Compound Interest Questions and Answers

Answer: B

When compounded annually, interest 

= 12000[1 + 20/100] - 12000 = Rs.2400 

When compounded semi-annually, interest

 = 12000[1 + 10/100]² - 12000 = Rs.2520 

Required difference = 2520 - 2400 = Rs.120

Answer: C

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

The jump from year 3 to year 4 is one year's growth on the year-3 amount.

\[ \frac{840}{800} = 1 + \frac{R}{100} = 1.05 \]

\[ R = 5\% \]

Equivalently, the Rs. 40 earned in the fourth year is 5% of the Rs. 800 standing at the start of it. No principal or time is needed - only the ratio of two consecutive amounts.

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 \( \text{SI}_{1} \times R \).

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