Compound Interest Questions and Answers

Answer: B

Simple interest of Rs. 40 over two years means Rs. 20 in each year. Compound interest exceeds it by Re. 1, and that excess is one year's interest on the first year's Rs. 20:

\[ 20 \times \frac{R}{100} = 1 \quad\Longrightarrow\quad R = 5\% \]

The principal never has to be found. This shortcut - the two-year excess is always one year's interest on one year's interest - solves the whole family of these questions.

Answer: D

Compound growth multiplies, so tripling twice gives a ninefold increase.

\[ \left(1 + \frac{R}{100}\right)^{3} = 3 \]

\[ \left(1 + \frac{R}{100}\right)^{6} = 3^{2} = 9 \]

So it takes 6 years, and the rate is never needed. Under simple interest the answer would be 12 years instead, because there the growth is additive rather than multiplicative.

Answer: C

\[ \left(1 + \frac{R}{100}\right)^{4} = 2 \]

Eight is \( 2^{3} \), so three doubling periods are needed:

\[ \left(1 + \frac{R}{100}\right)^{12} = 2^{3} = 8 \]

\[ 3 \times 4 = 12 \text{ years} \]

Multiplying the four years by eight would give 32 - the mistake of treating compound growth as if it were proportional to time.

Answer: B

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

The step from year 2 to year 3 is one year's growth.

\[ 1 + \frac{R}{100} = \frac{2662}{2420} = 1.1 \]

\[ R = 10\% \]

Equivalently, the Rs. 242 earned in the third year is 10% of the Rs. 2420 standing at its start. Neither the principal nor the total time is needed.

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 \( \left(\tfrac{21}{20}\right)^{2} \).

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.

Answer: D

The sum more than doubles once \( (1.25)^{n} > 2 \).

\[ (1.25)^{2} = 1.5625, \qquad (1.25)^{3} = 1.953125 \]

\[ (1.25)^{4} = 2.4414 \]

The third year still falls just short of doubling, so the answer is 4 years.

The near miss at year 3 - 1.953, only 2.3% below 2 - is exactly what the question is testing.

Answer: B

Over two years the gap between compound and simple interest is

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

\[ 72 = 5000 \times \frac{R^{2}}{10000} = 0.5R^{2} \]

\[ R^{2} = 144 \quad\Longrightarrow\quad R = 12\% \]

Checking: SI = Rs. 1200 and CI = \( 5000(1.2544 - 1) = 1272 \) - exactly Rs. 72 apart.

Answer: D

The rate \( 6\tfrac{2}{3}\% \) is \( \tfrac{1}{15} \), so the yearly factor is \( \tfrac{16}{15} \). Over three years:

\[ \text{CI} - \text{SI} = P\left[\left(\frac{16}{15}\right)^{3} - 1 - \frac{3}{15}\right] \]

\[ \left(\frac{16}{15}\right)^{3} = \frac{4096}{3375} = 1.213629 \]

\[ 1.213629 - 1.2 = 0.013629 \]

\[ P = \frac{184}{0.013629} = \text{Rs. } 13{,}500 \]

Turning the awkward percentage into the fraction \( \tfrac{1}{15} \) first keeps the cube exact and avoids rounding drift.

Answer: D

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

Compare the two schemes over the same two years.

Yearly: \( P(1.2)^{2} = 1.44P \)

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

\[ P(1.1)^{4} = 1.4641P \]

\[ 1.4641P - 1.44P = 0.0241P = 482 \]

\[ P = \frac{482}{0.0241} = \text{Rs. } 20{,}000 \]

More frequent rests always earn a little more on the same nominal rate; here the gain is 2.41% of the principal.