Practice MCQ Questions and Answer on Interest

41.

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 $\begin{aligned} = 8% - \frac{31}{4} % \\ \Leftrightarrow \frac{1}{4} % = 61.50 \\ \Leftrightarrow 100 % = 24600 \end{aligned}$

42.

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

43.

If the amount obtained by A by investing Rs. 9,100 for three years at a rate of 10% p.a. on simple interest is equal to the amount obtained by B by investing a certain sum of money for five year at a rate of 8% p.a. on simple interest, then 90% of the sum invested by B (in Rs.) is:

(A) 7,800
(B) 8,450
(C) 8,540
(D) 7,605

Solution: Amount obtained by A is equal to amount obtained by B Amount = Principal + Simple Interest $\begin{aligned} 9100 + \frac{9100 \times 10 \times 3}{100} = B + \frac{B \times 8 \times 5}{100} \\ 9100 + 2730 = \frac{140}{100} B \\ B = 11830 \times \frac{100}{140} \\ B = 8450 \end{aligned}$ 90% of invested amount by B $\begin{aligned} = 8450 \times \frac{90}{100} \\ = 7605 \end{aligned}$

44.

Mohan lent some amount of money at 9% simple interest and an equal amount of money at 10% simple interest each for two years. If his total interest was Rs. 760, what amount was lent in each case ?

(A) Rs. 1700
(B) Rs. 1800
(C) Rs. 1900
(D) Rs. 2000

Solution: Let the amount invested = Rs. P According to the questions, $\begin{aligned} \Leftrightarrow \frac{P \times 9 \times 2}{100} + \frac{P \times 10 \times 2}{100} = 760 \\ \Leftrightarrow \frac{18 P + 20P}{100} = 760 \\ \Leftrightarrow 38 P = 76000 \\ \Leftrightarrow P = 2000 \end{aligned}$ Alternate Total interest percent $\begin{aligned} = (9 \times 2) % + (10 \times 2) % \\ \Rightarrow 38 % = 760 \\ \Rightarrow 100 % = 2000 \end{aligned}$ Hence required principal = Rs. 2000

45.

On simple interest, a certain sum becomes Rs. 59,200 in 6 years and Rs. 72,000 in 10 years. If the rate of interest had been 2% more, then in how many years would the sum have become Rs. 76,000?

(A) 10
(B) 9
(C) 8
(D) 7

Solution: $\begin{matrix} P \overset{6 years}{\to} 59, 200 \\ P \overset{10 years}{\to} 72, 000 \end{matrix}$ $\begin{aligned} Difference 4 years \\ = (72, 000 - 59, 200) \\ = 12, 800 \\ 1 year = \frac{12, 800}{4} = 3, 200 \\ ∴ 6 years S .I . \\ = 3, 200 \times 6 = 19, 200 \\ P = 59, 200 - 19, 200 = 40, 000 \\ S .I . = \frac{P \times r \times t}{100} \\ 19, 200 = \frac{40, 000 \times r \times 6}{100} \\ r = 8 % \\ If r = 8 % + 2 % = 10 % \\ P = 40, 000 \\ t = ? \\ A = 76, 000 \\ ∴ S .I . = 36, 000 \\ 36, 000 = \frac{40, 000 \times 10 \times t}{100} \\ t = 9 years \end{aligned}$

46.

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. $\begin{aligned} Value of installment \\ = \frac{Principal \times 100}{Time \times 100 + (t_{n - 1} + t_{n - 2} . . .1) \times Rate %} \\ Principal = Rs . 848 \\ Rate = 4 % \\ Time = 4 year \\ Installment \\ = \frac{848 \times 100}{4 \times 100 + (3 + 2 + 1) \times 4} \\ = \frac{848 \times 100}{(400 + 24)} \\ = \frac{848 \times 100}{424} \\ = Rs . 200 \end{aligned}$

47.

The simple interest on Rs. 36000 for the period from 5^th January to 31^st May, 2013 at 9.5% per annum is = ?

(A) Rs. 1368
(B) Rs. 1338
(C) Rs. 1425
(D) Rs. 1400

Solution: Number of days = 26 + 28 + 31 + 30 + 30 + 31 = 146 days $\begin{aligned} \Rightarrow SI = \frac{P \times R \times T}{100} \\ \Rightarrow SI = \frac{36000 \times 9.5 \times 146}{100} \\ \Rightarrow SI = Rs . 1368 \end{aligned}$

48.

Simple interest on a certain sum one-fourth of the sum and the interest rate percentage per annum is 4 times the numbers of years. If the rate of interest increases by 2%, then what will be the simple interest (in Rs.) on Rs. 5,000 for 3 years?

(A) 300
(B) 1,500
(C) 2,000
(D) 1,800

Solution: $\frac{1}{4} \begin{matrix} \leftarrow S .I . \\ \leftarrow P \end{matrix}$ $\begin{aligned} Let P = 4 x and S .I . = x \\ Time = t years \\ Rate = 4 t % \\ ∴ S .I . = \frac{P \times R \times T}{100} \\ x = \frac{4 x \times 4 t \times t}{100} \\ t^{2} = \frac{100}{16} \\ t^{2} = \frac{25}{4} \\ ∴ t = \frac{5}{2} = 2.5 years \\ ∴ R = 4 \times 2.5 years = 10 % \\ New rate = 10 % + 2 % \\ P = Rs . 5000 \\ T = 3 years \\ S .I . = \frac{P \times R \times T}{100} \\ = \frac{5000 \times 12 \times 3}{100} \\ = Rs . 1800 \end{aligned}$

49.

Rs. 2,500, when invested for 8 years at a given rate of simple interest per year, amounted to Rs. 3,725 on maturity. What was the rate of simple interest that was paid per annum?

(A) 6%
(B) 6.125%
(C) 6.25%
(D) 5.875%

Solution: $\begin{aligned} Simple interest = 3725 - 2500 = 1225 \\ SI = \frac{P \times r \times t}{100} \\ 1225 = \frac{2500 \times r \times 8}{100} \\ r = \frac{49}{8} \\ r = 6.125 % \end{aligned}$

50.

Simple interest on Rs. 500 for 4 years at 6.25% per annum is equal to the simple interest on Rs. 400 at 5% per annum for a certain period of time. The period of time is = ?

(A) 4 years
(B) 5 years
(C) $6 \frac{1}{4}$ years
(D) $8 \frac{2}{3}$ years

Solution: $\begin{aligned} Let the required time = t years \\ According to the question, \\ \Leftrightarrow \frac{500 \times 4 \times 6.25}{100} = \frac{400 \times 5 \times t}{100} \\ \Leftrightarrow 5 \times 4 \times 625 = 400 \times 5 \times t \\ \Leftrightarrow t = \frac{625}{100} = \frac{25}{4} \\ \Leftrightarrow t = 6 \frac{1}{4} years \end{aligned}$

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Practice MCQ Questions and Answer on Interest

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