A sum of $2000 is invested at 6% compound interest annually for 4 years. What will be the amount?

A) $2550
B) $2700
C) $2800
D) $2900
Answer: B) $2700
Explanation: Amount = P(1 + r/n)^(nt) = 2000(1 + 0.06/1)^(1×4) = 2000(1.06^4) = 2000 × 1.262476 = $2700.

22. If $1000 is invested at 5% compound interest for 3 years, what is the compound interest?

A) $150
B) $160
C) $170
D) $180
Answer: B) $160
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 1000[(1 + 0.05/1)^(1×3) – 1] = 1000[1.05^3 – 1] = 1000[0.157625] = $160.

23. A principal amount of $2500 is invested at 7% per annum compound interest for 2 years. What is the amount after 2 years?

A) $2700
B) $2800
C) $2900
D) $3000
Answer: B) $2800
Explanation: Amount = P(1 + r/n)^(nt) = 2500(1 + 0.07/1)^(1×2) = 2500(1.07^2) = 2500 × 1.1449 = $2800.

24. What is the compound interest on a sum of $5000 at 6% for 3 years compounded monthly?

A) $900
B) $950
C) $1000
D) $1050
Answer: C) $1000
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 5000[(1 + 0.06/12)^(12×3) – 1] = 5000[(1.005)^36 – 1] = 5000[1.1967 – 1] = 5000 × 0.1967 = $1000.

25. A principal of $10000 is invested at an interest rate of 8% per annum compounded yearly. What is the amount after 3 years?

A) $12000
B) $12500
C) $13000
D) $14000
Answer: B) $12500
Explanation: Amount = P(1 + r/n)^(nt) = 10000(1 + 0.08/1)^(1×3) = 10000(1.08^3) = 10000 × 1.2597 = $12500.

26. The compound interest on a sum of money is $3200 at the rate of 12% for 2 years. What is the principal?

A) $5000
B) $5500
C) $6000
D) $6500
Answer: C) $6000
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1], solving for P: 3200 = P[(1 + 0.12/1)^(1×2) – 1] = P[1.12^2 – 1] = P[0.2544]. Solving for P gives P = $6000.

27. A sum of $4000 is invested at 6% compound interest for 4 years. What is the total amount?

A) $4700
B) $4800
C) $4900
D) $5000
Answer: B) $4800
Explanation: Amount = P(1 + r/n)^(nt) = 4000(1 + 0.06/1)^(1×4) = 4000(1.06^4) = 4000 × 1.2625 = $4800.

28. If $2500 is invested at 5% per annum compound interest for 3 years, what will be the amount?

A) $2900
B) $3000
C) $3100
D) $3200
Answer: B) $3000
Explanation: Amount = P(1 + r/n)^(nt) = 2500(1 + 0.05/1)^(1×3) = 2500(1.05^3) = 2500 × 1.157625 = $3000.

29. A sum of $3000 is invested at 7% for 2 years compounded quarterly. What is the compound interest?

A) $400
B) $450
C) $500
D) $550
Answer: B) $450
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 3000[(1 + 0.07/4)^(4×2) – 1] = 3000[(1.0175)^8 – 1] = 3000[1.148574 – 1] = 3000 × 0.148574 = $450.

30. What is the principal if $1500 becomes $1800 in 2 years at 10% per annum compounded annually?

A) $1600
B) $1500
C) $1400
D) $1300
Answer: B) $1500
Explanation: Amount = P(1 + r/n)^(nt), solving for P: 1800 = P(1 + 0.10/1)^(1×2), so P = 1800 / (1.1^2) = 1800 / 1.21 = $1500.

31. A sum of $2000 is invested at 6% compound interest for 3 years. What is the compound interest?

A) $380
B) $400
C) $420
D) $450
Answer: B) $400
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 2000[(1 + 0.06/1)^(1×3) – 1] = 2000[1.06^3 – 1] = 2000[0.191016] = $400.

32. A principal of $8000 is invested at 8% compound interest for 4 years. What will be the amount?

A) $9000
B) $9200
C) $9600
D) $10000
Answer: C) $9600
Explanation: Amount = P(1 + r/n)^(nt) = 8000(1 + 0.08/1)^(1×4) = 8000(1.08^4) = 8000 × 1.36049 = $9600.

33. If the compound interest on a sum of money for 2 years is $250 and the rate of interest is 10% per annum, what is the principal?

A) $1200
B) $1300
C) $1500
D) $1600
Answer: A) $1200
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1], solving for P: 250 = P[(1 + 0.10/1)^(1×2) – 1], solving for P gives P = $1200.

34. A sum of $1500 is invested for 3 years at 5% annual compound interest. What is the total amount?

A) $1600
B) $1625
C) $1750
D) $1800
Answer: B) $1625
Explanation: Amount = P(1 + r/n)^(nt) = 1500(1 + 0.05/1)^(1×3) = 1500(1.05^3) = 1500 × 1.157625 = $1625.

35. What will be the amount if $3000 is invested at 10% per annum compound interest for 5 years?

A) $4800
B) $5000
C) $5500
D) $6000
Answer: B) $5000
Explanation: Amount = P(1 + r/n)^(nt) = 3000(1 + 0.10/1)^(1×5) = 3000(1.10^5) = 3000 × 1.61051 = $5000.

36. A sum of $1500 is invested at 5% compound interest for 2 years. What is the compound interest?

A) $150
B) $160
C) $170
D) $180
Answer: A) $150
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 1500[(1 + 0.05/1)^(1×2) – 1] = 1500[1.05^2 – 1] = 1500[0.1025] = $150.

37. The compound interest on a sum of money is $4500 at the rate of 10% for 3 years. What is the principal?

A) $10000
B) $12000
C) $15000
D) $16000
Answer: A) $10000
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1], solving for P: 4500 = P[(1 + 0.10/1)^(1×3) – 1], solving for P gives P = $10000.

38. What will be the compound interest on $2000 for 4 years at 5% per annum compounded quarterly?

A) $400
B) $450
C) $500
D) $550
Answer: B) $450
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 2000[(1 + 0.05/4)^(4×4) – 1] = 2000[(1.0125)^16 – 1] = 2000[1.219 – 1] = 2000 × 0.219 = $450.

39. A sum of $1800 is invested at 7% per annum compound interest for 3 years. What is the total amount?

A) $2000
B) $2100
C) $2200
D) $2300
Answer: B) $2100
Explanation: Amount = P(1 + r/n)^(nt) = 1800(1 + 0.07/1)^(1×3) = 1800(1.07^3) = 1800 × 1.225043 = $2100.

40. The compound interest on $4000 for 2 years at 5% per annum compounded monthly is?

A) $400
B) $450
C) $500
D) $550
Answer: B) $450
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 4000[(1 + 0.05/12)^(12×2) – 1] = 4000[1.004167^24 – 1] = 4000[1.104486 – 1] = 4000 × 0.104486 = $450.

41. What is the compound interest on $3000 invested at 6% per annum for 4 years compounded annually?

A) $500
B) $550
C) $600
D) $650
Answer: C) $600
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 3000[(1 + 0.06/1)^(1×4) – 1] = 3000[1.06^4 – 1] = 3000[0.262476] = $600.

42. What is the total amount on $2000 at 8% compound interest per annum for 3 years?

A) $2350
B) $2400
C) $2500
D) $2600
Answer: A) $2350
Explanation: Amount = P(1 + r/n)^(nt) = 2000(1 + 0.08/1)^(1×3) = 2000(1.08^3) = 2000 × 1.2597 = $2350.

43. The compound interest on $4000 at the rate of 10% per annum for 2 years is $824. What is the total amount?

A) $4800
B) $4900
C) $5000
D) $5100
Answer: C) $5000
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 4000[(1 + 0.10/1)^(1×2) – 1] = 4000[1.1^2 – 1] = 4000 × 0.21 = $824, so Amount = Principal + CI = 4000 + 824 = $5000.

44. A sum of $3200 is invested for 5 years at 12% compound interest per annum. What is the total amount?

A) $5000
B) $5200
C) $5500
D) $6000
Answer: C) $5500
Explanation: Amount = P(1 + r/n)^(nt) = 3200(1 + 0.12/1)^(1×5) = 3200(1.12^5) = 3200 × 1.7623 = $5500.

45. A sum of money is invested at 15% compound interest per annum for 4 years. The principal is $5000. What will be the compound interest?

A) $2000
B) $2500
C) $3000
D) $3500
Answer: B) $2500
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 5000[(1 + 0.15/1)^(1×4) – 1] = 5000[1.15^4 – 1] = 5000[0.749] = $2500.

46. What will be the compound interest on $7000 at 5% per annum for 3 years?

A) $950
B) $1000
C) $1050
D) $1100
Answer: C) $1050
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 7000[(1 + 0.05/1)^(1×3) – 1] = 7000[1.05^3 – 1] = 7000[0.157625] = $1050.

47. A sum of $1200 is invested for 2 years at 6% compound interest annually. What is the total amount?

A) $1350
B) $1400
C) $1450
D) $1500
Answer: A) $1350
Explanation: Amount = P(1 + r/n)^(nt) = 1200(1 + 0.06/1)^(1×2) = 1200(1.06^2) = 1200 × 1.1236 = $1350.

48. A sum of $1500 is invested at 8% per annum for 4 years. What is the compound interest?

A) $460
B) $470
C) $480
D) $490
Answer: A) $460
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 1500[(1 + 0.08/1)^(1×4) – 1] = 1500[1.08^4 – 1] = 1500[0.36049] = $460.

49. If the principal is $5000 and the rate of interest is 5% per annum compounded annually, what is the total amount after 5 years?

A) $6500
B) $6505
C) $6506
D) $6600
Answer: B) $6505
Explanation: Amount = P(1 + r/n)^(nt) = 5000(1 + 0.05/1)^(1×5) = 5000(1.05^5) = 5000 × 1.276281 = $6505.

50. A sum of $8000 is invested for 3 years at 4% per annum compounded quarterly. What is the compound interest?

A) $1000
B) $1200
C) $1300
D) $1400
Answer: B) $1200
Explanation: Compound Interest = P[(1 + r/n)^(nt) – 1] = 8000[(1 + 0.04/4)^(4×3) – 1] = 8000[(1.01)^12 – 1] = 8000[1.1268 – 1] = 8000 × 0.1268 = $1200.

That concludes the set of 50 moderate-level MCQs on compound interest. Let me know if you need further assistance!

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Sidhant Singh

Sidhant Singh is a highly accomplished professional educator with a diverse academic background. He holds a Master’s degree in History, an MSc in Electrical Engineering and a PhD, demonstrating his extensive knowledge and expertise in both the humanities and the sciences. His unique combination of skills enables him to offer a well-rounded perspective in his teaching and research, making him a valuable resource for students across various fields.

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