Practice MCQ Questions and Answer on Interest

161.

What equal installment of annual payment will discharge a debt which is due as Rs. 848 at the end of 4 years at 4% per annum simple interest ?

(A) Rs. 212
(B) Rs. 200
(C) Rs. 250
(D) Rs. 225

Solution: Note : In such type of questions to save your valuable time follow the given below method. $\begin{aligned} Value of installment \\ = \frac{Principal \times 100}{Time \times 100 + (t_{n - 1} + t_{n - 2} . . .1) \times Rate %} \\ Principal = Rs . 848 \\ Rate = 4 % \\ Time = 4 year \\ Installment \\ = \frac{848 \times 100}{4 \times 100 + (3 + 2 + 1) \times 4} \\ = \frac{848 \times 100}{(400 + 24)} \\ = \frac{848 \times 100}{424} \\ = Rs . 200 \end{aligned}$

162.

An amount of Rs. 12,000 was borrowed at some rate of simple interest. After four months, Rs. 6,000 more were added to it and the rate of interest on the total principal was doubled from the previous rate. At the end of the year, Rs. 2,800 was paid as interest, calculate the rate of interest initially charged?

(A) 14%
(B) 16%
(C) 12%
(D) 10%

Solution: $\begin{aligned} Let, Rate = 12 x % per year \\ \frac{12000 \times 4 x}{100} + \frac{18000 \times 16 x}{100} = 2800 \\ 480 x + 2880 x = 2800 \\ 3360 x = 2800 \\ x = \frac{10}{12} \\ Rate = 12 x \\ = \frac{10}{12} \times 12 \\ = 10 % \end{aligned}$

163.

In a certain time, a sum becomes 4 times of itself on simple interest at the rate of 10% per annum. What is the rate of interest if the same sum becomes 7 times of itself in the same duration?

(A) 15%
(B) 10%
(C) 5%
(D) 20%

164.

If a certain sum of money becomes doubles at simple interest in 12 years, what would be the rate of interest per annum?

(A) $8 \frac{1}{3}$
(B) 10
(C) 12
(D) 14

Solution: Let, Principal, P = Rs. 100; Amount, A = Rs. 200; Time = 12 years; Interest = Rs. 100; Rate of interest $\begin{aligned} = \frac{Total Interest}{Given Time} \\ = \frac{100}{12} \\ = 8 \frac{1}{3} % \end{aligned}$

165.

A sum of money becomes $\frac{7}{6}$ of itself in 3 years at a certain rate of simple interest. The rate of interest per annum is ?

(A) $5 \frac{5}{9}$ %
(B) $6 \frac{5}{9}$ %
(C) 18 %
(D) 25 %

Solution: $\begin{aligned} Let principal = 6P \\ Hence Amount \\ = 6P \times \frac{7}{6} = 7 P \\ ∴ SI = 7P - 6P = P \\ Time = 3 years \\ SI = \frac{P \times T \times R}{100} \\ \Rightarrow P = \frac{6 P \times R \times 3}{100} \\ \Rightarrow R = \frac{100}{18} \\ = \frac{50}{9} \\ = 5 \frac{5}{9} % \end{aligned}$ Alternate: Note: In such type of questions to save your valuable time try to think like that. $\begin{matrix} \begin{matrix} Amount & Principal \end{matrix} \\ \underset{+ 1}{\underset{⏟}{7 6}} \end{matrix}$ $\begin{aligned} Required rate % \\ = \frac{1}{6} \times \frac{100}{3} \\ = 5 \frac{5}{9} % \end{aligned}$

166.

A person borrowed 1,200 at 8% per annual and Rs. 1,800 at 10% per annum, as simple interest for the same period. He had to pay Rs. 1380 in all as interest. Find the time? (in year)

(A) 4 years
(B) 5 years
(C) 6 years
(D) 10 years

Solution: $\begin{aligned} \frac{1200 \times t \times 8}{100} + \frac{1800 \times t \times 10}{100} = 1380 \\ 96 t + 180 t = 1380 \\ 276 t = 1380 \\ t = 5 \end{aligned}$

167.

A person borrowed 2,000 at 5% annual simple interest repayable in 3 equal annual installments. What will be the annual installment?

(A) $Rs . 730 \frac{10}{63}$
(B) $Rs . 840 \frac{9}{61}$
(C) $Rs . 640 \frac{11}{63}$
(D) $Rs . 250 \frac{10}{63}$

Solution: $\begin{aligned} Amount after 3 years \\ = 2000 + \frac{2000 \times 5 \times 3}{100} \\ = 2000 + 300 \\ = 2300 \\ EMI = \frac{A \times 100}{100 \times t + r t (\frac{t - 1}{2})} \\ = \frac{2300 \times 100}{100 \times 3 + 5 \times 3 (\frac{3 - 1}{2})} \\ = \frac{2300 \times 100}{315} \\ = \frac{2300 \times 20}{63} \\ = \frac{46000}{63} \\ = 730 \frac{10}{63} \end{aligned}$

168.

In what time will the simple interest on Rs. 1750 at 9% per annum be the same as that on Rs. 2500 at 10.5% per annum in 4 years?

(A) 6 years and 8 months
(B) 7 years and 3 months
(C) 6 years
(D) 7 years and 6 months

Solution: $\begin{aligned} Let time is T years . \\ According to questions, \\ \frac{1750 \times 9 \times T}{100} = \frac{(2500 \times 10.5 \times 4)}{100} \\ O r, T = \frac{2500 \times 10.5 \times 4}{1750 \times 9} \\ O r, T = 6.66 = 6 years and 8 months \end{aligned}$

169.

If a sum doubles in 16 years, how much will it be in 8 years ?

(A) $1 \frac{1}{2}$ times
(B) $1 \frac{1}{3}$ times
(C) $1 \frac{1}{4}$ times
(D) $1 \frac{3}{4}$ times

Solution: Let Sum = Rs. x. Then, S.I. = Rs. x, Time = 16 years $\begin{aligned} ∴ Rate = (\frac{100 \times x}{x \times 16}) % = \frac{25}{4} % = 6 \frac{1}{4} % \\ Now, sum = Rs . x, \\ Time = 8 years \\ Rate = 6 \frac{1}{4} % \\ ∴ S .I . = Rs . (\frac{x \times 25 \times 8}{100 \times 4}) \\ = Rs . \frac{x}{2} \\ So, \\ Amount = Rs . (x + \frac{x}{2}) \\ = Rs . \frac{3 x}{2} \\ = 1 \frac{1}{2} times \end{aligned}$

170.

Kruti took a loan at simple interest rate of 6 p.c.p.a. in the first year and it increased by 1.5 p.c.p.a. every year. If she pays Rs. 8190 as interest at the end of 3 years, what was her loan amount ?

(A) Rs. 35400
(B) Rs. 36000
(C) Rs. 36800
(D) Cannot be determined

Solution: Answer & Solution Answer: Option E Solution: Let the loan amount be Rs. x $\begin{aligned} Then, \\ \Rightarrow \frac{6 x}{100} + \frac{7.5 x}{100} + \frac{9 x}{100} = 8190 \\ \Rightarrow 22.5 x = 819000 \\ \Rightarrow x = 36400 \end{aligned}$

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Practice MCQ Questions and Answer on Interest

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