Practice MCQ Questions and Answer on Compound Interest

71.

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

72.

A certain sum of money yields Rs. 1261 as compound interest for three years at 5% per annum. The sum is ?

(A) Rs. 9000
(B) Rs. 8400
(C) Rs. 7500
(D) Rs. 8000

Solution: Let the principal be x Rs. Now, $\begin{aligned} C . I . = P [{(1 + \frac{R}{100})}^{T} - 1] \\ \Rightarrow 1261 = x [{(1 + \frac{5}{100})}^{3} - 1] \\ \Rightarrow 1261 = x (\frac{9261}{8000} - 1) \\ \Rightarrow 1261 = x (\frac{9261 - 8000}{8000}) \\ \Rightarrow 1261 = \frac{1261 x}{8000} \\ \Rightarrow x = \frac{1261 \times 8000}{1261} \\ = Rs . 8000 \end{aligned}$

73.

The compound interest on Rs. 5000 for 3 years at 10% p.a. will amount to = ?

(A) Rs. 1654
(B) Rs. 1655
(C) Rs. 1600
(D) Rs. 1565

Solution: $\begin{aligned} Principal = Rs. 5000 \\ Time = 3 years \\ Rate = 10 % = \frac{1}{10} \\ Principal Amount \\ 10 11 \\ 10 11 \\ 10 11 \\ \underset{331 units}{\underset{⏟}{\overset{―}{1000 1331}}} \\ \Rightarrow 1000 units = Rs 5000 \\ \Rightarrow 1 units = Rs 5 \\ \Rightarrow 331 units = 331 \times 5 \\ = Rs. 1655 \end{aligned}$

74.

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

75.

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

76.

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

77.

What is the difference (in Rs.) between the interests on Rs. 50,000 for one year at 8% per annum compounded half yearly and yearly?

(A) 80
(B) 50
(C) 70
(D) 100

Solution: $\begin{aligned} 1 year CI = 1 year SI \\ SI = \frac{P \times R \times T}{100} \\ = \frac{50000 \times 8 \times 1}{100} \\ = 4000 \\ Compound interest at half yearly \\ Rate = \frac{8}{2} = 4 % \\ Time = 1 \times 2 = 2 years \\ A = P {(1 + \frac{R}{100})}^{n} \\ A = 50000 {(1 + \frac{4}{100})}^{2} \\ A = 50000 \times \frac{104}{100} \times \frac{104}{100} \\ A = 5 \times 10816 \\ A = 54080 \\ CI = 54080 - 50000 = 4080 \\ Difference between the interest of on year \\ = 4080 - 4000 \\ = 80 \end{aligned}$

78.

A sum amount to Rs. 7,562 in 4 years and to Rs. 8,469.44 in 5 years at a certain rate percent per annum, when the interest is compounded yearly. The rate of interest is:

(A) 8%
(B) 12%
(C) 20%
(D) 15%

Solution: $\begin{aligned} 7562________8469.44 \\ Rate of interest = \frac{907.44}{7562} \times 100 = 12 % \end{aligned}$

79.

In how many years will Rs. 2000 amounts to Rs. 2420 at 10% per annum compound interest?

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

Solution: $\begin{aligned} Principal = Rs. 2000 \\ Amount = Rs. 2420 \\ Rate = 10% \\ By using formula, \\ \Rightarrow 2420 = 2000 {(1 + \frac{10}{100})}^{n} \\ \Rightarrow \frac{2420}{2000} = {(1 + \frac{10}{100})}^{n} \\ \Rightarrow \frac{121}{100} = {(\frac{11}{10})}^{n} \\ \Rightarrow {(\frac{11}{10})}^{2} = {(\frac{11}{10})}^{n} \\ \Rightarrow n = 2 \\ Hence, \\ required time = 2 years \end{aligned}$ Alternative Note : In such type of questions to save your valuable time follow the given below method. Principal : Amount Ratio → 2000 : 2420 100 : 121 $Rate = 10% = \frac{1}{10}$ Principal : Amount 1st year → 10 : 11 2nd year → 10 : 11 Ratio → 100 : 121 Note : Now after 2nd year both the principal and amount will be in the same ratio. Hence required time = 2 years

80.

What will be the compound interest accrued on an amount of Rs.10000 @ 20 p.c.p.a in 2 years if the interest is compounded half - yearly?

(A) Rs. 4400
(B) Rs. 4600
(C) Rs. 4641
(D) Rs. 4680

Solution: $\begin{aligned} P = Rs .10000, \\ R = 20 % p . a . \\ = 10 % per half year \\ T = 2 years = 4 half years \\ Amount \\ = Rs . [10000 \times {(1 + \frac{10}{100})}^{4}] \\ = Rs . (10000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}) \\ = Rs .14641 \\ ∴ C .I . = Rs . (14641 - 10000) \\ = Rs . 4641 \end{aligned}$

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Practice MCQ Questions and Answer on Compound Interest

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