Compound Interest Questions and Answers

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.

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.

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.

Answer: A

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

Take the common sum as \( P \). Over 3 years at 8%:

\[ \text{SI} = P \times 0.08 \times 3 = 0.24P \]

\[ \text{CI} = P\left[(1.08)^{3} - 1\right] = P(1.259712 - 1) = 0.259712P \]

\[ \frac{\text{SI}}{\text{CI}} = \frac{0.24}{0.259712} = \frac{240000}{259712} = \frac{1875}{2029} \]

The sum cancels, so the ratio holds for any principal. Compound interest is always the larger of the two beyond the first year, so the ratio must be less than 1.

Answer: C

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

Quarterly compounding quarters the rate and counts quarters instead of years.

\[ \text{rate} = \frac{20}{4} = 5\% \text{ per quarter}, \qquad n = \frac{9}{3} = 3 \text{ quarters} \]

\[ A = 1600 \times (1.05)^{3} = 1600 \times 1.157625 = \text{Rs. } 1852.20 \]

\[ \text{CI} = 1852.20 - 1600 = \text{Rs. } 252.20 \]

Nine months is three quarters, not nine periods - using \( n = 9 \) would wildly overstate the interest.

Answer: A

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

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

\[ 1 = P(0.04)^{2} = 0.0016P \quad\Longrightarrow\quad P = \text{Rs. } 625 \]

Checking: simple interest is Rs. 50 and compound interest Rs. 51. The Re. 1 gap is 4% of the Rs. 25 earned in the first year, which is exactly what the squared-rate formula encodes.

Answer: B

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

Reckoned half-yearly, the 10% becomes 5% per half-year over two periods.

\[ \text{CI} = 1200\left[(1.05)^{2} - 1\right] = 1200(1.1025 - 1) = \text{Rs. } 123 \]

\[ \text{SI} = 1200 \times 0.10 \times 1 = \text{Rs. } 120 \]

\[ \text{difference} = 123 - 120 = \text{Rs. } 3 \]

With annual compounding over a single year the two would be identical; it is the half-yearly rest that creates any gap at all.

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.

Answer: D

The population grows at 25% a year, so past figures are found by dividing, not subtracting.

\[ P_{-3} = \frac{100000}{(1.25)^{3}} = \frac{100000}{1.953125} = 51{,}200 \]

\[ P_{-2} = \frac{100000}{(1.25)^{2}} = \frac{100000}{1.5625} = 64{,}000 \]

\[ \text{difference} = 64000 - 51200 = 12{,}800 \]

Reducing by 25% a year instead of dividing by 1.25 would give 42,187 and 56,250 - a quite different pair. Growth of 25% is undone by a cut of 20%, not 25%.

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.