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Compound Interest Questions and Answers

Important Instructions
101.

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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102.

Find the amount on Rs.5000 in 2 years, the rate of interest being 4% per first year and 5% for the second year?

Answer: C

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

The rates differ by year, so multiply the two factors rather than raising one to a power.

\[ A = 5000 \times 1.04 \times 1.05 \]

\[ = 5000 \times 1.092 = \text{Rs. } 5460 \]

The question asks for the amount, so the Rs. 460 of interest is only an intermediate value - and it is offered as the first distractor.

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103.

In 3 years by compound interest, a sum becomes Rs. 900. But in 4 years by compound interest, the same sum becomes Rs. 1000. What is the sum and the rate of interest?

Answer: B

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

The ratio of consecutive amounts gives the rate directly.

\[ 1 + \frac{R}{100} = \frac{1000}{900} = \frac{10}{9} = 1.1111 \quad\Longrightarrow\quad R = 11.11\% \]

Now work back three years from the Rs. 900:

\[ P = \frac{900}{(10/9)^{3}} = 900 \times \frac{729}{1000} = \text{Rs. } 656.10 \]

The rate \( \tfrac{1}{9} \) is what makes the cube \( \tfrac{729}{1000} \) exact, so the principal comes out to the paisa.

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104.

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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105.

At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years?

Answer: A

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

\[ \left(1 + \frac{R}{100}\right)^{2} = \frac{1348.32}{1200} = 1.1236 \]

\[ 1 + \frac{R}{100} = \sqrt{1.1236} = 1.06 \]

\[ R = 6\% \]

Averaging the total growth of 12.36% over two years to get 6.18% is close but wrong; the square root, not the halving, is what inverts two years of compounding.

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106.

A property decreases in value every year at the rate of 6 1/4% of its value at the beginning of the year its value at the end of 3 years was Rs.21093. Find its value at the beginning of the first year?

Answer: A

A yearly fall of \( 6\tfrac{1}{4}\% = \tfrac{1}{16} \) leaves \( \tfrac{15}{16} \) of the value each year.

\[ P \times \left(\frac{15}{16}\right)^{3} = 21093 \]

\[ P = 21093 \times \frac{16^{3}}{15^{3}} = 21093 \times \frac{4096}{3375} \]

\[ = \text{Rs. } 25{,}600.24 \]

Depreciation is compound too, so the value must be divided by the yearly factor - adding back three lots of 6.25% would give a different and smaller figure.

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107.

Find the sum lend at C.I. at 5 p.c per annum will amount to Rs.441 in 2 years?

Answer: B

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

\[ P \times (1.05)^{2} = 441 \]

\[ P = \frac{441}{1.1025} = \text{Rs. } 400 \]

The numbers are chosen so that \( 441 = 21^{2} \) and \( 400 = 20^{2} \), making the ratio exactly \( \left(\tfrac{21}{20}\right)^{2} \).

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108.

A sum of money becomes Rs 13,380 in 3 years and Rs 20,070 in 6 years at compound interest. The initial sum is?

Answer: C

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

The same growth factor applies to each three-year stretch, so compare the two amounts.

\[ \frac{20070}{13380} = 1.5 \]

That 1.5 is the three-year multiplier, and it also carries the sum from the start to year 3:

\[ P \times 1.5 = 13380 \quad\Longrightarrow\quad P = \text{Rs. } 8920 \]

The annual rate works out at \( \sqrt[3]{1.5} - 1 = 14.47\% \), but it is never needed - the three-year factor does all the work.

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109.

Find the compound interest and the amount on Rs.8000 at 5% per annum for 3 years when C.I is reckoned yearly?

Answer: A

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

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

\[ \text{CI} = 9261 - 8000 = \text{Rs. } 1261 \]

Simple interest would be \( 8000 \times 0.15 = 1200 \); the extra Rs. 61 is the interest earned on earlier interest.

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110.

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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