Compound Interest Questions and Answers

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: 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.

\[ \text{rate} = \frac{4}{2} = 2\% \text{ per half-year}, \qquad n = 2 \times 2 = 4 \]

\[ A = 25000 \times (1.02)^{4} = 25000 \times 1.08243216 = 27{,}060.804 \]

\[ \text{CI} = 27060.804 - 25000 = \text{Rs. } 2060.804 \]

Annual compounding at 4% would give only Rs. 2040 - compounding more often always earns a little more.

Answer: D

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

Find the rate from the simple-interest statement first.

\[ 7200 = \frac{12000 \times R \times 5}{100} = 600R \quad\Longrightarrow\quad R = 12\% \]

Now compound the same sum for two years:

\[ \text{CI} = 12000\left[(1.12)^{2} - 1\right] = 12000(1.2544 - 1) \]

\[ = 12000 \times 0.2544 = \text{Rs. } 3052.80 \]

Two years of simple interest would be Rs. 2880; the extra Rs. 172.80 is 12% of the first year's Rs. 1440.

Answer: C

Each year's interest is the previous year's grown by the rate, because the balance itself grows by that factor.

\[ 1 + \frac{R}{100} = \frac{1440}{1200} = 1.2 \]

\[ R = 20\% \text{ p.a.} \]

The difference of Rs. 240 between the two years is exactly 20% of the earlier Rs. 1200 - the interest earned on that interest. Neither the principal nor the first year's interest is needed.

Answer: C

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

\[ A = 8000 \times (1.05)^{2} = 8000 \times 1.1025 = \text{Rs. } 8820 \]

The interest is Rs. 820, of which Rs. 800 is simple interest and Rs. 20 is the second year's 5% on the first year's Rs. 400 - which is why Rs. 8800 is not the answer.

Answer: B

Apply the two changes as multipliers in turn.

\[ P \times 1.10 \times 0.90 = 0.99P \]

\[ 0.99P = 7920 \quad\Longrightarrow\quad P = 8000 \]

A rise of 10% followed by a fall of 10% does not return to the start: the fall acts on the larger figure, leaving a net loss of \( \left(\tfrac{10}{100}\right)^{2} = 1\% \). The order of the two makes no difference to that.

Answer: B

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

The step from year 2 to year 3 is one year's growth.

\[ 1 + \frac{R}{100} = \frac{2662}{2420} = 1.1 \]

\[ R = 10\% \]

Equivalently, the Rs. 242 earned in the third year is 10% of the Rs. 2420 standing at its start. Neither the principal nor the total time is needed.

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