Practice MCQ Questions and Answer on Compound Interest

41.

A bank offers 5% compound interest calculated on half-yearly basis. A customer deposits Rs. 1600 each on 1^st January and 1^st July of a year. At the end of the year, the amount he would have gained by way of interest is:

(A) Rs. 120
(B) Rs. 121
(C) Rs. 122
(D) Rs. 123

Solution: $\begin{aligned} Amount \\ = 1600 \times {(1 + \frac{5}{2 \times 100})}^{2} + 1600 \times (1 + \frac{5}{2 \times 100}) \\ = 1600 \times \frac{41}{40} \times \frac{41}{40} + 1600 \times \frac{41}{40} \\ = 1600 \times \frac{41}{40} (\frac{41}{40} + 1) \\ = \frac{1600 \times 41 \times 81}{40 \times 40} \\ = R s . 3321 \\ ∴ C . I . = R s . (3321 - 3200) \\ = R s . 121 \end{aligned}$

42.

A loan is to be returned in two equal yearly instalments. If the rate of interest is 10% p.a., compounded annually, and each instalment if Rs. 5,808, then the total interest charged in this scheme is:

(A) Rs. 1,563
(B) Rs. 1,536
(C) Rs. 1,632
(D) Rs. 1,602

Solution: $\begin{aligned} 5808 \times \frac{10}{11} \times \frac{21}{11} \\ Principal = 10080 \\ Amount = 5808 \times 2 = 11616 \\ Interest = 1536 \end{aligned}$

43.

An amount of Rs 6000 lent at 5% per annum compounded interest for 2 years will become = ?

(A) Rs. 600
(B) Rs. 6600
(C) Rs. 6610
(D) Rs. 6615

Solution: $\begin{aligned} Amount = 6000 {(1 + \frac{5}{100})}^{2} \\ \Rightarrow Amount = 6000 \times \frac{21}{20} \times \frac{21}{20} \\ \Rightarrow Amount = Rs. 6615 \end{aligned}$

44.

Vipul and Manish invested the sum on Rs. 15000 and Rs. 20000 at the rate of 20 percent p.a and 30 percent p.a. respectively on compound interest (compounding annually). If time period is 3 years for both, then what will be the total compound interest earned by Vipul and Manish?

(A) Rs. 29460
(B) Rs. 32480
(C) Rs. 31688
(D) Rs. 34860

Solution: $\begin{aligned} 15000 [{(1 + \frac{20}{100})}^{3} - 1] + 20000 [{(1 + \frac{30}{100})}^{3} - 1] \\ = 15000 (\frac{216}{125} - 1) + 20000 (\frac{2197}{1000} - 1) \\ = 15000 \times \frac{91}{125} + 20000 \times \frac{1197}{1000} \\ = 10920 + 23940 \\ = Rs . 34860 \end{aligned}$

45.

If the compound interest on a certain sum for two years at 12% per annum is Rs. 2544, the simple interest on it at the same rate for 2 years will be = ?

(A) Rs. 2400
(B) Rs. 2500
(C) Rs. 2480
(D) Rs. 2440

Solution: $\begin{aligned} Rate = 12 % \\ Time = 2 years \\ Effective rate of CI for 2 years \\ = 12 + 12 + \frac{12 \times 12}{100} \\ = 25.44 % \\ Effective rate of SI for 2 years \\ = 12 \times 2 = 24 % \\ According to question, \\ Required SI \\ = \frac{2544}{25.44} \times 24 = Rs. 2400 \\ Required sum \\ = Rs. 2400 \end{aligned}$

46.

Simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years at 10% per annum. The sum placed on simple interest is:

(A) Rs. 1550
(B) Rs. 1650
(C) Rs. 1750
(D) Rs. 2000

Solution: $\begin{aligned} C .I . = R s . [4000 \times {(1 + \frac{10}{100})}^{2} - 4000] \\ = R s . (4000 \times \frac{11}{10} \times \frac{11}{10} - 4000) \\ = R s . 840 \\ ∴ Sum = R s . (\frac{420 \times 100}{3 \times 8}) \\ = R s . 1750 \end{aligned}$

47.

At what percent is the annual compound interest rate taken such that a certain sum amounts to its 16 times in 4 years?

(A) 175%
(B) 100%
(C) 200%
(D) 75%

Solution: $\Rightarrow Rate = \frac{1}{1} \times 100 = 100 Answer$

48.

Ramesh borrowed Rs. 1,20,000 interest free from Aman. If he pays back at 6.25% of this amount quarterly and has already paid Rs. 30,000, then for how many months does he need to pay back his loan for the rest of the amount?

(A) 36 months
(B) 24 months
(C) 42 months
(D) 38 months

Solution: Borrowed amount = 120000 First return amount = 30000 Remaining amount = 90000 Return in 3 month = 120000 × 6.25% = 120000 × $\frac{1}{16}$ = 12 × 25 × 25

49.

A sum of money becomes Rs. 11,880 after 4 years and Rs. 17,820 after 6 years on compound interest, if the interest is compounded annually. What is the half of the sum (in Rs.)?

(A) 2,410
(B) 2,640
(C) 2,530
(D) 2,750

Solution: $\begin{aligned} A = P {(1 + \frac{R}{100})}^{n} \\ 11880 = P {(1 + \frac{R}{100})}^{4} . . . . . (i) \\ 17820 = P {(1 + \frac{R}{100})}^{6} . . . . . (ii) \\ Equation (ii) divide by (i) \\ \frac{17820}{11880} = {(1 + \frac{R}{100})}^{2} \\ \frac{3}{2} = {(1 + \frac{R}{100})}^{2} . . . . . Put in equation (i) \\ 11880 = P {(1 + \frac{R}{100})}^{4} \\ 11880 = P {(1 + \frac{R}{100})}^{2} {(1 + \frac{R}{100})}^{2} \\ 11880 = P \times \frac{3}{2} \times \frac{3}{2} \\ P = 11880 \times \frac{2}{3} \times \frac{2}{3} \\ P = 5280 \\ Half part of sum = \frac{1}{2} \times 5280 = 2640 \end{aligned}$

50.

On a certain sum of money the compound interest for 2 years is Rs. 282.15 and the simple interest for the same period of time is Rs. 270. The rate of interest per annum is = ?

(A) 6.07%
(B) 10%
(C) 9%
(D) 12.15%

Solution: $\begin{aligned} CI for 2 years = Rs. 282 .15 \\ SI for 2 year = Rs. 270 \\ SI for 1 year = \frac{270}{2} \\ = Rs. 135 \\ Difference between CI and SI \\ = (282.15 - 270) \\ = Rs. 12 .15 \\ Required rate % \\ = \frac{12.15}{135} \times 100 = 9 % \end{aligned}$ Note : Always remainder for first year CI and SI will be same.

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Practice MCQ Questions and Answer on Compound Interest

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