Compound Interest Questions and Answers

Answer: A

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

Track the debt in two stages.

First two years:

\[ 4000 \times (1.05)^{2} = 4000 \times 1.1025 = \text{Rs. } 4410 \]

He repays Rs. 2210, leaving Rs. 2200.

Next two years:

\[ 2200 \times (1.05)^{2} = 2200 \times 1.1025 = \text{Rs. } 2425.50 \]

\[ \text{total repaid} = 2210 + 2425.50 = \text{Rs. } 4635.50 \]

\[ \text{interest} = 4635.50 - 4000 = \text{Rs. } 635.50 \]

The part-payment reduces the balance that compounds thereafter, so it must be subtracted before the second stage begins.

Answer: D

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

\[ \text{SI} = 1000 \times 0.10 \times 4 = \text{Rs. } 400 \]

\[ \text{CI} = 1000\left[(1.1)^{4} - 1\right] = 1000(1.4641 - 1) = \text{Rs. } 464.10 \]

\[ \text{difference} = 464.10 - 400 = \text{Rs. } 64.10 \]

The gap widens sharply with time: over three years it is only Rs. 31, and over two years Rs. 10. That growth is the compounding acting on ever more accumulated interest.

Answer: A

Discount both installments back to the present at 20%.

\[ \frac{x}{1.2} + \frac{x}{(1.2)^{2}} = 11000 \]

\[ x\left(\frac{5}{6} + \frac{25}{36}\right) = x \times \frac{55}{36} = 11000 \]

\[ x = \frac{11000 \times 36}{55} = \text{Rs. } 7200 \]

Checking: after one year the debt is \( 11000 \times 1.2 = 13200 \); paying 7200 leaves 6000, which grows to 7200 by the second year and is cleared exactly.

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.

\[ 50 = \frac{P \times 5 \times 2}{100} = 0.1P \quad\Longrightarrow\quad P = \text{Rs. } 500 \]

\[ \text{CI} = 500\left[(1.05)^{2} - 1\right] = 500(1.1025 - 1) \]

\[ = \text{Rs. } 51.25 \]

The Rs. 1.25 excess is 5% of the first year's Rs. 25 of interest - the only difference compounding makes over two years.

Answer: D

Compound growth multiplies, so tripling twice gives a ninefold increase.

\[ \left(1 + \frac{R}{100}\right)^{3} = 3 \]

\[ \left(1 + \frac{R}{100}\right)^{6} = 3^{2} = 9 \]

So it takes 6 years, and the rate is never needed. Under simple interest the answer would be 12 years instead, because there the growth is additive rather than multiplicative.

Answer: B

Over two years the gap between compound and simple interest is

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

\[ 72 = 5000 \times \frac{R^{2}}{10000} = 0.5R^{2} \]

\[ R^{2} = 144 \quad\Longrightarrow\quad R = 12\% \]

Checking: SI = Rs. 1200 and CI = \( 5000(1.2544 - 1) = 1272 \) - exactly Rs. 72 apart.

Answer: B

Let the sum be Rs.P

 P{ [ 1 + 8/100]² - 1 } = 2828.80 

P(8/100)(2 + 8/100) = 2828.80 [a² - b² = (a - b) ( a + b)] P = 2828.80 / (0.08)(2.08) = 1360/0.08 = 17000 Principal + Interest = Rs. 19828.80

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 nine months is three quarters.

\[ \text{rate} = \frac{16}{4} = 4\% \text{ per quarter}, \qquad n = 3 \]

\[ A = 15625 \times (1.04)^{3} = 15625 \times 1.124864 = \text{Rs. } 17{,}576 \]

\[ \text{CI} = 17576 - 15625 = \text{Rs. } 1951 \]

The principal is \( 25^{3} \) and the amount \( 26^{3} \), since the factor is \( \left(\tfrac{26}{25}\right)^{3} \) - which is why both come out as whole numbers.

Answer: B

Compound growth multiplies, so express 64 as a power of the known growth factor.

\[ \left(1 + \frac{R}{100}\right)^{6} = 4 = 2^{2} \quad\Longrightarrow\quad \left(1 + \frac{R}{100}\right)^{3} = 2 \]

So the money doubles every 3 years. Since \( 64 = 2^{6} \), six doublings are needed:

\[ 6 \times 3 = 18 \text{ years} \]

Equivalently \( 64 = 4^{3} \), and three quadruplings take \( 3 \times 6 = 18 \) years - the same answer by a different route.

Answer: A

Over two years the gap between compound and simple interest is

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

\[ 15 = P(0.10)^{2} = 0.01P \]

\[ P = \text{Rs. } 1500 \]

Checking: SI = Rs. 300 and CI = \( 1500(1.21 - 1) = 315 \) - a Rs. 15 gap, which is 10% of the first year's Rs. 150 of interest.