Practice MCQ Questions and Answer on Compound Interest

71.

A man invests Rs 4000 for 3 years at compound interest. After one year the money amounts to Rs. 4320. What will be the amount (to the nearest rupee) due at the end of 3 years ?

(A) Rs. 4939
(B) Rs. 5039
(C) Rs. 5789
(D) Rs. 6129

Solution: $\begin{aligned} Le the rate be R % p .a . \\ Then, \\ 4000 (1 + \frac{R}{100}) = 4320 \\ \Rightarrow 1 + \frac{R}{100} = \frac{4320}{4000} = \frac{108}{100} \\ \Rightarrow \frac{R}{100} = \frac{8}{100} \\ \Rightarrow R = 8 \\ ∴ Amount after 3 yeras \\ = Rs . [4000 + {(1 + \frac{8}{100})}^{3}] \\ = Rs . (4000 \times \frac{27}{25} \times \frac{27}{25} \times \frac{27}{25}) \\ = Rs . (\frac{629856}{125}) \\ = Rs . 5038.848 \approx 5039 \end{aligned}$

72.

A sum of Rs. 15,000 is lent at 16% p.a. compound interest. What is the difference between the compound interest for the second year and the third year?

(A) Rs. 544
(B) Rs. 445.44
(C) Rs. 445.88
(D) Rs. 548

73.

A sum amounts to Rs. 18,600 after 3 years and to Rs. 27,900 after 6 years, at a certain rate percent p.a, when the interest is compounded annually. The sum is:

(A) Rs. 11,800
(B) Rs. 12,400
(C) Rs. 14,400
(D) Rs. 14,600

Solution: $\begin{aligned} P \times \frac{27, 900}{18, 600} = 18, 600 \\ P = \frac{18, 600 \times 186}{279} \\ P = 12, 400 \end{aligned}$

74.

A sum of money becomes eight times in 3 years, If the rate is compounded annually. In how much time will the same amount at the same compound rate become sixteen times ?

(A) 6 years
(B) 4 years
(C) 8 years
(D) 5 years

Solution: $\begin{aligned} Let principal = P \\ C a s e (I) \\ Time = 3 years, \\ Amount = 8P \\ \Rightarrow 8 P = P {(1 + \frac{R}{100})}^{3} \\ \Rightarrow {(2)}^{3} = {(1 + \frac{R}{100})}^{3} \\ Taking cube root of both sides, \\ \Rightarrow 2 = (1 + \frac{R}{100}) \\ \Rightarrow R = 100 % \\ C a s e (I I) \\ Let after t years it will be 16 times \\ \Rightarrow 16 P = P {(1 + \frac{R}{100})}^{t} \\ \Rightarrow 16 = {(2)}^{t} \\ \Rightarrow {(2)}^{4} = {(2)}^{t} \\ \Rightarrow t = 4 \\ Hence required time \\ (t) = 4 years \end{aligned}$

75.

At what rate percent per annum of compound interest, will a sum of money become four times of itself in two years ?

(A) 100%
(B) 75%
(C) 50%
(D) 20%

Solution: $\begin{aligned} Principal Amount \\ 1 \to 4 \\ \Rightarrow 4 = 1 {(1 + \frac{r}{100})}^{2} \\ \Rightarrow 4 = {(1 + \frac{r}{100})}^{2} \\ \Rightarrow r = 100 % \\ Alternate \\ Principal Amount \\ \sqrt[2]{1} \to \sqrt[2]{4} \\ 1 \to 2 \\ \Rightarrow Rate of interest \\ = \frac{(2 - 1)}{1} \times 100 = 100 % \end{aligned}$

76.

A sum of money placed at compound interest double itself in 4 years. In how many years will it amount to four times itself ?

(A) 12 years
(B) 13 years
(C) 8 years
(D) 16 years

Solution: $\begin{aligned} Let, \\ Principal = R s . 100 % \\ Amount = R s . 200 \\ Rate = r % \\ Time = 4 years \\ Now, \\ A = P \times {[1 + (\frac{r}{100})]}^{n} \\ 200 = 100 \times {[1 + (\frac{r}{100})]}^{4} \\ 2 = {[1 + (\frac{r}{100})]}^{4} - - - - - - (i) \\ If sum become 8 times in the time n years \\ then, \\ 4 = {(1 + (\frac{r}{100}))}^{n} \\ 2^{2} = {(1 + (\frac{r}{100}))}^{n} - - - - - - (i i) \\ Using eqn (i) i n (i i), we get \\ {({[1 + (\frac{r}{100})]}^{4})}^{2} = {(1 + (\frac{r}{100}))}^{n} \\ {[1 + (\frac{r}{100})]}^{8} = {(1 + (\frac{r}{100}))}^{n} \\ Thus, n = 8 years . \end{aligned}$

77.

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

78.

Shashi had a certain amount of money. He invested $\frac{2}{3}$ of the total money in scheme A for 6 years and rest of the money he invested in scheme B for 2 years. Scheme A offers simple interest at a rate of 12% p.a. and scheme B offers compound interest ( compound annually) at a rate of 10% p.a. If the total interest obtained from both the schemes is Rs. 2750. What was the total amount invested by him in scheme A and scheme B together ? (Approximate value)

(A) Rs. 4500
(B) Rs. 4200
(C) Rs. 4050
(D) Rs. 5000

Solution: Let the total sum of money invested by Shashi be Rs. x In scheme A money invested at simple interest for 6 years at a rate of 12% p.a. $∴ \frac{2}{3} of x \times \frac{12 \times 6}{100} = \frac{48 x}{100} . . . . (i)$ In scheme B money at compound interest for 2 year at a rate of 10% p.a. $\begin{aligned} \frac{x}{3} {(1 + \frac{10}{100})}^{2} - \frac{x}{3} \\ \Rightarrow \frac{x}{3} {(1 + \frac{10}{100})}^{2} - \frac{x}{3} = \frac{7 x}{100} \end{aligned}$ According to given information we get $\begin{aligned} \Rightarrow \frac{48 x}{100} + \frac{7 x}{100} = 2750 \\ \Rightarrow 55 x = 2750 \times 100 \\ \Rightarrow x = \frac{2750 \times 100}{55} \\ \Rightarrow x = R s . 5000 \end{aligned}$

79.

A certain sum amounts to Rs. 4,205.55 at 15% p.a. in $2 \frac{2}{5}$ ÃÂÃÂ years, interest compounded yearly. The sum is:

(A) Rs. 2,700
(B) Rs. 3,000
(C) Rs. 3,500
(D) Rs. 3,200

80.

Rs.2000 amounts to Rs. 2226.05 in 2 years at compound interest. What will be the rate of interest ?

(A) 5%
(B) 5.25%
(C) 5.5%
(D) 6%

Solution: $\begin{aligned} Let the rate be R % p .a . \\ Then, \\ 2000 {(1 + \frac{R}{100})}^{2} = 2226.05 \\ \Rightarrow {(1 + \frac{R}{100})}^{2} = \frac{222605}{200000} \\ \Rightarrow {(1 + \frac{R}{100})}^{2} = \frac{44521}{40000} \\ \Rightarrow {(1 + \frac{R}{100})}^{2} = {(\frac{221}{200})}^{2} \\ \Rightarrow 1 + \frac{R}{100} = \frac{211}{200} \\ \Rightarrow \frac{R}{100} = \frac{11}{200} \\ \Rightarrow R = \frac{11}{2} % \\ \Rightarrow R = 5.5 % \end{aligned}$

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Practice MCQ Questions and Answer on Compound Interest

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