Practice MCQ Questions and Answer on Interest

141.

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

142.

In a certain duration of time, a sum become 2 times itself at the rate of 5% per annum simple interest. What will be the rate of interest if the same sum becomes 5 times itself in the same duration?

(A) 20%
(B) 16%
(C) 10%
(D) 18%

143.

If Rs. 12000 is divided into two parts such that the simple interest on the first part for 3 years at 12% per annum is equal to the simple interest on the second part for $4 \frac{1}{2}$ years at 16% per annum, the greater part is = ?

(A) Rs. 8000
(B) Rs. 6000
(C) Rs. 7000
(D) Rs. 7500

Solution: $\begin{aligned} Let the first part = Rs . x \\ ∴ Hence second part \\ = Rs . (12000 - x) \\ According to the question, \\ \Rightarrow \frac{x \times 12 \times 3}{100} = \frac{(12000 - x) \times 9 \times 16}{2 \times 100} \\ \Rightarrow 36 x = 72 (12000 - x) \\ \Rightarrow x = 24000 - 2 x \\ \Rightarrow 3 x = 24000 \\ \Rightarrow x = Rs . 8000 \\ 1^{st} part = Rs . 8000 \\ 2^{nd} part \\ = Rs . (12000 - 8000) \\ = Rs . 4000 \\ Hence maximum part \\ = Rs . 8000 \end{aligned}$ Alternate Note : In such type of questions to save your valuable time follow the given below method. Let two parts P1 and P2 respectively According to the question, $\begin{aligned} \Rightarrow P_{1} \times \frac{36}{100} \times 1 = P_{2} \times \frac{9}{2} \times \frac{16}{100} \times 1 \\ \Rightarrow P_{1} \times 4 = 8 P_{2} \\ \Rightarrow \frac{P_{1}}{P_{2}} = \frac{2}{1} \\ \Rightarrow P_{1} : P_{2} = 2 : 1 \\ Hence greater part \\ = \frac{12000}{(2 + 1)} \times 2 \\ = Rs . 8000 \end{aligned}$

144.

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

145.

Brajesh had Rs. 8,600 which he invested in two parts. Simple interest received on the first part at 15% p.a. in 4 years is equal to the simple interest received on the second part at 20% p.a. in 3 years. Find the difference in the two parts.

(A) Rs. 0
(B) Rs. 210
(C) Rs. 190
(D) Rs. 290

Solution: Difference between both part $\begin{aligned} = (4 \times 15 % - 3 \times 20 %) of 8600 \\ = 0 % of 8600 \\ = 0 Answer \end{aligned}$

146.

Simple interest on a certain amount is $\frac{9}{16}$ of the principal. If the numbers representing the rate of interest in percent and time in years be equal, then time, for which the principal is lent out, is -

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

Solution: $\begin{aligned} Let sum = x . \\ Then, \\ S .I . = \frac{9}{16} x \\ Let rate = R % and \\ time = R years \\ ∴ (\frac{x \times R \times R}{100}) = \frac{9 x}{16} \\ \Rightarrow R^{2} = \frac{900}{16} \\ R = \frac{30}{4} = 7 \frac{1}{2} \\ Hence, \\ time = 7 \frac{1}{2} years \end{aligned}$

147.

In what time will Rs. 3300 becomes Rs. 3399 at 6% per annum interest compounded half-yearly?

(A) 6 months
(B) 1 year
(C) $1 \frac{1}{2}$ years
(D) 3 months

Solution: $\begin{aligned} P = R s . 3300 \\ A = R s . 3399 \\ R = 6 % per annum \\ Let the time be n years . \\ Compound interest is taken half - yearly . \\ A = P \times {[1 + (\frac{R}{2} \times 100)]}^{2 n} \\ 3399 = 3300 {(1 + \frac{3}{100})}^{2 n} \\ {(1.03)}^{2 n} = \frac{3399}{3300} \\ {(1.03)}^{2 n} = {(1.03)}^{1} \\ T h u s, 2 n = 1 y e a r \\ n = \frac{1}{2} year = 6 months \end{aligned}$

148.

Find the simple interest on Rs. 2,700 for 8 months at 5 paisa per rupee per month?

(A) Rs. 950
(B) Rs. 720
(C) Rs. 540
(D) Rs. 1,080

Solution: $\begin{aligned} Rate = \frac{5 paisa}{1 rupee} = 5 % \\ Time = 8 months \\ S .I . = \frac{2700 \times 5 \times 8}{100} = 1080 \end{aligned}$

149.

A person borrows Rs. 5000 for 2 year at 4% p.a. simple interest. He immediately lends it to another person at $6 \frac{1}{4}$ % p.a. for 2 years. Find his gain in the transaction per year.

(A) Rs. 112.50
(B) Rs. 125
(C) Rs. 150
(D) Rs. 167.50

Solution: $\begin{aligned} Gain in 2 years \\ = (5000 \times \frac{25}{4} \times \frac{2}{100}) - (\frac{5000 \times 4 \times 2}{100}) \\ = Rs . (625 - 400) \\ = Rs . 225 \\ ∴ Gain 1 year = Rs . (\frac{225}{2}) \\ = Rs . 112.50 \end{aligned}$

150.

Rs. 2100 is lent at compound interest of 5% per annum for 2 years. Find the amount after two years.

(A) Rs. 2300
(B) Rs. 2315.25
(C) Rs. 2310
(D) Rs. 2320

Solution: $\begin{aligned} We can use formula of compound interest \\ A = P \times {[1 + (\frac{r}{100})]}^{n} \\ A = 2100 \times {[1 + (\frac{5}{100})]}^{2} \\ A = 2100 \times {[\frac{105}{100}]}^{2} \\ A = \frac{2100 \times 11025}{10000} \\ Hence, Amount A = R s . 2315.25 \end{aligned}$

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Practice MCQ Questions and Answer on Interest

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