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Compound Interest Questions and Answers

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111.

The simple interest accrued on an amount of Rs.49500 at he end of three is Rs.17820. What would be the compound interest accured on the same amount at teh same rate in the same period? (Round off your answer to two decimal places)

Answer: A

Let the rate of interest be R% p.a. 

(49500)(3)(R)/100 = 17820 

495R = 5940 

R = 12 

CI on Rs.49500 at 12% p.a. for three years 

= 49500{[1 + 12/100]³ - 1} = 69543.936 - 49500 = Rs.20043.94 

when rounded off to two decimal places.

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112.

The compound interest accrued on an amount of Rs.44000 at the end of two years is Rs.1193.60. What would be the simple interest accrued on the same amount at the same rate in the same period?

Answer: C

Let the rate of interest be R% p.a. 

4400{[1 + R/100]² - 1} = 11193.60 

[1 + R/100]² = (44000 + 11193.60)/44000

[1 + R/100]² = 1 + 2544/1000 = 1 + 159/625 

[1 + R/100]² = 784/625 = (28/25)2 

1 + R/100 = 28/25 

R/100 = 3/25 

Therefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100 

=Rs.10560

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113.

A person got Rs.48 more when he invested a certain sum at compound interest instead of simple interest for two years at 8% p.a. Find the sum?

Answer: A

Over two years the gap between compound and simple interest is

\[ \text{CI} - \text{SI} = P\left(\frac{R}{100}\right)^{2} \]

\[ 48 = P(0.08)^{2} = 0.0064P \]

\[ P = \frac{48}{0.0064} = \text{Rs. } 7500 \]

Checking: SI = Rs. 1200 and CI = \( 7500(1.1664 - 1) = 1248 \) - exactly Rs. 48 more.

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114.

The compound and the simple interests on a certain sum at the same rate of interest for two years are Rs.11730 and Rs.10200 respectively. Find the sum.

Answer: B

The simple interest for the first year is 10200/2 is Rs.5100 and compound interest for first year also is Rs.5100. The compound interest for second year on Rs.5100 for one year 

So rate of the interest = (100 * 1530)/ (5100 * 1) = 30% p.a. 

So P = (100 * 10200)/ (30 * 2) = Rs.17000

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115.

What is the difference between the compound interest on Rs.12000 at 20% p.a. for one year when compounded yearly and half yearly?

Answer: B

When compounded annually, interest 

= 12000[1 + 20/100] - 12000 = Rs.2400 

When compounded semi-annually, interest

 = 12000[1 + 10/100]² - 12000 = Rs.2520 

Required difference = 2520 - 2400 = Rs.120

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116.

The simple interest on Rs. 12000 at a certain rate of interest in five years is Rs. 7200. Find the compound interest on the same amount for two years at the same rate of interest.

Answer: D

\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]

Find the rate from the simple-interest statement first.

\[ 7200 = \frac{12000 \times R \times 5}{100} = 600R \quad\Longrightarrow\quad R = 12\% \]

Now compound the same sum for two years:

\[ \text{CI} = 12000\left[(1.12)^{2} - 1\right] = 12000(1.2544 - 1) \]

\[ = 12000 \times 0.2544 = \text{Rs. } 3052.80 \]

Two years of simple interest would be Rs. 2880; the extra Rs. 172.80 is 12% of the first year's Rs. 1440.

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117.

Ravi invested certain amount for two rates of simple interests at 6% p.a. and 7% p.a. What is the ratio of Ravi's investments if the interests from those investments are equal?

Answer: D

Let the two investments be \( P_{1} \) at 6% and \( P_{2} \) at 7%, over the same period \( n \). Equal interest means

\[ \frac{P_{1} \times 6 \times n}{100} = \frac{P_{2} \times 7 \times n}{100} \]

\[ 6P_{1} = 7P_{2} \quad\Longrightarrow\quad \frac{P_{1}}{P_{2}} = \frac{7}{6} \]

The investments are in the inverse ratio of the rates: the lower rate needs the larger sum to yield the same return. The time cancels, so it need not be known.

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118.

Siddharth wants to borrow Rs.6000 at rate of interest 6% p.a. at S.I and lend the same amount at C.I at same rate of interest for two years. What would be his income in the above transaction?

Answer: A

He pays simple interest and receives compound interest on the same sum at the same rate, so his profit is the gap between the two.

\[ \text{SI paid} = 6000 \times 0.06 \times 2 = \text{Rs. } 720 \]

\[ \text{CI received} = 6000\left[(1.06)^{2} - 1\right] = 6000(1.1236 - 1) = \text{Rs. } 741.60 \]

\[ \text{income} = 741.60 - 720 = \text{Rs. } 21.60 \]

The shortcut gives it in one line: \( P\left(\tfrac{R}{100}\right)^{2} = 6000 \times 0.0036 = 21.60 \).

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119.

Albert invested Rs. 8000 in a scheme for 2 years at compound interest rate 5% p.a. How much amount will Albert get on maturity of the fixed deposit?

Answer: D

\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]

\[ A = 8000 \times (1.05)^{2} = 8000 \times 1.1025 = \text{Rs. } 8820 \]

Rs. 8820 is not among the choices, so the answer is none of these.

The Rs. 8800 on offer is what simple interest would give (Rs. 800 of interest); the extra Rs. 20 is the second year's 5% on the first year's Rs. 400.

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120.

What will be the compound interest on a sum of Rs. 25,000 after 3 years at the rate of 12% p.a.?

Answer: C

\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]

\[ \text{CI} = 25000\left[(1.12)^{3} - 1\right] \]

\[ (1.12)^{3} = 1.404928 \]

\[ \text{CI} = 25000(0.404928) = \text{Rs. } 10{,}123.20 \]

Simple interest would be \( 25000 \times 0.36 = 9000 \), so the compounding adds a little over Rs. 1100 - which is why the 9000-ish option is offered.

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