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Compound Interest Questions and Answers

Important Instructions
91.

Find the sum The difference between the compound and S.I. on a certain sum of money for 2 years at 10% per annum is Rs.15 of money?

Answer: A

Over two years the gap between compound and simple interest is

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

\[ 15 = P(0.10)^{2} = 0.01P \]

\[ P = \text{Rs. } 1500 \]

Checking: SI = Rs. 300 and CI = \( 1500(1.21 - 1) = 315 \) - a Rs. 15 gap, which is 10% of the first year's Rs. 150 of interest.

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92.

The difference between the compound interest compounded annually and simple interest for 2 years at 20% per annum is Rs.144. Find the principal?

Answer: C

Over two years the gap between compound and simple interest is

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

\[ 144 = P(0.20)^{2} = 0.04P \]

\[ P = \frac{144}{0.04} = \text{Rs. } 3600 \]

Checking: SI = Rs. 1440 and CI = \( 3600(1.44 - 1) = 1584 \), differing by Rs. 144 - which is 20% of the first year's Rs. 720.

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93.

What sum of money put at C.I amounts in 2 years to Rs.8820 and in 3 years to Rs.9261?

Answer: A

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

The step from year 2 to year 3 gives the rate.

\[ 1 + \frac{R}{100} = \frac{9261}{8820} = 1.05 \quad\Longrightarrow\quad R = 5\% \]

Now work back two years from Rs. 8820:

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

Getting the rate from two consecutive amounts first is what makes the principal a single division away.

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94.

Find the amount on Rs.8000 in 9 months at 20% per annum, if the interest being compounded quarterly?

Answer: B

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

Quarterly compounding at 20% per annum gives 5% per quarter, and nine months is three quarters.

\[ A = 8000 \times (1.05)^{3} = 8000 \times 1.157625 = \text{Rs. } 9261 \]

The question asks for the amount, not the interest - the Rs. 1261 on offer is the interest alone.

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95.

Every year an amount increases by 1/8th of itself. How much will it be after two years if its present value is Rs.64000?

Answer: A

An increase of \( \tfrac{1}{8} \) each year multiplies the value by \( \tfrac{9}{8} \).

\[ A = 64000 \times \left(\frac{9}{8}\right)^{2} = 64000 \times \frac{81}{64} \]

\[ = 1000 \times 81 = \text{Rs. } 81{,}000 \]

The 64 in the principal cancels the \( 8^{2} \) in the denominator exactly, which is why the answer is so clean. Adding \( \tfrac{1}{8} \) of the original twice would give only Rs. 80,000.

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96.

Rajan borrowed Rs.4000 at 5% p.a compound interest. After 2 years, he repaid Rs.2210 and after 2 more year, the balance with interest. What was the total amount that he paid as interest?

Answer: A

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

Track the debt in two stages.

First two years:

\[ 4000 \times (1.05)^{2} = 4000 \times 1.1025 = \text{Rs. } 4410 \]

He repays Rs. 2210, leaving Rs. 2200.

Next two years:

\[ 2200 \times (1.05)^{2} = 2200 \times 1.1025 = \text{Rs. } 2425.50 \]

\[ \text{total repaid} = 2210 + 2425.50 = \text{Rs. } 4635.50 \]

\[ \text{interest} = 4635.50 - 4000 = \text{Rs. } 635.50 \]

The part-payment reduces the balance that compounds thereafter, so it must be subtracted before the second stage begins.

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97.

Indu gave Bindu Rs.1250 on compound interest for 2 years at 4% per annum. How much loss would Indu has suffered had she given it to Bindu for 2 years at 4% per annum simple interest?

Answer: B

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

\[ \text{CI} = 1250\left[(1.04)^{2} - 1\right] = 1250(1.0816 - 1) = \text{Rs. } 102 \]

\[ \text{SI} = 1250 \times 0.04 \times 2 = \text{Rs. } 100 \]

\[ \text{loss} = 102 - 100 = \text{Rs. } 2 \]

The same figure comes from the shortcut \( P\left(\tfrac{R}{100}\right)^{2} = 1250 \times 0.0016 = 2 \).

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98.

In one year, the population, of a village increased by 10% and in the next year, it decreased by 10%. If at the end of 2nd year, the population was 7920, what was it in the beginning?

Answer: B

Apply the two changes as multipliers in turn.

\[ P \times 1.10 \times 0.90 = 0.99P \]

\[ 0.99P = 7920 \quad\Longrightarrow\quad P = 8000 \]

A rise of 10% followed by a fall of 10% does not return to the start: the fall acts on the larger figure, leaving a net loss of \( \left(\tfrac{10}{100}\right)^{2} = 1\% \). The order of the two makes no difference to that.

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99.

How much more would Rs.20000 fetch, after two years, if it is put at 20% p.a. compound interest payable half yearly than if is put at 20% p.a. compound interest payable yearly?

Answer: A

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

Work out the two schemes separately over two years.

Yearly:

\[ 20000 \times (1.2)^{2} = 20000 \times 1.44 = \text{Rs. } 28{,}800 \]

Half-yearly - 10% per half-year over four periods:

\[ 20000 \times (1.1)^{4} = 20000 \times 1.4641 = \text{Rs. } 29{,}282 \]

\[ \text{difference} = 29282 - 28800 = \text{Rs. } 482 \]

The nominal rate is the same in both; only the frequency of the rests differs, and more frequent rests always earn more.

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100.

A money lender lends Rs.2000 for six months at 20% p.a. rate. It the interest is reckoned quarterly than find the amount given after time limit.

Answer: A

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

Reckoned quarterly, 20% per annum is 5% per quarter, and six months is two quarters.

\[ A = 2000 \times (1.05)^{2} = 2000 \times 1.1025 = \text{Rs. } 2205 \]

Simple interest for the same six months would give Rs. 2200; the extra Rs. 5 is the second quarter's 5% on the first quarter's Rs. 100.

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