Compound Interest Questions and Answers

Answer: A

He pays simple interest and receives compound interest on the same sum at the same rate, so his profit is the gap between the two.

\[ \text{SI paid} = 6000 \times 0.06 \times 2 = \text{Rs. } 720 \]

\[ \text{CI received} = 6000\left[(1.06)^{2} - 1\right] = 6000(1.1236 - 1) = \text{Rs. } 741.60 \]

\[ \text{income} = 741.60 - 720 = \text{Rs. } 21.60 \]

The shortcut gives it in one line: \( P\left(\tfrac{R}{100}\right)^{2} = 6000 \times 0.0036 = 21.60 \).

Answer: B

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

The rates differ by year, so multiply the two growth factors rather than raising one to a power. The second year's rate is \( 10 + 2 = 12\% \).

\[ A = 77500 \times 1.10 \times 1.12 \]

\[ = 77500 \times 1.232 = 95{,}480 \]

\[ \text{CI} = 95480 - 77500 = \text{Rs. } 17{,}980 \]

Averaging the rates to 11% and squaring would give Rs. 17,975.98 - close, but not the same, because compounding is not linear in the rate.

Answer: C

The two sources of growth act in the same year, so combine them into a single annual rate.

\[ 4\% + 1\% = 5\% \text{ per annum} \]

Population growth compounds, so over two years:

\[ (1.05)^{2} = 1.1025 \]

\[ \text{increase} = 10.25\% \]

Simply doubling the 5% would give 10%; the extra 0.25% is the second year's growth on the first year's addition.

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.

\[ \text{rate} = 5\% \text{ per half-year}, \qquad n = 2 \]

\[ A = 15000 \times (1.05)^{2} = 15000 \times 1.1025 = \text{Rs. } 16{,}537.50 \]

Annual compounding at 10% would give Rs. 16,500; the extra Rs. 37.50 is the second half-year's 5% on the first half-year's Rs. 750.

Answer: C

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

\[ \text{CI} = 25000\left[(1.12)^{3} - 1\right] \]

\[ (1.12)^{3} = 1.404928 \]

\[ \text{CI} = 25000(0.404928) = \text{Rs. } 10{,}123.20 \]

Simple interest would be \( 25000 \times 0.36 = 9000 \), so the compounding adds a little over Rs. 1100 - which is why the 9000-ish option is offered.

Answer: D

Let the rate of interest be R% p.a. 

17400[1 + R/100]² = 17400 + 1783.50 

[1 + R/100]² = (17400 + 1783.50)/17400 

= 1 + 1025/10000 = 1 + 41/400 

= 441/400 = [21/20]² 

[1 + R/100] = 21/20 

R/100 = 1/20 

Therefore R = 5

Answer: A

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

\[ \text{SI} = 1000 \times 0.10 \times 3 = \text{Rs. } 300 \]

\[ \text{CI} = 1000\left[(1.1)^{3} - 1\right] = 1000(1.331 - 1) = \text{Rs. } 331 \]

\[ \text{difference} = 331 - 300 = \text{Rs. } 31 \]

Using the three-year formula \( P\left[\left(1+\tfrac{R}{100}\right)^{3} - 1 - \tfrac{3R}{100}\right] = 1000(1.331 - 1.3) \) gives the same Rs. 31 in one step.

Answer: D

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

Recover the principal from the compound interest. With \( 12\tfrac{1}{2}\% = \tfrac{1}{8} \), the yearly factor is \( \tfrac{9}{8} \).

\[ 510 = P\left[\left(\frac{9}{8}\right)^{2} - 1\right] = P\left(\frac{81}{64} - 1\right) = \frac{17P}{64} \]

\[ P = \frac{510 \times 64}{17} = \text{Rs. } 1920 \]

\[ \text{SI} = 1920 \times 0.125 \times 2 = \text{Rs. } 480 \]

The Rs. 30 by which the compound interest exceeds it is \( P\left(\tfrac{R}{100}\right)^{2} = 1920 \times \tfrac{1}{64} \) - a useful check on the principal.

Answer: A

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

\[ A = 30000 + 4347 = \text{Rs. } 34{,}347 \]

\[ \left(1.07\right)^{n} = \frac{34347}{30000} = 1.1449 \]

\[ (1.07)^{2} = 1.1449 \quad\Longrightarrow\quad n = 2 \text{ years} \]

Recognising 1.1449 as \( 1.07^{2} \) is quicker than taking logarithms; the answer options are small integers precisely so that trial of the first few powers settles it.

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 first.

\[ 400 = \frac{P \times 10 \times 4}{100} = 0.4P \quad\Longrightarrow\quad P = \text{Rs. } 1000 \]

\[ \text{CI} = 1000\left[(1.1)^{4} - 1\right] = 1000(1.4641 - 1) \]

\[ = \text{Rs. } 464.10 \]

The Rs. 64.10 by which compound exceeds simple interest is all the interest earned on earlier interest.