Practice MCQ Questions and Answer on Compound Interest

101.

The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is:

(A) 3
(B) 4
(C) 5
(D) 6

Solution: $\begin{aligned} P {(1 + \frac{20}{100})}^{n} > 2 P \Rightarrow {(\frac{6}{5})}^{n} > 2 \\ Now, (\frac{6}{5} \times \frac{6}{5} \times \frac{6}{5} \times \frac{6}{5}) > 2 \\ S o, n = 4 years \end{aligned}$

102.

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

103.

A sum becomes Rs. 2916 in 2 years at 8% per annum compound interest. The simple interest at 9% per annum for 3 years on the same amount will be = ?

(A) Rs. 600
(B) Rs. 675
(C) Rs. 650
(D) Rs. 625

Solution: $\begin{aligned} Amount = Rs . 2916 \\ Time = 2 years \\ Rate = 8 % \\ Effective rate % CI for 2 years \\ = 8 + 8 + \frac{8 \times 8}{100} = 16.64 % \\ Required sum \\ = \frac{2916}{(100 + 16.64)} \times 100 \\ = Rs . 2500 \\ Required simple interest \\ = \frac{2500 \times 9 \times 3}{100} \\ = Rs . 675 \end{aligned}$

104.

The compound interest on a certain sum of money at 5% per annum for 2 years is Rs 246. The simple interest on the same sum for 3 years at 6% per annum is = ?

(A) Rs. 435
(B) Rs. 450
(C) Rs. 430
(D) Rs. 432

Solution: $\begin{aligned} Effective rate of CI for 2 years \\ = 5 + 5 + \frac{5 \times 5}{100} \\ = 10.25 % \\ Effective rate of SI for 3 years \\ = 6 \times 3 = 18% \\ According to question \\ Required SI \\ = \frac{246}{10.25} \times 18 \\ = Rs. 432 \end{aligned}$

105.

The compound interest on a certain sum of money for 2 years at 10% per annum is Rs. 525.The simple interest on the same sum of money for double the time at half the rate percent per annum is ?

(A) Rs. 1000
(B) Rs. 500
(C) Rs. 200
(D) Rs. 800

Solution: Let the sum of money be rs. P Then, $\begin{aligned} \Rightarrow [P {(1 + \frac{R}{100})}^{t} - P] = C .I . \\ \Rightarrow [P {(1 + \frac{10}{100})}^{2} - P] = 525 \\ \Rightarrow P {(\frac{11}{10})}^{2} - 1 = 525 \\ \Rightarrow P (\frac{121}{100} - 1) = 525 \\ \Rightarrow P (\frac{21}{100}) = 525 \\ \Rightarrow P = \frac{525 \times 100}{21} \\ \Rightarrow P = Rs . 2500 \\ ∴ Sum of money \\ = Rs . 2500 \end{aligned}$ Simple interest on the same sum Rs. 2500 for 4 (double the time) years at 5% (half the rate of percent per annum) is So, $\begin{aligned} S .I . = Rs . (\frac{2500 \times 5 \times 4}{100}) \\ = Rs . 500 \end{aligned}$

106.

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

107.

If the difference between the compound interest and simple interest on a certain sum at the rate of 5% per annum for 2 years is Rs. 20, then the sum is = ?

(A) Rs. 2000
(B) Rs. 4000
(C) Rs. 6000
(D) Rs. 8000

Solution: Let the sum be = 400x Simple Interest = $\frac{P T R}{100}$ S.I = $\frac{400 x \times 2 \times 5}{100}$ = 40x for Compond Interest 5% = $\frac{1}{20}$ C.I = 20x + 20x + x = 41x Difference = 41x - 40x = 20 i.e. x = 20 ∴ Sum is = 400 × 20 = 8000

108.

A man gets a simple interest on Rs. 1000 on a certain principal at the rate of 5 p.c.p.a. in 4 years. What compound interest will the man get on twice the principal in 2 years at the same rate ?

(A) Rs. 1000
(B) Rs. 1005
(C) Rs. 10125
(D) Rs. 11025

Solution: Answer & Solution Answer: Option E Solution: $\begin{aligned} Principal \\ = Rs . (\frac{100 \times 1000}{5 \times 4}) \\ = Rs . 5 000 \\ Now, P = Rs . 10000, \\ T = 2 years, \\ R = 5% \\ Amount \\ = Rs . [10000 \times {(1 + \frac{5}{100})}^{2}] \\ = Rs . (10000 \times \frac{21}{20} \times \frac{21}{20}) \\ = Rs . 11025 \\ ∴ C .I . = (11025 - 10000) \\ = Rs . 1025 \end{aligned}$

109.

The difference between compound interest and simple interest on a certain sum of money for 2 years at 5% per annum is Rs. 41. What is the sum of money ?

(A) Rs. 7200
(B) Rs. 9600
(C) Rs. 16400
(D) Rs. 8400

Solution: $\begin{aligned} \Rightarrow C .I . - S .I . = 41 \\ \Rightarrow C .I . - S .I . = P {(\frac{r}{100})}^{2} \\ \Rightarrow 41 = P (\frac{25}{10000}) \\ \Rightarrow P = 16400 \end{aligned}$

110.

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

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Practice MCQ Questions and Answer on Compound Interest

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