Interest - Aptitude MCQ Questions and Solutions with Explanations

51.

Tushar borrowed a sum of Rs. 12000 at 15% per annum from a money - lender on 13^th January, 1987 and return the amount on 8^th June, 1987 to clear his debt. Then the amount paid by Tushar to the money - lender to clear his debt was = ?

(A) Rs. 15000
(B) Rs. 11394
(C) Rs. 12720
(D) Rs. 13650

Solution: Time = 18 + 28 + 31 + 30 + 31 + 8 = 146 days $\begin{aligned} Simple Interest \\ = \frac{12000 \times 15 \times 146}{365 \times 100} \\ = Rs . 720 \\ ∴ Amount will be \\ = Rs . (12000 + 720) \\ = Rs . 12720 \end{aligned}$

52.

A sum of Rs. 1550 was lent partly at 5% and partly at 8% p.a. simple interest. The total interest received after 3 years was Rs. 300. The ratio of the money lent at 5% to that lent at 8% is:

(A) 5 : 8
(B) 8 : 5
(C) 16 : 15
(D) 31 : 6

Solution: Let the sum lent at 5% be Rs. x and that lent 8% be Rs. (1550 - x). Then, $\begin{aligned} (\frac{x \times 5 \times 3}{100}) + [\frac{(1550 - x) \times 8 \times 3}{100}] = 300 \\ \Leftrightarrow 15 x - 24 x + (1550 \times 24) = 30000 \\ \Leftrightarrow 9 x = 7200 \\ \Leftrightarrow x = 800. \\ ∴ Required ratio = 800 : 750 \\ = 16 : 15 \end{aligned}$

53.

Rs. 1200 is lent out at 5% per annum simple interest for 3 years. Find the amount after 3 years.

(A) Rs. 1380
(B) Rs. 1290
(C) Rs. 1470
(D) Rs.1200

Solution: $\begin{aligned} A = P + I \\ A = 1200 + \frac{P T R}{100} \\ A = 1200 + \frac{1200 \times 5 \times 3}{100} \\ Amount, A = R s . 1380 \end{aligned}$

54.

Tushar borrowed a sum of Rs. 12000 at 15% per annum from a money - lender on 13^th January, 1987 and return the amount on 8^th June, 1987 to clear his debt. Then the amount paid by Tushar to the money - lender to clear his debt was = ?

(A) Rs. 15000
(B) Rs. 11394
(C) Rs. 12720
(D) Rs. 13650

Solution: Time = 18 + 28 + 31 + 30 + 31 + 8 = 146 days $\begin{aligned} Simple Interest \\ = \frac{12000 \times 15 \times 146}{365 \times 100} \\ = Rs . 720 \\ ∴ Amount will be \\ = Rs . (12000 + 720) \\ = Rs . 12720 \end{aligned}$

55.

How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?

(A) 3.5 years
(B) 4 years
(C) 4.5 years
(D) 5 years

Solution: $\begin{aligned} Time = (\frac{100 \times 81}{450 \times 4.5}) years \\ = 4 years \end{aligned}$

56.

A sum of Rs. 1600 gives a simple interest of Rs. 252 in 2 years and 3 months. The rate of interest per annum is = ?

(A) $5 \frac{1}{2}$ %
(B) 8 %
(C) 7 %
(D) 6 %

Solution: $\begin{aligned} Time = 2 years 3 months \\ = 2 + \frac{3}{12} = \frac{9}{2} years \\ We know SI = \frac{P \times T \times R}{100} \\ P = Rs 1600, \\ T = \frac{9}{4} years, \\ SI = Rs 252 \\ Put values in the formula , \\ \Rightarrow 252 = \frac{1600 \times R \times 9}{100} \\ \Rightarrow 252 = 36 R \\ \Rightarrow R = \frac{252}{36} = 7 % \end{aligned}$

57.

A sum becomes its double in 10 years. Find the annual rate of simple interest.

(A) 8%
(B) 5%
(C) 10%
(D) 20%

Solution: Let Sum be Rs. x and Let S.I. = Rs. x $\begin{aligned} Time = 10 years \\ ∴ Rate = \frac{S .I . \times 100}{Principal \times Time} \\ = \frac{x \times 100}{x \times 10} \\ = 10 % per annum . \end{aligned}$

58.

The simple interest on a sum of money is $\frac{4}{9}$ of the principal and the number of years is equal to the rate percent per annum. The rate per annum is = ?

(A) 5%
(B) $6 \frac{2}{3}$ %
(C) 6%
(D) $7 \frac{1}{5}$ %

Solution: $\begin{aligned} Let the principal \\ = 9 units \\ Hence simple interest \\ = \frac{4}{9} \times 9 = 4 units \\ Let, \\ Rate of interest = R% \\ R = T (given) \\ By using formula \\ SI = \frac{P \times T \times R}{100} \\ 4 = \frac{9 \times R \times R}{100} \\ R^{2} = \frac{400}{9} \\ R = \frac{20}{3} = 6 \frac{2}{3} % \end{aligned}$

59.

A lends Rs. 2500 to B and a certain sum to C at the same time at 7% p.a. simple interest. If after 4 years, A altogether receives Rs. 1120 as interest from B and C, then the sum lent to C is :

(A) Rs. 700
(B) Rs. 1500
(C) Rs. 4000
(D) Rs. 6500

Solution: Let the sum lent to C be Rs. x. Then, $\begin{aligned} (\frac{2500 \times 7 \times 4}{100}) + (\frac{x \times 7 \times 4}{100}) = 1120 \\ \Rightarrow \frac{7 x}{25} = (1120 - 700) \\ \Rightarrow x = (\frac{420 \times 25}{7}) \\ = 1500. \end{aligned}$

60.

A computer is available for Rs. 39000 cash or Rs. 17000 as cash down payment followed by five monthly instalments of Rs. 4800 each. What is the rate of interest under the instalment plan?

(A) 35.71 % p.a.
(B) 36.71 % p.a.
(C) 37.71 % p.a.
(D) 38.71 % p.a.

Solution: Total cost of the computer = Rs. 39000 Down payment = Rs. 17000 Balance = Rs. (39000 - 17000) = Rs. 22000. Let the rate of interest be R% p.a. Amount of Rs. 22000 for 5 months $\begin{aligned} = Rs . (22000 + 22000 \times \frac{5}{12} \times \frac{R}{100}) \\ = Rs . (22000 + \frac{275 R}{3}) \end{aligned}$ The customer pays the shopkeeper Rs. 4800 after 1 month, Rs. 4800 after 2 months, ...... and Rs. 4800 after 5 months. Thus, the shopkeeper keeps Rs. 4800 for 4 months, Rs. 4800 for 3 months, Rs. 4800 for 2 months, Rs. 4800 for 1 months and Rs. 4800 at the end. ∴ sum of the amounts of these installments = (Rs. 4800 + S.I. on Rs 4800 for 4 months) + (Rs. 4800 + S.I. on Rs. 4800 for 3 months) + ...... + (Rs. 4800 + S.I. on Rs. 4800 for 1 month) + Rs. 4800 = Rs. (4800 × 5) + S.I. on Rs. 4800 for (4 + 3 + 2 + 1) months = Rs. 24000 + S.I. on Rs. 4800 for 10 months $= Rs . (24000 + 4800 \times R \times \frac{10}{12} \times \frac{1}{100}) =$ $Rs . (24000 + 40 R)$ $\begin{aligned} ∴ 22000 + \frac{275 R}{3} = 24000 + 40 R \\ \Rightarrow \frac{155}{3} = 2000 \\ \Rightarrow R = \frac{2000 \times 3}{155} \\ = 38.71 % p .a . \end{aligned}$

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