Compound Interest Questions and Answers

Answer: D

The effective rate is the single annual rate that produces the same growth as the nominal rate does with its compounding.

Half-yearly payment at a nominal 6% means 3% every six months:

\[ (1.03)^{2} = 1.0609 \]

\[ \text{effective rate} = 1.0609 - 1 = 6.09\% \]

The extra 0.09% over the nominal 6% is the second half-year's 3% earned on the first half-year's 3% - the whole point of the effective-rate idea.

Answer: D

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

Over two years the gap between the two is exactly one year's interest on the first year's interest:

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

\[ 50 = P(0.04)^{2} = 0.0016P \]

\[ P = \frac{50}{0.0016} = \text{Rs. } 31{,}250 \]

Checking: SI = \( 31250 \times 0.08 = 2500 \); CI = \( 31250(1.0816 - 1) = 2550 \) - exactly Rs. 50 apart.

Answer: B

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

\[ A = 8000 \times (1.1)^{2} = 8000 \times 1.21 = \text{Rs. } 9680 \]

The interest is Rs. 1680, of which Rs. 1600 is simple interest and Rs. 80 is the second year's 10% on the first year's Rs. 800.

Answer: A

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

Reckoned quarterly, 20% per annum is 5% per quarter, and six months is two quarters.

\[ A = 2000 \times (1.05)^{2} = 2000 \times 1.1025 = \text{Rs. } 2205 \]

Simple interest for the same six months would give Rs. 2200; the extra Rs. 5 is the second quarter's 5% on the first quarter's Rs. 100.

Answer: C

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

The rates differ by year, so multiply the two factors rather than raising one to a power.

\[ A = 5000 \times 1.04 \times 1.05 \]

\[ = 5000 \times 1.092 = \text{Rs. } 5460 \]

The question asks for the amount, so the Rs. 460 of interest is only an intermediate value - and it is offered as the first distractor.

Answer: B

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

The ratio of consecutive amounts gives the rate directly.

\[ 1 + \frac{R}{100} = \frac{1000}{900} = \frac{10}{9} = 1.1111 \quad\Longrightarrow\quad R = 11.11\% \]

Now work back three years from the Rs. 900:

\[ P = \frac{900}{(10/9)^{3}} = 900 \times \frac{729}{1000} = \text{Rs. } 656.10 \]

The rate \( \tfrac{1}{9} \) is what makes the cube \( \tfrac{729}{1000} \) exact, so the principal comes out to the paisa.

Answer: B

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

\[ \text{CI} = 1250\left[(1.04)^{2} - 1\right] = 1250(1.0816 - 1) = \text{Rs. } 102 \]

\[ \text{SI} = 1250 \times 0.04 \times 2 = \text{Rs. } 100 \]

\[ \text{loss} = 102 - 100 = \text{Rs. } 2 \]

The same figure comes from the shortcut \( P\left(\tfrac{R}{100}\right)^{2} = 1250 \times 0.0016 = 2 \).

Answer: A

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

\[ \left(1 + \frac{R}{100}\right)^{2} = \frac{1348.32}{1200} = 1.1236 \]

\[ 1 + \frac{R}{100} = \sqrt{1.1236} = 1.06 \]

\[ R = 6\% \]

Averaging the total growth of 12.36% over two years to get 6.18% is close but wrong; the square root, not the halving, is what inverts two years of compounding.

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