Compound interest means interest on interest. It is a favourite in banking and aptitude exams.
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Q1. The compound interest on a certain sum for 3 years at 40% per annum, compounded annually, is Rs 4360. Find the sum.
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Answer: D
Let Principal = P, Rate R = 40% per annum, Time T = 3 years, compounded annually. Amount = P(1+R/100)^T = P x (7/5)^T. So C.I. = Amount – P = P x [(7/5)^3 – 1]. For T=3, (7/5)^3 – 1 = 218/125. Given C.I. = Rs 4360, so P x 218/125 = 4360. So P = 4360 x 125/218 = Rs 2500. Check: Amount = P + C.I. = 2500 + 4360 = Rs 6860, matching P x (7/5)^3.
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Q2. The compound interest on a certain sum for 3 years at 50% per annum, compounded annually, is Rs 228. Find the sum.
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Answer: C
Let Principal = P, Rate R = 50% per annum, Time T = 3 years, compounded annually. Amount = P(1+R/100)^T = P x (3/2)^T. So C.I. = Amount – P = P x [(3/2)^3 – 1]. For T=3, (3/2)^3 – 1 = 19/8. Given C.I. = Rs 228, so P x 19/8 = 228. So P = 228 x 8/19 = Rs 96. Check: Amount = P + C.I. = 96 + 228 = Rs 324, matching P x (3/2)^3.
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Q3. The compound interest on a certain sum for 2 years at 40% per annum, compounded annually, is Rs 24. Find the sum.
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Answer: B
Let Principal = P, Rate R = 40% per annum, Time T = 2 years, compounded annually. Amount = P(1+R/100)^T = P x (7/5)^T. So C.I. = Amount – P = P x [(7/5)^2 – 1]. For T=2, (7/5)^2 – 1 = 24/25. Given C.I. = Rs 24, so P x 24/25 = 24. So P = 24 x 25/24 = Rs 25. Check: Amount = P + C.I. = 25 + 24 = Rs 49, matching P x (7/5)^2.
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Q4. The compound interest on a certain sum for 2 years at 10% per annum, compounded annually, is Rs 420. Find the sum.
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Answer: C
Let Principal = P, Rate R = 10% per annum, Time T = 2 years, compounded annually. Amount = P(1+R/100)^T = P x (11/10)^T. So C.I. = Amount – P = P x [(11/10)^2 – 1]. For T=2, (11/10)^2 – 1 = 21/100. Given C.I. = Rs 420, so P x 21/100 = 420. So P = 420 x 100/21 = Rs 2000. Check: Amount = P + C.I. = 2000 + 420 = Rs 2420, matching P x (11/10)^2.
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Q5. The compound interest on a certain sum for 3 years at 12% per annum, compounded annually, is Rs 63270. Find the sum.
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Answer: A
Let Principal = P, Rate R = 12% per annum, Time T = 3 years, compounded annually. Amount = P(1+R/100)^T = P x (28/25)^T. So C.I. = Amount – P = P x [(28/25)^3 – 1]. For T=3, (28/25)^3 – 1 = 6327/15625. Given C.I. = Rs 63270, so P x 6327/15625 = 63270. So P = 63270 x 15625/6327 = Rs 156250. Check: Amount = P + C.I. = 156250 + 63270 = Rs 219520, matching P x (28/25)^3.
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Q6. The compound interest on a certain sum for 2 years at 12% per annum, compounded annually, is Rs 2226. Find the sum.
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Answer: D
Let Principal = P, Rate R = 12% per annum, Time T = 2 years, compounded annually. Amount = P(1+R/100)^T = P x (28/25)^T. So C.I. = Amount – P = P x [(28/25)^2 – 1]. For T=2, (28/25)^2 – 1 = 159/625. Given C.I. = Rs 2226, so P x 159/625 = 2226. So P = 2226 x 625/159 = Rs 8750. Check: Amount = P + C.I. = 8750 + 2226 = Rs 10976, matching P x (28/25)^2.
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Q7. The compound interest on a certain sum for 2 years at 40% per annum, compounded annually, is Rs 144. Find the sum.
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Answer: B
Let Principal = P, Rate R = 40% per annum, Time T = 2 years, compounded annually. Amount = P(1+R/100)^T = P x (7/5)^T. So C.I. = Amount – P = P x [(7/5)^2 – 1]. For T=2, (7/5)^2 – 1 = 24/25. Given C.I. = Rs 144, so P x 24/25 = 144. So P = 144 x 25/24 = Rs 150. Check: Amount = P + C.I. = 150 + 144 = Rs 294, matching P x (7/5)^2.
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Q8. The compound interest on a certain sum for 2 years at 16% per annum, compounded annually, is Rs 2160. Find the sum.
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Answer: A
Let Principal = P, Rate R = 16% per annum, Time T = 2 years, compounded annually. Amount = P(1+R/100)^T = P x (29/25)^T. So C.I. = Amount – P = P x [(29/25)^2 – 1]. For T=2, (29/25)^2 – 1 = 216/625. Given C.I. = Rs 2160, so P x 216/625 = 2160. So P = 2160 x 625/216 = Rs 6250. Check: Amount = P + C.I. = 6250 + 2160 = Rs 8410, matching P x (29/25)^2.
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Q9. The compound interest on a certain sum for 2 years at 15% per annum, compounded annually, is Rs 129. Find the sum.
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Answer: B
Let Principal = P, Rate R = 15% per annum, Time T = 2 years, compounded annually. Amount = P(1+R/100)^T = P x (23/20)^T. So C.I. = Amount – P = P x [(23/20)^2 – 1]. For T=2, (23/20)^2 – 1 = 129/400. Given C.I. = Rs 129, so P x 129/400 = 129. So P = 129 x 400/129 = Rs 400. Check: Amount = P + C.I. = 400 + 129 = Rs 529, matching P x (23/20)^2.
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Q10. A sum of Rs 1200, invested at compound interest, amounts to Rs 1323 in 2 years. Find the rate of interest per annum.
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Answer: D
Let Principal P = Rs 1200, Amount after 2 years = Rs 1323. Using Amount = P(1+R/100)^2, (1+R/100)^2 = Amount/P = 1323/1200 = 441/400. Taking the square root, 1+R/100 = 21/20. So R/100 = 21/20 – 1 = 1/20. So R = 100/20 = 5%. Check: P(1+R/100)^2 = 1200 x (21/20)^2 = Rs 1323, matching the given amount.
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Q11. A sum of Rs 6250, invested at compound interest, amounts to Rs 6760 in 2 years. Find the rate of interest per annum.
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Answer: C
Let Principal P = Rs 6250, Amount after 2 years = Rs 6760. Using Amount = P(1+R/100)^2, (1+R/100)^2 = Amount/P = 6760/6250 = 676/625. Taking the square root, 1+R/100 = 26/25. So R/100 = 26/25 – 1 = 1/25. So R = 100/25 = 4%. Check: P(1+R/100)^2 = 6250 x (26/25)^2 = Rs 6760, matching the given amount.
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Q12. A sum of Rs 8750, invested at compound interest, amounts to Rs 10206 in 2 years. Find the rate of interest per annum.
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Answer: C
Let Principal P = Rs 8750, Amount after 2 years = Rs 10206. Using Amount = P(1+R/100)^2, (1+R/100)^2 = Amount/P = 10206/8750 = 729/625. Taking the square root, 1+R/100 = 27/25. So R/100 = 27/25 – 1 = 2/25. So R = 200/25 = 8%. Check: P(1+R/100)^2 = 8750 x (27/25)^2 = Rs 10206, matching the given amount.
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Q13. A sum of Rs 1250, invested at compound interest, amounts to Rs 1458 in 2 years. Find the rate of interest per annum.
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Answer: D
Let Principal P = Rs 1250, Amount after 2 years = Rs 1458. Using Amount = P(1+R/100)^2, (1+R/100)^2 = Amount/P = 1458/1250 = 729/625. Taking the square root, 1+R/100 = 27/25. So R/100 = 27/25 – 1 = 2/25. So R = 200/25 = 8%. Check: P(1+R/100)^2 = 1250 x (27/25)^2 = Rs 1458, matching the given amount.
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Q14. A sum of Rs 2500, invested at compound interest, amounts to Rs 3364 in 2 years. Find the rate of interest per annum.
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Answer: A
Let Principal P = Rs 2500, Amount after 2 years = Rs 3364. Using Amount = P(1+R/100)^2, (1+R/100)^2 = Amount/P = 3364/2500 = 841/625. Taking the square root, 1+R/100 = 29/25. So R/100 = 29/25 – 1 = 4/25. So R = 400/25 = 16%. Check: P(1+R/100)^2 = 2500 x (29/25)^2 = Rs 3364, matching the given amount.
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Q15. A sum of Rs 80, invested at compound interest, amounts to Rs 125 in 2 years. Find the rate of interest per annum.
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Answer: B
Let Principal P = Rs 80, Amount after 2 years = Rs 125. Using Amount = P(1+R/100)^2, (1+R/100)^2 = Amount/P = 125/80 = 25/16. Taking the square root, 1+R/100 = 5/4. So R/100 = 5/4 – 1 = 1/4. So R = 100/4 = 25%. Check: P(1+R/100)^2 = 80 x (5/4)^2 = Rs 125, matching the given amount.
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Q16. A sum of Rs 6800, invested at compound interest, amounts to Rs 7497 in 2 years. Find the rate of interest per annum.
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Answer: D
Let Principal P = Rs 6800, Amount after 2 years = Rs 7497. Using Amount = P(1+R/100)^2, (1+R/100)^2 = Amount/P = 7497/6800 = 441/400. Taking the square root, 1+R/100 = 21/20. So R/100 = 21/20 – 1 = 1/20. So R = 100/20 = 5%. Check: P(1+R/100)^2 = 6800 x (21/20)^2 = Rs 7497, matching the given amount.
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Q17. A sum of Rs 240, invested at compound interest, amounts to Rs 375 in 2 years. Find the rate of interest per annum.
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Answer: A
Let Principal P = Rs 240, Amount after 2 years = Rs 375. Using Amount = P(1+R/100)^2, (1+R/100)^2 = Amount/P = 375/240 = 25/16. Taking the square root, 1+R/100 = 5/4. So R/100 = 5/4 – 1 = 1/4. So R = 100/4 = 25%. Check: P(1+R/100)^2 = 240 x (5/4)^2 = Rs 375, matching the given amount.
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Q18. A sum of Rs 3200, invested at compound interest, amounts to Rs 4232 in 2 years. Find the rate of interest per annum.
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Answer: C
Let Principal P = Rs 3200, Amount after 2 years = Rs 4232. Using Amount = P(1+R/100)^2, (1+R/100)^2 = Amount/P = 4232/3200 = 529/400. Taking the square root, 1+R/100 = 23/20. So R/100 = 23/20 – 1 = 3/20. So R = 300/20 = 15%. Check: P(1+R/100)^2 = 3200 x (23/20)^2 = Rs 4232, matching the given amount.
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Q19. A sum of Rs 272, invested at compound interest, amounts to Rs 425 in 2 years. Find the rate of interest per annum.
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Answer: B
Let Principal P = Rs 272, Amount after 2 years = Rs 425. Using Amount = P(1+R/100)^2, (1+R/100)^2 = Amount/P = 425/272 = 25/16. Taking the square root, 1+R/100 = 5/4. So R/100 = 5/4 – 1 = 1/4. So R = 100/4 = 25%. Check: P(1+R/100)^2 = 272 x (5/4)^2 = Rs 425, matching the given amount.
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Q20. The difference between the compound interest and the simple interest on a certain sum for 3 years at 20% per annum is Rs 30720. Find the sum.
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Answer: B
Let the sum be P, Rate R = 20% per annum, Time = 3 years. For 3 years, the standard result is C.I. – S.I. = P x (R/100)^2 x (3 + R/100). Given C.I. – S.I. = Rs 30720, so 30720 = P x (20/100)^2 x (320/100). So P = (30720 x 1000000) / (400 x 320) = Rs 240000. Check: P x (R/100)^2 x (3+R/100) = 240000 x (20/100)^2 x (320/100) = Rs 30720, matching the given difference.
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Q21. The difference between the compound interest and the simple interest on a certain sum for 3 years at 40% per annum is Rs 161024. Find the sum.
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Answer: D
Let the sum be P, Rate R = 40% per annum, Time = 3 years. For 3 years, the standard result is C.I. – S.I. = P x (R/100)^2 x (3 + R/100). Given C.I. – S.I. = Rs 161024, so 161024 = P x (40/100)^2 x (340/100). So P = (161024 x 1000000) / (1600 x 340) = Rs 296000. Check: P x (R/100)^2 x (3+R/100) = 296000 x (40/100)^2 x (340/100) = Rs 161024, matching the given difference.
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Q22. The difference between the compound interest and the simple interest on a certain sum for 3 years at 50% per annum is Rs 273000. Find the sum.
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Answer: C
Let the sum be P, Rate R = 50% per annum, Time = 3 years. For 3 years, the standard result is C.I. – S.I. = P x (R/100)^2 x (3 + R/100). Given C.I. – S.I. = Rs 273000, so 273000 = P x (50/100)^2 x (350/100). So P = (273000 x 1000000) / (2500 x 350) = Rs 312000. Check: P x (R/100)^2 x (3+R/100) = 312000 x (50/100)^2 x (350/100) = Rs 273000, matching the given difference.
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Q23. The difference between the compound interest and the simple interest on a certain sum for 3 years at 15% per annum is Rs 7371. Find the sum.
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Answer: A
Let the sum be P, Rate R = 15% per annum, Time = 3 years. For 3 years, the standard result is C.I. – S.I. = P x (R/100)^2 x (3 + R/100). Given C.I. – S.I. = Rs 7371, so 7371 = P x (15/100)^2 x (315/100). So P = (7371 x 1000000) / (225 x 315) = Rs 104000. Check: P x (R/100)^2 x (3+R/100) = 104000 x (15/100)^2 x (315/100) = Rs 7371, matching the given difference.
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Q24. The difference between the compound interest and the simple interest on a certain sum for 3 years at 20% per annum is Rs 17408. Find the sum.
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Answer: D
Let the sum be P, Rate R = 20% per annum, Time = 3 years. For 3 years, the standard result is C.I. – S.I. = P x (R/100)^2 x (3 + R/100). Given C.I. – S.I. = Rs 17408, so 17408 = P x (20/100)^2 x (320/100). So P = (17408 x 1000000) / (400 x 320) = Rs 136000. Check: P x (R/100)^2 x (3+R/100) = 136000 x (20/100)^2 x (320/100) = Rs 17408, matching the given difference.
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Q25. The difference between the compound interest and the simple interest on a certain sum for 3 years at 10% per annum is Rs 7440. Find the sum.
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Answer: A
Let the sum be P, Rate R = 10% per annum, Time = 3 years. For 3 years, the standard result is C.I. – S.I. = P x (R/100)^2 x (3 + R/100). Given C.I. – S.I. = Rs 7440, so 7440 = P x (10/100)^2 x (310/100). So P = (7440 x 1000000) / (100 x 310) = Rs 240000. Check: P x (R/100)^2 x (3+R/100) = 240000 x (10/100)^2 x (310/100) = Rs 7440, matching the given difference.
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