Practice MCQ Questions and Answer on Interest

131.

Rahul borrowed a sum of Rs. 1150 from Amit at the simple interest rate of 6 p.c.p.a. for 3 Years. He then added some more money to the borrowed sum and lent it to Sachin for the same time at 9 p.c.p.a simple interest. If Rahul gains Rs. 274.95 by way of interest on borrowed sum as well as his own amount from the whole transaction, then what is the sum lent by him to Sachin ?

(A) Rs. 1200
(B) Rs. 1285
(C) Rs. 1690
(D) Rs. 1785

Solution: Let the money added by Rahul be Rs. x Then, $\Rightarrow \frac{(1150 + x) \times 9 \times 3}{100} -$ $\frac{1150 \times 6 \times 3}{100} =$ $274.95$ ⇒ 1150 × 27 + 27x - 1150 × 18 = 27495 ⇒ 27x + 1150 × (27 - 18) = 27495 ⇒ 27x = 27495 - 10350 ⇒ 27x = 17145 ⇒ x = 635 So, sum lent by Rahul to Sachin = Rs. ( 1150 + 635 ) = Rs. 1785

132.

If the rate increases by 2%, the simple interest received on a sum of money increases by Rs. 108. If the time period is increased by 2 years, the simple interest on the same sum increases by Rs. 180. The sum is:

(A) Rs. 1800
(B) Rs. 3600
(C) Rs. 5400
(D) Data inadequate

Solution: $\begin{aligned} Let the sum be Rs . x \\ Rate be R % p .a . \\ Time be T years . \\ Then, \\ [\frac{x \times (R \times 2) \times T}{100}] - (\frac{x \times R \times T}{100}) = 108 \\ \Leftrightarrow 2 x T = 10800 . . . . . . . . (i) \\ A n d, \\ [\frac{x \times R \times (T + 2)}{100}] - (\frac{x \times R \times T}{100}) = 108 \\ \Leftrightarrow 2 x R = 18000 . . . . . . . (i i) \end{aligned}$ Clearly, from (i) and (ii), we cannot the find the value of x. So, the data is inadequate.

133.

A man buys a TV priced at Rs. 16000. He pays Rs. 4000 at once and the rest after 15 months on which he is charges a simple interest at the rate of 12% per year. The total amount he pays for TV is = ?

(A) Rs. 18200
(B) Rs. 17200
(C) Rs. 17800
(D) Rs. 16800

Solution: Total price of TV = Rs. 16000 Initial payment = Rs. 4000 Remaining amount = Rs. 12000 Simple interest in 15 months for Rs. 12000 $\begin{aligned} \Rightarrow S .I . = \frac{P \times R \times T}{100} \\ \Rightarrow S .I . = \frac{12000 \times 12 \times 15}{100 \times 12} \\ \Rightarrow S .I . = Rs . 1800 \end{aligned}$ ⇒ With S.I. total amount to be paid for principal amount Rs. 12000 = Rs. (12000 + 1800) = Rs. 13800 = Therefore, total amount he pays for the TV is = 4000 + 13800 = Rs. 17800

134.

If the simple interest for 6 years be equal to 30% of the principal, it will be equal to the principal after = ?

(A) 20 years
(B) 30 years
(C) 10 years
(D) 22 years

Solution: $\begin{aligned} Let principal = 10P \\ Interest = 10P \times \frac{30}{100} \\ = 3P \\ According to the question, \\ Case (I) \\ \Rightarrow 3 P = \frac{10P \times R \times 6}{100} \\ \Rightarrow R = 5 % \\ Case (II) \\ Interest = Principal = 10P \\ \Rightarrow 10P = \frac{10P \times 5 \times t}{100} \\ \Rightarrow t = 20 years \end{aligned}$

135.

A sum of Rs. 10 is lent to be returned in 11 monthly instalments of Rs. 1 each, interest being simple. The rate of interest is:

(A) $9 \frac{1}{11}$ %
(B) 10%
(C) 11%
(D) $21 \frac{9}{11}$ %

Solution: ⇒ Rs. 10 + S.I. on Rs. 10 for 11 months = Rs. 11 + S.I. on Rs. 1 for (1 + 2 + 3 + 4 + ........... + 10) months ⇒ Rs. 10 + S.I. on Rs. 1 for 110 months = Rs. 11 + S.I. on Rs. 1 for 55 months S.I. on Rs. 1 for 55 months = Rs. 1 $\begin{aligned} ∴ Rate = (\frac{100 \times 12}{1 \times 55}) % \\ = 21 \frac{9}{11} % \end{aligned}$

136.

What will be the ratio of simple interest earned by certain amount at the same rate of interest for 6 years and that for 9 years?

(A) 1 : 3
(B) 1 : 4
(C) 2 : 3
(D) Data inadequate

Solution: Let the principal be P and rate of interest be R%. Therefore Required ratio $\begin{aligned} = \frac{\frac{P \times R \times 6}{100}}{\frac{P \times R \times 9}{100}} \\ = \frac{6 P R}{6 P R} \\ = \frac{6}{9} \\ = 2 : 3 \end{aligned}$

137.

A sum was invested at simple interest at a certain interest for 2 years. It would have fetched Rs. 60 more had it been invested at 2% higher rate. What was the sum?

(A) Rs. 1500
(B) Rs. 1300
(C) Rs. 2500
(D) Rs. 1000

Solution: Let the rate be R at which Principal P is invested for 2 years. According to question, {Interest at Rate (R + 2)}% - (interest at rate R%) = Rs. 60 $\frac{(P \times 2 \times (R + 2))}{100} -$ $\frac{(P \times 2 \times R)}{100}$ $= 60$ $\begin{aligned} \frac{2 P R + 4 P - 2 P R}{100} = 60 \\ 4 P = 60 \times 100 \\ Or, P = \frac{60 \times 100}{4} \\ Hence, P = Rs . 1500 \end{aligned}$

138.

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

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

Solution: Let the sum at 8% be Rs. x Let the sum at 5% be 1550 - x According to the given condition, $\begin{aligned} \frac{x \times 8 \times 3}{100} + \frac{(1550 - x) \times 5 \times 3}{100} = 300 \\ \Rightarrow \frac{24 x + (1550 - x) \times 15}{100} = 300 \\ \Rightarrow \frac{24 x + 23250 - 15 x}{100} = 300 \\ \Rightarrow 24 x - 15 x = 30000 - 23250 \\ \Rightarrow 24 x - 15 x = 6750 \\ \Rightarrow 9 x = 6750 \\ \Rightarrow x = \frac{6750}{9} = 750 \\ \frac{Money lent at 5 %}{Money lent at 8 %} \\ = \frac{1550 - 750}{750} \\ = \frac{800}{750} \\ = \frac{16}{15} \\ = 16 : 15 \end{aligned}$

139.

In a certain time, the ratio of a certain principal and interest obtained from it are in the ratio 10 : 3 at 10% interest per annum. The number of years for which the money was invested is = ?

(A) 1 year
(B) 3 year
(C) 5 year
(D) 7 year

Solution: According to the question, Principal Interest 10 3 $\begin{aligned} Rate % = 10 % \\ Time = \frac{3}{10} \times \frac{100}{10} \\ = 3 years \end{aligned}$

140.

A sum of Rs. 210 was taken as a loan. This is to be paid back in two equal installments. If the rate of the interest be 10% compounded annually, then the value of each installment is:

(A) Rs. 121
(B) Rs. 127
(C) Rs. 210
(D) Rs. 225

Solution: Let X = equal installment at the end of one year( rate% annually) . Now 1st year, P =210, Interest = $\frac{PTR}{100}$ = 210 * 0.1 = 21. Let X is to be paid as an equal installment. Now, at the beginning of 2nd year, P = 210 + 21 - X, Interest at the end of 2nd year, = (231 - X) * 0.1 = 23.1 - 0.1X. Hence,total installment, 2X = 210 + 21 + 23.1 - 0.1X, X = $\frac{254 .1}{2.1}$ = 121.

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Practice MCQ Questions and Answer on Interest

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