Practice MCQ Questions and Answer on Interest

31.

Jhon invested a sum of money at an annual simple interest rate of 10%. At the end of four years the amount invested plus interest earned was Rs. 770. The amount invested was = ?

(A) Rs. 650
(B) Rs. 350
(C) Rs. 550
(D) Rs. 500

Solution: Let the amount invested = Rs. P According to the question, $\begin{aligned} P + \frac{P \times 10 \times 4}{100} = 770 \\ \Rightarrow P + \frac{4 P}{10} = 770 \\ \Rightarrow \frac{14 P}{10} = 770 \\ \Rightarrow P = \frac{770 \times 10}{14} \\ \Rightarrow P = Rs 550 \end{aligned}$ Hence, required invested amount = Rs. 550 Alternate $\begin{aligned} 10 % = \frac{1 \to Interest}{10 \to Principal} \\ Interest in 4 years \\ = 1 \times 4 \\ = 4 \\ Amount = \\ = (Interest + Principal) \\ = 4 + 10 \\ = 14 \\ According to the question, \\ 14 units = 770 \\ 1 unit = \frac{770}{14} \\ 10 units = \frac{770}{14} \times 10 \\ = Rs . 550 \\ The amount invested \\ = Rs . 550 \end{aligned}$

32.

In how many years will the simple interest on a sum of money be equal to the principal at the rate of $16 \frac{2}{3}$ % per annum ?

(A) 4 years
(B) 5 years
(C) 6 years
(D) 8 years

Solution: $\begin{aligned} 16 \frac{2}{3} = \frac{1 \to Interest}{6 \to Principal} \\ Let principal = 6 \\ Interest = 6 \\ Time = t years \\ By using formula \\ 6 = \frac{6 \times 50 \times t}{3 \times 100} \\ \Rightarrow t = 6 years \end{aligned}$ Alternate Note : In such type of questions to save your valuable time think like the given way. $\begin{aligned} Rate % \\ = 16 \frac{2}{3} % = \frac{1 \to Interest}{6 \to Principal} \\ Represent for 1 years \\ According to the question, \\ Principal = Interest \\ 6 = 1 \times 6 \\ Hence, \\ Time = 1 \times 6 \\ = 6 years \end{aligned}$ Note : If interest will be six times then time will also be six times.

33.

The principal which gives Rs 1 interest per day at a rate of 5% simple interest per annum is = ?

(A) Rs. 5000
(B) Rs. 35500
(C) Rs. 7300
(D) Rs. 3650

Solution: $\begin{aligned} According to the question, \\ Interest = Rs . 1 per day \\ ∴ Interest in one year \\ = 1 \times 365 = Rs . 365 \\ ∴ S .I . = \frac{P \times T \times R}{100} \\ \Rightarrow 365 = \frac{P \times 5 \times 1}{100} \\ \Rightarrow P = \frac{365 \times 100}{5} \\ \Rightarrow P = Rs . 7300 \end{aligned}$

34.

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 $= Rs . (24000 + 4800 \times R \times \frac{10}{12} \times \frac{1}{100}) =$ $Rs . (24000 + 40 R)$ $\begin{aligned} ∴ 22000 + \frac{275 R}{3} = 24000 + 40 R \\ \Rightarrow \frac{155}{3} = 2000 \\ \Rightarrow R = \frac{2000 \times 3}{155} \\ = 38.71 % p .a . \end{aligned}$

35.

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, $\begin{aligned} (\frac{2500 \times 7 \times 4}{100}) + (\frac{x \times 7 \times 4}{100}) = 1120 \\ \Rightarrow \frac{7 x}{25} = (1120 - 700) \\ \Rightarrow x = (\frac{420 \times 25}{7}) \\ = 1500. \end{aligned}$

36.

Rs. 6000 becomes Rs. 7200 in 4 years. If the rate becomes 1.5 times of itself, the amount of the same principal in 5 years will be = ?

(A) Rs. 8000
(B) Rs. 8250
(C) Rs. 9250
(D) Rs. 9000

Solution: $\begin{aligned} Principal Amount \\ \underset{+ 1200}{\underset{⏟}{6000 7200}} \\ By using formula, \\ Rate % \\ = \frac{1200}{6000} \times \frac{100}{4} \\ = 5 % \\ New rate % \\ = 5 \times \frac{3}{2} = 7 .5 % \\ Interest after 5 years \\ = \frac{6000 \times 7.5 \times 5}{100} \\ = Rs . 2250 \\ Hence, \\ amount \\ = Rs . (6000 + 2250) \\ = Rs 8250 \end{aligned}$

37.

A sum becomes 4 times at simple interest in 10 years. What is the rate of interest?

(A) 10%
(B) 20%
(C) 30%
(D) 40%

Solution: $\begin{aligned} 1^{st} Method : \\ Let rate is R % \\ Now, \\ P = 100, \\ A = 400, \\ I = 400 - 100 = 300, \\ Time, T = 10 years \\ I = \frac{P T R}{100} \\ Or, R = \frac{100 \times I}{P T} \\ Or, R = \frac{100 \times 300}{100 \times 10} \\ Hence, R = 30 % \end{aligned}$ 2nd Method : Here, the sum become 4 times that means 100, become 400. Rate of such question is given by $R = \frac{interest}{time} = \frac{300}{10} = 30 %$ 3rd Method : Here, 300% of rise in the sum so $\begin{aligned} 100 - - - 300 % ↑ - - - 400; \\ R = \frac{total percentage rise}{given time} \\ = \frac{300 %}{10} \\ = 30 % \end{aligned}$

38.

A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to 8 times?

(A) 9 years
(B) 8 years
(C) 27 years
(D) 12 years

Solution: $\begin{aligned} Let, \\ Principal = R s . 100 \\ Amount = R s . 200 \\ Rate = r % \\ Time = 4 years \\ Now, \\ A = P \times {[1 + (\frac{r}{100})]}^{n} \\ 200 = 100 \times {[1 + (\frac{r}{100})]}^{4} \\ 2 = {[1 + (\frac{r}{100})]}^{4} - - - - - - (i) \\ If sum become 8 times in the time n years \\ then, \\ 8 = {(1 + (\frac{r}{100}))}^{n} \\ 2^{3} = {(1 + (\frac{r}{100}))}^{n} - - - - - - (i i) \\ Using eqn (i) i n (i i), we get \\ {({[1 + (\frac{r}{100})]}^{4})}^{3} = {(1 + (\frac{r}{100}))}^{n} \\ {[1 + (\frac{r}{100})]}^{12} = {(1 + (\frac{r}{100}))}^{n} \\ Thus, n = 12 years . \end{aligned}$

39.

Divide Rs. 6000 into two parts so that simple interest on the first part for 2 years at 6% p.a. may be equal to the simple interest on the second part for 3 years at 8% p.a.

(A) Rs. 4000, Rs. 2000
(B) Rs. 5000, Rs. 1000
(C) Rs. 3000, Rs. 3000
(D) None of these

Solution: $\begin{aligned} Let 1^{s t} part is x and 2^{n d} part is (6000 - x) \\ According to question, \\ \frac{x \times 2 \times 6}{100} = \frac{(6000 - x) \times 3 \times 8}{100} \\ 12 x = 144000 - 24 x \\ O r, 36 x = 144000 \\ O r, x = \frac{144000}{36} = R s . 4000 \\ 1^{s t} part = R s . 4000 \\ 2^{n d} part = R s . 2000 \end{aligned}$

40.

Mr. Dutta desired to deposit his retirement benefit of Rs. 3 lacs partly to a post office and partly to a bank at 10% and 6% simple interests respectively. If his monthly income was Rs. 2000, then the difference of his deposits in the post office and in the bank was = ?

(A) Rs. 100000
(B) Rs. 40000
(C) Rs. 50000
(D) Rs. Nil

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Practice MCQ Questions and Answer on Interest

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