Practice MCQ Questions and Answer on Interest

71.

What sum will amount to Rs. 7000 in 5 years at $3 \frac{1}{3}$ % simple interest ?

(A) Rs. 6300
(B) Rs. 6500
(C) Rs. 6000
(D) Rs. 5000

Solution: $\begin{aligned} Amount Rs 7000 \\ Total interest in 5 years \\ = 5 \times \frac{10}{3} % = \frac{50}{3} % = \frac{1}{6} \end{aligned}$ Principal Amount 6 (6 + 1) ↓ × 1000 ↓ × 1000 6000 7000 Hence required principal = Rs. 6000

72.

A sum of Rs. 1440 is lent out in three parts in such away that the interests on first part at 2% for 3 years, second part at 3% for 4 years and third part at 4% for 5 years are equal. Then the difference between the largest and the smallest sum is -

(A) Rs. 200
(B) Rs. 400
(C) Rs. 460
(D) Rs. 560

Solution: Let the parts be Rs. x, Rs. y and Rs. [1440 - (x + y)]. Then, $\begin{aligned} = \frac{x \times 2 \times 3}{100} = \frac{y \times 3 \times 4}{100} \\ = \frac{[1440 - (x + y)] \times 4 \times 5}{100} \\ ∴ 6 x = 12 y o r x = 2 y . \\ S o, \frac{x \times 2 \times 3}{100} = \frac{[1440 - (x + y)] \times 4 \times 5}{100} \\ \Leftrightarrow 12 y = (1440 - 3 y) \times 20 \\ \Rightarrow 72 y = 28800 \\ \Rightarrow y = 400 \\ First part = x = 2 y = Rs . 800, \\ Second part = Rs . 400 \\ Third part \\ = Rs . [1440 - (800 + 400)] \\ = Rs . 240. \\ ∴ Required difference \\ = Rs . (800 - 240) \\ = Rs .560 \end{aligned}$

73.

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

74.

A sum amounts to Rs. 14,395.20 at 9.25% p.a. simple interest in 5.4 years. What will be the simple interest on the same sum at 8.6% p.a. in 4.5 years?

(A) Rs. 3,715.20
(B) Rs. 3,627
(C) Rs. 3,797.76
(D) Rs. 3,672

Solution: $\begin{aligned} \frac{P \times 5.4 \times 9.25}{100} = (14395.20 - P) \\ P = 9600 \\ \frac{9600 \times 8.6 \times 4.5}{100} = SI \\ SI = 3715.20 \end{aligned}$

75.

A some of money lent out at simple interest amount to Rs. 720 after 2 years and Rs. 1020 after a further period of 5 years. Find the principal ?

(A) Rs. 6000
(B) Rs. 600
(C) Rs. 1740
(D) Rs. 120

76.

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

77.

A person deposited Rs. 15,600 in a fixed deposit at 10% per annum simple interest. After every second year he adds his interest earned to the principal. The interest at the end of 4 years is:

(A) Rs. 6,655
(B) Rs. 6,864
(C) Rs. 3,975
(D) Rs. 3,744

Solution: $\begin{aligned} S .I . for 2 years \\ = \frac{15600 \times 2 \times 10}{100} \\ = 3120 \\ Principal for next 2 years \\ = 15600 + 3120 \\ = 18720 \\ S .I . for next 2 years \\ = \frac{18720 \times 10 \times 2}{100} \\ = 3744 \\ Total S .I . \\ = 3120 + 3744 \\ = 6864 \end{aligned}$

78.

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 $\begin{aligned} \Rightarrow SI = \frac{P \times R \times T}{100} \\ \Rightarrow SI = \frac{36000 \times 9.5 \times 146}{100} \\ \Rightarrow SI = Rs . 1368 \end{aligned}$

79.

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

80.

A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest?

(A) 3.6%
(B) 4.5%
(C) 5%
(D) 6%

Solution: Answer & Solution Answer: Option E Solution: Let the original rate be R%. Then, new rate = (2R)%. Note: Here, original rate is for 1 year(s); the new rate is for only 4 months i.e. $\frac{1}{3}$ year(s). $\begin{aligned} ∴ \frac{725 \times R \times 1}{100} + \frac{362.50 \times 2 R \times 1}{100 \times 3} \\ = 33.50 \\ \Rightarrow (2175 + 725) R = 33.50 \times 100 \times 3 \\ \Rightarrow (2175 + 725) R = 10050 \\ \Rightarrow (2900) R = 10050 \\ \Rightarrow R = \frac{10050}{2900} = 3.46 \\ ∴ Original rate = 3.46 % \end{aligned}$

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Practice MCQ Questions and Answer on Interest

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