Practice MCQ Questions and Answer on Compound Interest

51.

The compound interest on a sum of Rs. 20,000 at 15% p.a., for $2 \frac{2}{3}$ ÃÂÃÂ years, interest compounded yearly is:

(A) Rs. 9,098
(B) Rs. 8.896
(C) Rs. 9,000
(D) Rs. 9,095

52.

A sum becomes Rs.1352 in 2 years at 4% per annum compound interest. The sum is = ?

(A) Rs. 1225
(B) Rs. 1270
(C) Rs. 1245
(D) Rs. 1250

Solution: Let the sum be Rs. x $\begin{aligned} ∴ 1352 = x {(1 + \frac{4}{100})}^{2} \\ \Rightarrow 1352 = x {(1 + \frac{1}{25})}^{2} \\ \Rightarrow 1352 = x {(\frac{26}{25})}^{2} \\ \Rightarrow x = \frac{1352 \times 25 \times 25}{26 \times 26} \\ \Rightarrow x = Rs . 1250 \end{aligned}$

53.

A loan has to be returned in two equal yearly instalments each of Rs. 44,100. If the rate of interest is 5% p.a., compounded annually, then the total interest paid is:

(A) Rs. 5,840
(B) Rs. 6,000
(C) Rs. 6,200
(D) Rs. 6,280

Solution: 5% = $\frac{1}{20}$ $\begin{matrix} Principal & Installment \\ I^{st} year \to & 20_{\times 21} & : & 21_{\times 21} \\ I I^{nd} year \to & 400 & : & 441 & \to 44100 \end{matrix}$ 1 unit → 100 Principal = 400 + 420 = 820 Total installment = 441 × 2 = 882 Interest = 882 - 820 = 62 × 100 [1 unit = 100] = 6200

54.

The compound interest on Rs. 30000 at 7% per annum is Rs. 4347. The period (in years) is = ?

(A) 2 years
(B) $2 \frac{1}{2}$ years
(C) 3 years
(D) 4 years

Solution: $\begin{aligned} Amount = Rs . (30000 - 4347) \\ Amount = Rs . 34347 \\ Let the time be n years \\ Then, \\ 30000 {(1 + \frac{7}{100})}^{n} = 34347 \\ \Leftrightarrow {(\frac{107}{100})}^{n} = \frac{34347}{30000} \\ \Leftrightarrow {(\frac{107}{100})}^{n} = \frac{11449}{10000} = {(\frac{107}{100})}^{2} \\ ∴ n = 2 years \end{aligned}$

55.

If the rate of interest be 4% per annum for first year, 5% per annum foe second year and 6% per annum for third year, then the compound interest of Rs.10000 for three years will be ?

(A) Rs. 1575.20
(B) Rs. 1600
(C) Rs. 1625.80
(D) Rs. 2000

Solution: $\begin{aligned} = Rs .10000 [(1 + \frac{4}{100}) (1 + \frac{5}{100}) (1 + \frac{6}{100})] \\ = Rs . (10000 \times \frac{26}{25} \times \frac{21}{20} \times \frac{53}{50}) \\ = Rs . (\frac{57876}{5}) = Rs .11575 .20 \\ C .I . = Rs . (11575.20 - 10000) \\ = Rs .1575 .20 \end{aligned}$

56.

A sum of Rs. 12000 deposited at compound interest become double after 5 years. After 20 years it will become ?

(A) Rs. 96000
(B) Rs. 120000
(C) Rs. 124000
(D) Rs. 192000

Solution: $\begin{aligned} 12000 \times {(1 + \frac{R}{100})}^{5} = 24000 \\ \Rightarrow {(1 + \frac{R}{100})}^{5} = 2 \\ ∴ {[{(1 + \frac{R}{100})}^{5}]}^{4} = 2^{4} = 16 \\ \Rightarrow {(1 + \frac{R}{100})}^{20} = 16 \\ \Rightarrow P {(1 + \frac{R}{100})}^{20} = 16P \\ \Rightarrow 12000 {(1 + \frac{R}{100})}^{20} = 16 \times 12000 \\ \Rightarrow 12000 {(1 + \frac{R}{100})}^{20} = 192000 \end{aligned}$

57.

The compound interest on Rs. 30000 at 7% per annum for a certain time is Rs. 4347. The times is = ?

(A) 3 years
(B) 4 years
(C) 2 years
(D) 2.5 years

Solution: $\begin{aligned} Principal = Rs . 30000 \\ CI = Rs 4347 \\ Rate = 7 % \\ By using formula, \\ \Rightarrow (30000 + 4347) = 30000 {(1 + \frac{7}{100})}^{t} \\ \Rightarrow 34347 = 30000 {(1 + \frac{7}{100})}^{t} \\ \Rightarrow \frac{34347}{30000} = {(\frac{107}{100})}^{t} \\ \Rightarrow (\frac{11449}{10000}) = {(\frac{107}{100})}^{t} \\ \Rightarrow {(\frac{107}{100})}^{2} = {(\frac{107}{100})}^{t} \\ \Rightarrow t = 2 years \end{aligned}$

58.

A sum of Rs. 8000 will amount to Rs. 8820 in 2 years if the interest is calculated every year. The rate of compound interest is = ?

(A) 6%
(B) 7%
(C) 3%
(D) 5%

Solution: $\begin{aligned} Principal = Rs 8000 \\ Amount = Rs 8820 \\ Let Rate = R \\ Time = 2 years \\ By using formula, \\ \Rightarrow 8820 = 8000 {(1 + \frac{R}{100})}^{2} \\ \Rightarrow \frac{8820}{8000} = {(1 + \frac{R}{100})}^{2} \\ \Rightarrow \frac{441}{400} = {(1 + \frac{R}{100})}^{2} \\ Taking square root of both sides, \\ \Rightarrow \frac{21}{20} = (1 + \frac{R}{100}) \\ \Rightarrow R = 5 % \end{aligned}$

59.

The compound interest on Rs. 30,000 at 7% per annum is Rs. 4347. The period (in years) is:

(A) 2
(B) $2 \frac{1}{2}$
(C) 3
(D) 4

Solution: $\begin{aligned} Amount = R s . (30000 + 4347) \\ = R s . 34347 \\ Let the time be n years \\ Then, 30000 {(1 + \frac{7}{100})}^{n} = 34347 \\ \Rightarrow {(\frac{107}{100})}^{n} = \frac{34347}{30000} = \frac{11449}{10000} = {(\frac{107}{100})}^{2} \\ ∴ n = 2 years \end{aligned}$

60.

The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Rs. 1. The sum (in Rs.) is:

(A) 625
(B) 630
(C) 640
(D) 650

Solution: $\begin{aligned} Let the sum be R s . x . Then, \\ C .I . = x {(1 + \frac{4}{100})}^{2} - x \\ = \frac{676}{625} x - x \\ = \frac{51}{625} x \\ S .I . = \frac{x \times 4 \times 2}{100} \\ = \frac{2 x}{25} \\ ∴ \frac{51 x}{625} - \frac{2 x}{25} = 1 \\ \Rightarrow x = 625 \end{aligned}$

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Practice MCQ Questions and Answer on Compound Interest

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