Compound Interest Questions and Answers

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

Over two years the gap between compound and simple interest is

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

\[ 96 = 15000 \times \frac{R^{2}}{10000} = 1.5R^{2} \]

\[ R^{2} = 64 \quad\Longrightarrow\quad R = 8\% \]

The rate is fully determined, so "cannot be determined" is a trap: for a two-year period the difference, the principal and the rate are locked together by that single relation.

Answer: B

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

Recover the principal from the compound interest.

\[ 525 = P\left[(1.1)^{2} - 1\right] = 0.21P \quad\Longrightarrow\quad P = \text{Rs. } 2500 \]

Double the time is 4 years and half the rate is 5%:

\[ \text{SI} = \frac{2500 \times 5 \times 4}{100} = \text{Rs. } 500 \]

Doubling the time while halving the rate leaves simple interest unchanged, since \( Rn \) is the same - the answer would be identical for 2 years at 10%.

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: D

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

\[ A = 7500 \times (1.04)^{2} = 7500 \times 1.0816 = \text{Rs. } 8112 \]

The interest is Rs. 612, of which Rs. 600 is simple interest and Rs. 12 is the second year's 4% on the first year's Rs. 300 - which is why Rs. 8100 is offered as a near miss.

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.

\[ \text{rate} = 2\% \text{ per half-year}, \qquad n = 1\tfrac{1}{2} \times 2 = 3 \]

\[ A = 5000 \times (1.02)^{3} = 5000 \times 1.061208 = \text{Rs. } 5306.04 \]

\[ \text{CI} = \text{Rs. } 306.04 \]

The one-and-a-half years become three whole half-year periods, which is exactly why half-yearly rests are convenient here.

Answer: D

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

\[ (1.05)^{n} = \frac{9261}{8000} = 1.157625 \]

Testing the small powers:

\[ (1.05)^{2} = 1.1025, \qquad (1.05)^{3} = 1.157625 \]

\[ n = 3 \text{ years} \]

The numbers 8000 and 9261 are \( 20^{3} \) and \( 21^{3} \), so the ratio is \( \left(\tfrac{21}{20}\right)^{3} \) - the cube is visible without any multiplication.

Answer: A

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

\[ \text{CI} = 20000\left[(1.1)^{3} - 1\right] = 20000(1.331 - 1) \]

\[ = 20000 \times 0.331 = \text{Rs. } 6620 \]

Simple interest would be Rs. 6000; the extra Rs. 620 is the compounding. The multiplier \( 1.1^{3} = 1.331 \) is worth memorising, as it appears constantly in these questions.

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

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

Recover the principal from the simple interest.

\[ 80 = \frac{P \times 4 \times 2}{100} = 0.08P \quad\Longrightarrow\quad P = \text{Rs. } 1000 \]

\[ \text{CI} = 1000\left[(1.04)^{2} - 1\right] = 1000(1.0816 - 1) = \text{Rs. } 81.60 \]

The Rs. 1.60 excess is 4% of the first year's Rs. 40 of interest - the whole of what compounding adds over two years.