Practice MCQ Questions and Answer on Interest

121.

Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?

(A) 3.6
(B) 6
(C) 18
(D) Cannot be determined

Solution: $\begin{aligned} Let rate = R % and \\ Time = R years \\ Then, \frac{1200 \times R \times R}{100} = 432 \\ \Rightarrow 12 R^{2} = 432 \\ \Rightarrow R^{2} = 36 \\ \Rightarrow R = 6 \end{aligned}$

122.

The simple interest on a sum of money is $\frac{1}{4}$ of the principal and the number of years is equal to rate percent per annum. The rate percent is = ?

(A) 2.5%
(B) 5%
(C) 7.5%
(D) 10%

Solution: $\begin{aligned} Principal Interest \\ \underset{}{\underset{⏟}{4 P P}} \\ Time = Rate % (given) \\ Now by using formula , \\ P = \frac{4 P \times R \times R}{100} \\ \Rightarrow R^{2} = \frac{100}{4} \\ \Rightarrow R = \frac{10}{2} \\ \Rightarrow R = 5 % \end{aligned}$

123.

A money lender finds that due to fall in the annual rate of interest from 8% to $7 \frac{3}{4} %,$ ÃÂÃÂ his yearly income diminishes by Rs. 61.50. His capital is = ?

(A) Rs. 22400
(B) Rs. 23800
(C) Rs. 24600
(D) Rs. 26000

Solution: Let initial capital = 100 units According to the question, $\begin{aligned} \frac{1}{4} unit = Rs . 61 .50 \\ 1 unit = Rs . 61 .50 \times 4 \\ = Rs . 246 \\ 100 units = Rs . 24600 \end{aligned}$ Hence, Required capital = Rs. 24600 Alternate Difference in percentage $\begin{aligned} = 8% - \frac{31}{4} % \\ \Leftrightarrow \frac{1}{4} % = 61.50 \\ \Leftrightarrow 100 % = 24600 \end{aligned}$

124.

If the simple interest on a sum of money for 15 months at $7 \frac{1}{2}$ % per annum exceeds the simple interest on the same sum for 8 months at $12 \frac{1}{2}$ % per annum by Rs. 32.50, then the sum of money ( In Rs.) is ?

(A) Rs. 312
(B) Rs. 312.50
(C) Rs. 3120
(D) Rs. 3120.50

Solution: $\begin{aligned} T_{1} = 15 months \\ = \frac{15}{12} y e a r s \\ R_{1} = 7 \frac{1}{2} % = \frac{15}{2} % \\ T_{2} = 8 months \\ = \frac{8}{12} y e a r s \\ R_{2} = 12 \frac{1}{2} % = \frac{25}{2} % \\ Let the principal = P \\ According to the question, \\ \Leftrightarrow \frac{P \times 15 \times 15}{12 \times 2 \times 100} - \frac{P \times 25 \times 8}{12 \times 2 \times 100} = 32.50 \\ \Rightarrow \frac{225 P}{2400} - \frac{200 P}{2400} = 32.50 \\ \Rightarrow \frac{25 P}{2400} = 32.50 \\ \Rightarrow P = Rs 3120 \\ Hence required principal \\ = Rs . 3120 \end{aligned}$

125.

The simple interest on a certain sum of money at the rate of 5% p.a. for 8 years is Rs. 840. At what rate of interest the same account of interest can be received on the same sum after 5 years?

(A) 6%
(B) 8%
(C) 9%
(D) 10%

Solution: $\begin{aligned} S .I . = Rs . 840 \\ R = 5 % \\ T = 8 years \\ Principal \\ = Rs . (\frac{100 \times 840}{5 \times 8}) \\ = Rs . 2100 \\ Now, \\ P = Rs . 2100 \\ S .I . = Rs . 840 \\ T = 5 years \\ ∴ Rate \\ = (\frac{100 \times 840}{2100 \times 5}) % \\ = 8 % \end{aligned}$

126.

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

127.

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

128.

Vishwas borrowed a total amount of Rs. 30000, part of it on simple interest rate of 12 p.c.p.a. and remaining on simple interest rate of 10 p.c.p.a. If at the end of 2 year she paid in all Rs. 36480 to settle the loan amount, what was the amount borrowed at 12 p.c.p.a ?

(A) Rs. 12000
(B) Rs. 16000
(C) Rs. 17500
(D) Rs. 18000

Solution: Let the sum borrowed at 12% p.a. be Rs. x and that borrowed at 10% p.a. be Rs. (30000 - x) S.I. at the end of 2 years = Rs. (36480 - 30000) = Rs. 6480 $∴ (\frac{x \times 12 \times 2}{100}) +$ $[\frac{(30000 - x) \times 10 \times 2}{100}]$ $= 6480$ $\begin{aligned} \Leftrightarrow 24 x + 600000 - 20 x = 648000 \\ \Leftrightarrow 4 x = 48000 \\ \Leftrightarrow x = 12000 \end{aligned}$

129.

A mobile phone is available for Rs. 25,000 or Rs. 5,200 down payment, followed by 4 equal monthly instalments. If the rate of interest is 25% p.a. simple interest, calculate the amount of each instalment.

(A) Rs. 5,020
(B) Rs. 4,950
(C) Rs. 5,362
(D) Rs. 5,200

Solution: $\begin{aligned} Cash payment = 25000 \\ Down payment = 5200 \\ Remaining = 19800 \\ Installment (x) = 4 \end{aligned}$ $\begin{matrix} \Rightarrow & 19800 & - x \\ 19800 & - 2 x \\ \underset{___________}{19800} & \underset{_________}{- 3 x} \\ 79200 & - 6 x \end{matrix}$ $\begin{aligned} \Rightarrow Interest = 4 x - 19800 \\ \Rightarrow \frac{(79200 - 6 x) \times 25}{12 \times 100} = 4 x - 19800 \\ \Rightarrow 1920 x - 50400 = 79200 - 60 x \\ \Rightarrow 1980 x = 1029600 \\ \Rightarrow x = 5200 \end{aligned}$

130.

There is a decrease of 10% yearly on an article. If this article was bought 3 years ago and present cost is Rs. 5,832 then what was the cost of article at buying time?

(A) Rs. 7,200
(B) Rs. 7,862
(C) Rs. 8,000
(D) Rs. 8,500

Solution: $A = P {(1 - \frac{R}{100})}^{n}$ Where A = Value of goods after n years P = Initial Price R = Rate of depriciation $\begin{aligned} ∴ P = \frac{5832}{{(1 - \frac{10}{100})}^{3}} \\ \Rightarrow P = \frac{5832}{{(1 - \frac{1}{10})}^{3}} \\ \Rightarrow P = \frac{5832}{{(\frac{9}{10})}^{3}} \\ \Rightarrow P = 5832 \times \frac{10}{9} \times \frac{10}{9} \times \frac{10}{9} \\ \Rightarrow P = 8000 \end{aligned}$

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Practice MCQ Questions and Answer on Interest

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