Compound Interest Questions and Answers

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: 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.

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.

Answer: D

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

Over three years,

\[ \text{CI} - \text{SI} = P\left[\left(1 + \frac{R}{100}\right)^{3} - 1 - \frac{3R}{100}\right] \]

\[ = P\left[(1.05)^{3} - 1 - 0.15\right] = P(1.157625 - 1.15) = 0.007625P \]

\[ 122 = 0.007625P \quad\Longrightarrow\quad P = \text{Rs. } 16{,}000 \]

Checking: SI = 2400 and CI = \( 16000 \times 0.157625 = 2522 \) - a difference of exactly Rs. 122.

Answer: C

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

Half-yearly compounding halves the rate and doubles the number of periods.

\[ \text{rate} = \frac{4}{2} = 2\% \text{ per half-year}, \qquad n = 2 \times 2 = 4 \]

\[ A = 25000 \times (1.02)^{4} = 25000 \times 1.08243216 = 27{,}060.804 \]

\[ \text{CI} = 27060.804 - 25000 = \text{Rs. } 2060.804 \]

Annual compounding at 4% would give only Rs. 2040 - compounding more often always earns a little more.

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.

Answer: C

Each year's interest is the previous year's grown by the rate, because the balance itself grows by that factor.

\[ 1 + \frac{R}{100} = \frac{1440}{1200} = 1.2 \]

\[ R = 20\% \text{ p.a.} \]

The difference of Rs. 240 between the two years is exactly 20% of the earlier Rs. 1200 - the interest earned on that interest. Neither the principal nor the first year's interest is needed.

Answer: C

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

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

The interest is Rs. 820, of which Rs. 800 is simple interest and Rs. 20 is the second year's 5% on the first year's Rs. 400 - which is why Rs. 8800 is not the answer.

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.

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.