Compound Interest Questions and Answers

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: 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: 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: B

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

P is 13 and so waits 12 years; Q is 14 and waits 11. Both end with the same amount:

\[ P_{\text{share}}(1.04)^{12} = Q_{\text{share}}(1.04)^{11} \]

\[ Q_{\text{share}} = P_{\text{share}} \times 1.04 \]

The younger boy's money compounds for an extra year, so he needs less of it. Their shares are therefore in the ratio \( 25 : 26 \):

\[ P_{\text{share}} + 1.04\,P_{\text{share}} = 25500 \quad\Longrightarrow\quad 2.04\,P_{\text{share}} = 25500 \]

\[ P_{\text{share}} = 12{,}500, \qquad Q_{\text{share}} = \text{Rs. } 13{,}000 \]

Checking: \( 12500 \times 1.04 = 13000 \), so after Q's extra year the two balances coincide.

Answer: D

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

Prakash's money compounds for 17 years and Prashant's for 19, and both end equal:

\[ x(1.04)^{17} = y(1.04)^{19} \quad\Longrightarrow\quad x = y(1.04)^{2} = 1.0816\,y \]

Prashant's money has two extra years to grow, so he needs the smaller share.

\[ 1.0816y + y = 1301 \quad\Longrightarrow\quad 2.0816y = 1301 \]

\[ y = \text{Rs. } 625 \]

Prakash therefore gets Rs. 676, and indeed \( 625 \times 1.0816 = 676 \).

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: C

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

Recover the principal from the compound interest first.

\[ \text{CI} = P\left[(1.03)^{2} - 1\right] = P(0.0609) = 101.50 \]

\[ P = \frac{101.50}{0.0609} = \text{Rs. } 1666.67 \]

\[ \text{SI} = 1666.67 \times 0.03 \times 2 = \text{Rs. } 100 \]

The Rs. 1.50 gap between the two is interest on the first year's Rs. 50 of interest - which is the whole of what compounding adds over two years.

Answer: D

Compound growth is multiplicative, so doubling twice is the same as quadrupling.

\[ \left(1 + \frac{R}{100}\right)^{9} = 2 \]

\[ \left(1 + \frac{R}{100}\right)^{18} = \left[\left(1 + \frac{R}{100}\right)^{9}\right]^{2} = 2^{2} = 4 \]

So it takes 18 years. The rate never needs to be found.

This multiplicative behaviour is what separates compound from simple interest: under simple interest, doubling in 9 years means quadrupling in 27, since there the growth is additive.

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: B

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

Simple interest of Rs. 40 over two years means Rs. 20 each year. Compound interest exceeds it by Re. 1, and that excess is one year's interest on the first year's Rs. 20:

\[ 20 \times \frac{R}{100} = 1 \quad\Longrightarrow\quad R = 5\% \]

Now use the simple interest to find the principal:

\[ \frac{P \times 5 \times 2}{100} = 40 \quad\Longrightarrow\quad P = \text{Rs. } 400 \]

Reading the two-year gap as a rate directly is the standard shortcut for these, and it works because over two years the excess is always \( \text{SI}_{1} \times R \).