Practice MCQ Questions and Answer on Compound Interest

81.

What will be the difference between S.I. and C.I. on a sum of Rs. 15000 for 2 years at the same rate of interest of $12 \frac{1}{2}$ % per annum ?

(A) Rs. 230.550
(B) Rs. 234.375
(C) Rs. 250.129
(D) Rs. 324.357

Solution: $\begin{aligned} S .I . = Rs . (15000 \times \frac{25}{2} \times 2 \times \frac{1}{100}) \\ = Rs . 3750 \\ C .I . = Rs . [15000 {(1 + \frac{25}{2 \times 100})}^{2} - 15000] \\ = Rs . (15000 \times \frac{9}{8} \times \frac{9}{8} - 15000) \\ = Rs . (18948.375 - 15000) \\ = Rs . 3984.375 \\ Difference \\ = Rs . (3984.375 - 3750) \\ = Rs . 234.375 \end{aligned}$

82.

Under the Rural Housing Scheme, the Delhi Development Authority (DDA) allotted a house to Kamal Raj for Rs. 126100. This payment is to be made in three equal annual instalments. If the money is reckoned at 5% per annum compound interest, then how much is to be paid by Kamal Raj in each instalment ?

(A) Rs. 45205
(B) Rs. 46305
(C) Rs. 47405
(D) Rs. 48505

Solution: Let the value of each instalment be Rs. x Then, (P.W. of Rs. x due 1 year hence) + (P.W. of Rs. x due 2 year hence) + (P.W. of Rs. x due 3 year hence) = 126100 $\begin{aligned} \Rightarrow \frac{x}{(1 + \frac{5}{100})} + \frac{x}{{(1 + \frac{5}{100})}^{2}} + \frac{x}{{(1 + \frac{5}{100})}^{3}} = 126100 \\ \Rightarrow \frac{20 x}{21} + \frac{400 x}{441} + \frac{8000 x}{9261} = 126100 \\ \Rightarrow \frac{8820 x + 8400 x + 8000 x}{9261} = 126100 \\ \Rightarrow \frac{25220 x}{9261} = 126100 \\ \Rightarrow x = (\frac{126100 \times 9261}{25220}) \\ \Rightarrow x = 46305 \end{aligned}$

83.

Mr. Duggal invested Rs. 20000 with rate of interest @ 20 p.c.p.a. The interest was compounded half - yearly for first one year ans in the next year it was compounded yearly. What will be the total interest earned at the end of 2 year ?

(A) Rs. 8040
(B) Rs. 8800
(C) Rs. 9040
(D) Rs. 9800

Solution: $\begin{aligned} Amount \\ = Rs . [20000 {(1 + \frac{10}{100})}^{2} (1 + \frac{20}{100})] \\ = Rs . (20000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{6}{5}) \\ = Rs .29040 \\ C .I . = Rs . (29040 - 20000) \\ = Rs . 9040 \end{aligned}$

84.

The difference between CI and SI on a certain sum of money for 3 years at 5% p.a. is Rs. 122. Find the sum invested = ?

(A) Rs. 10000
(B) Rs. 12000
(C) Rs. 16000
(D) Rs. 20000

Solution: $\begin{aligned} Rate of interest = 5 % p .c .p .a . \\ If time 3 years than CI - SI \\ = P [{(\frac{R}{100})}^{3} + 3 {(\frac{R}{100})}^{2}] \\ \Rightarrow 122 = P [{(\frac{5}{100})}^{3} + 3 {(\frac{5}{100})}^{2}] \\ \Rightarrow 122 = P (\frac{125}{1000000} + \frac{75}{10000}) \\ \Rightarrow 122 = P [\frac{125 + 7500}{1000000}] \\ \Rightarrow 122 = P [\frac{7525}{1000000}] \\ \Rightarrow P = \frac{122 \times 1000000}{7625} \\ \Rightarrow P = Rs . 16000 \end{aligned}$

85.

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

86.

If the simple interest on a sum of money for 2 years at 5% per annum is Rs. 50, what is the compound interest on the same at the same rate and for the same time?

(A) Rs. 51.25
(B) Rs. 52
(C) Rs. 54.25
(D) Rs. 60

Solution: $\begin{aligned} Sum = R s . (\frac{50 \times 100}{2 \times 5}) \\ = R s . 500 \\ Amount = R s . [500 \times {(1 + \frac{5}{100})}^{2}] \\ = R s . (500 \times \frac{21}{20} \times \frac{21}{20}) \\ = R s . 551.25 \\ ∴ C .I . = R s . (551.25 - 500) \\ = R s . 51.25 \end{aligned}$

87.

A sum of Rs. 18,000 is lent at 10% p.a. compound interest, compounded annually. What is the difference between the compound interest for 3^rd year and 4^th year?

(A) Rs. 215.40
(B) Rs. 217.80
(C) Rs. 220.60
(D) Rs. 221.80

88.

The compound interest accrued on an amount of Rs. 25500 at the end of 3 years is Rs. 8440.50. What would be the simple interest accrued on the same amount at the same rate in the same period ?

(A) Rs. 4650
(B) Rs. 5650
(C) Rs. 6650
(D) Rs. 7650

Solution: $\begin{aligned} Let the rate be R % p .a . \\ then, \\ 25500 {(1 + \frac{R}{100})}^{3} \\ = 25500 + 8440.50 \\ = 33940.50 \end{aligned}$ $\Rightarrow {(1 + \frac{R}{100})}^{3} = \frac{33940.50}{25500} =$ $\frac{1331}{1000} =$ ${(\frac{11}{10})}^{3}$ $\begin{aligned} \Rightarrow 1 + \frac{R}{100} = \frac{11}{10} \\ \Rightarrow \frac{R}{100} = \frac{1}{10} \\ \Rightarrow R = 10 % \\ S . I . = R s . (\frac{25500 \times 10 \times 3}{100}) \\ = Rs . 7650 \end{aligned}$

89.

The compound interest on Rs. 4,000 after 3 year is Rs. 630.50. Then the rate of interest compounded yearly is:

(A) 7%
(B) 5%
(C) 8%
(D) 6%

Solution: $\begin{aligned} 4630.50 = 4000 {(1 + \frac{R}{100})}^{3} \\ 9261 = 8000 {(1 + \frac{R}{100})}^{3} \\ \frac{9261}{8000} = {(1 + \frac{R}{100})}^{3} \\ {(\frac{21}{20})}^{3} = {(1 + \frac{R}{100})}^{3} \\ \frac{21}{20} = (1 + \frac{R}{100}) \\ \frac{1}{20} = \frac{R}{100} \\ R = 5 \end{aligned}$

90.

In how many years will a sum of Rs. 320 amount to Rs. 405 if interest is compounded at 12.5% per annum?

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

Solution: $\begin{matrix} 320 & : & 405 \\ \sqrt{64} & : & \sqrt{81} \\ 8 & : & 9 \end{matrix}$ $\begin{aligned} 12 \frac{1}{2} % \to \frac{1}{8} \\ In one year = 8 : 9 \\ Time = 2 years \end{aligned}$

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Practice MCQ Questions and Answer on Compound Interest

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