Compound Interest problems
- 1. If the interest of a certain sum for the first year at 8% p.a. compound interest is Rs. 48. What will be the interest for the second years?
Options- A. Rs. 52.55
- B. Rs. 51.84
- C. Rs. 58.60
- D. Rs. 53.04 Discuss
Correct Answer: Rs. 51.84
Explanation:
S.I. or C.I. for first year are always equal
Principle = (S.I. x 100)/(Rate x Time)
= (48 x 100)/(8 x 1) = Rs. 600
Interest for second year = Amount of second year - Amount of the first year
= 600 (1 +8/100)2 - 600(1 +8/100)
= 600(27/25)2 - 600(27/25)
= 600(27/25) {27/25 - 1}
= 600(27/25) (2/25)
= Rs. 51.84
- 2. If the compound interest of a certain sum of money for 2 years at 5 % p.a. is Rs. 328. what will be the simple interest on the same time and at the same rate?
Options- A. Rs. 320
- B. Rs. 322
- C. Rs. 325
- D. Rs. 326 Discuss
Correct Answer: Rs. 320
Explanation:
? 328 = Principal {(1 +5/100)2 - 1}
? 328 = Principal (441/400 - 1)
? 328 =Principal 41/400
? Principal = (328 x 400)/41 = Rs. 3200
? Simple interest = (3200 x 5 x 2) / 100 = Rs. 320
- 3. Find the least number of complete years in which a sum of money put out at 20 percent compound interest will be more than doubled?
Options- A. 2 years
- B. 3 years
- C. 4 years
- D. Data inadequate Discuss
Correct Answer: 4 years
Explanation:
Here P(1 + 20/10)t > 2P
? (6/5)t > 2
? (6/5)3 = 1.728, and
(6/5)4 = 2.0736
By trial (6/5) x (6/5) x (6/5) x (6/5) > 2
? The required time is 4 years.
- 4. A sum of money put out at compound interest amounts in 2 years to Rs. 578.40 and in 3 years to Rs. 614.55. Find the rate of interest?
Options- A. 6%
- B. 61/4%
- C. 61/2%
- D. 63/4% Discuss
Correct Answer: 61/4%
Explanation:
Clearly the difference between Rs. 578.40 and Rs.614.55 is the interest on Rs. 578.40 for 1 year.
? Interest on Rs. 578.40 for 1 year = Rs. 614.55 - Rs. 578.40 = Rs. 36.15
? Interest on Rs.100 for 1 year = Rs. (36.15 x 100)/578.40
= Rs. 3615/57840 x 100/1
= Rs. 6.25
= Rs. 61/4
? The required rate is .61/4 percent.
- 5. The difference between simple interest and compound interest on Rs. 600 for 1 years at 10% per annum,reckoned half yearly is?
Options- A. Nil
- B. Rs. 6.60
- C. Rs. 4.40
- D. Rs. 1.50 Discuss
Correct Answer: Rs. 1.50
Explanation:
S.I. = Rs. (600 x 5 x 2)/100 = Rs.60
C.I.= Rs. [600 x (1 + 5/100)2 - 600] = Rs. 61.50
? Requred Difference = Rs. (61.50 - 60) = Rs.1.50
- 6. The value of a machine depreciates every year at the rate of 10% on its value at the beginning of that years. If the present value of the machine is Rs. 729. Its worth 3 years ago was?
Options- A. Rs. 947.10
- B. Rs. 800
- C. Rs. 1000
- D. Rs. 750.87 Discuss
Correct Answer: Rs. 1000
Explanation:
? P (1 - 10/100)3 = 729
? P = Rs.(729 x 10 x 10 x 10)/(9 x 9 x 9) = Rs. 1000
- 7. A tree increases annually by 1/8 th of its height. By how much will it increase after 2 years, if it stands today 64 cm high?
Options- A. 72 cm
- B. 74 cm
- C. 75 cm
- D. 81 cm Discuss
Correct Answer: 81 cm
Explanation:
Increase% = (1/8) x 100% = 12.5%
Height after 2 years =64 x {1 +25/(2 x 100)}2
= 64 x 9/8 x 9/8
= 81 cm
- 8. A loan was repaid in two annual instalments of Rs. 121 each. if the rate of interest be 10% per annum. compounded annually, the sum borrowed was?
Options- A. Rs. 200
- B. Rs. 210
- C. Rs. 217.80
- D. Rs. 280 Discuss
Correct Answer: Rs. 210
Explanation:
Principal = (P.W. of Rs. 121 due 1 year later) + (P.W. of Rs. 121 due 2 years later)
= Rs. [ 121 /(1 + 10/100) + (121 / (1 + 10/100)2]
= Rs. 210
- 9. A sum of Rs. 12000 deposited at compound interest becomes doubles after 5 years. After 20 years it will become?
Options- A. 120000
- B. 192000
- C. 124000
- D. 96000 Discuss
Correct Answer: 192000
Explanation:
? 2P = P(1 +r/100)5
? (1 +r/100)5 = 2
? (1 + r/100)20 = 24 = 16
Thus, P(1 +r/100)20 = 16P
= Rs. (12000 x 16)
= Rs. 192000
- 10. A loan was repaid in two annual instalment of ? 1089 each. If the rate of interest be 10% per annum compounded annually, the sum borrowed was?
Options- A. ? 1840
- B. ? 1890
- C. ? 1850
- D. ? 1860 Discuss
Correct Answer: ? 1890
Explanation:
Principle = ? 1089
? P/{(1 + 10/100) + (1 + 10/100)2 } = 1089
? P = {(1089 x 10/11) + (1089 x 100/121)} = (990 + 900) = ? 1890
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