Compound Interest Questions and Answers

Answer: D

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

Recover the principal from the compound interest. With \( 12\tfrac{1}{2}\% = \tfrac{1}{8} \), the yearly factor is \( \tfrac{9}{8} \).

\[ 510 = P\left[\left(\frac{9}{8}\right)^{2} - 1\right] = P\left(\frac{81}{64} - 1\right) = \frac{17P}{64} \]

\[ P = \frac{510 \times 64}{17} = \text{Rs. } 1920 \]

\[ \text{SI} = 1920 \times 0.125 \times 2 = \text{Rs. } 480 \]

The Rs. 30 by which the compound interest exceeds it is \( P\left(\tfrac{R}{100}\right)^{2} = 1920 \times \tfrac{1}{64} \) - a useful check on the principal.

Answer: A

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

\[ A = 30000 + 4347 = \text{Rs. } 34{,}347 \]

\[ \left(1.07\right)^{n} = \frac{34347}{30000} = 1.1449 \]

\[ (1.07)^{2} = 1.1449 \quad\Longrightarrow\quad n = 2 \text{ years} \]

Recognising 1.1449 as \( 1.07^{2} \) is quicker than taking logarithms; the answer options are small integers precisely so that trial of the first few powers settles it.

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 first.

\[ 400 = \frac{P \times 10 \times 4}{100} = 0.4P \quad\Longrightarrow\quad P = \text{Rs. } 1000 \]

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

\[ = \text{Rs. } 464.10 \]

The Rs. 64.10 by which compound exceeds simple interest is all the interest earned on earlier interest.

Answer: A

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

\[ A = 30000 + 4347 = \text{Rs. } 34{,}347 \]

\[ \left(1.07\right)^{n} = \frac{34347}{30000} = 1.1449 \]

\[ (1.07)^{2} = 1.1449 \quad\Longrightarrow\quad n = 2 \text{ years} \]

Recognising 1.1449 as \( 1.07^{2} \) is quicker than taking logarithms; the answer options are small integers precisely so that trial of the first few powers settles it.

Answer: C

Over two years the gap between compound and simple interest is

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

\[ 144 = P(0.20)^{2} = 0.04P \]

\[ P = \frac{144}{0.04} = \text{Rs. } 3600 \]

Checking: SI = Rs. 1440 and CI = \( 3600(1.44 - 1) = 1584 \), differing by Rs. 144 - which is 20% of the first year's Rs. 720.

Answer: C

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

Each year's saving is lent as it is made, so by the end of the third year the three deposits have been earning for 3, 2 and 1 years respectively.

\[ A = 200(1.05)^{3} + 200(1.05)^{2} + 200(1.05) \]

\[ = 200(1.157625) + 200(1.1025) + 200(1.05) \]

\[ = 231.53 + 220.50 + 210.00 = \text{Rs. } 662.02 \]

The deposits cannot be pooled and compounded together, because each has been invested for a different length of time - treating all Rs. 600 as present from the start would give Rs. 694.58.

Answer: B

Each installment is discounted back to today at 20%, and the three present values must add to the loan.

\[ \frac{x}{1.2} + \frac{x}{(1.2)^{2}} + \frac{x}{(1.2)^{3}} = 11000 \]

\[ x\left(0.833333 + 0.694444 + 0.578704\right) = 11000 \]

\[ 2.106481\,x = 11000 \quad\Longrightarrow\quad x = \text{Rs. } 5221.97 \]

Simply dividing Rs. 11,000 by three gives Rs. 3666.67 - the first option - but that ignores the interest accruing between installments.

Answer: B

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

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

The principal never has to be found. This shortcut - the two-year excess is always one year's interest on one year's interest - solves the whole family of these questions.

Answer: D

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

\[ A = 50000 \times (1.1)^{3} \]

\[ (1.1)^{3} = 1.331 \quad\Longrightarrow\quad A = 50000 \times 1.331 = 66{,}550 \]

Adding 10% three times over would suggest 65,000 under simple growth; the extra 1,550 is the compounding - growth on the growth already added.

Answer: D

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

For a period of whole years plus a part year, compound over the whole years and then apply simple interest for the remainder.

\[ A = 5000 \times (1.30)^{3} \times \left(1 + 0.30 \times \tfrac{10}{12}\right) \]

\[ (1.30)^{3} = 2.197 \quad\Longrightarrow\quad 5000 \times 2.197 = 10{,}985 \]

\[ 10985 \times 1.25 = 13{,}731.25 \]

\[ \text{CI} = 13731.25 - 5000 = \text{Rs. } 8731.25 \]

The ten months contribute \( 0.30 \times \tfrac{10}{12} = 0.25 \), so the final factor is 1.25 - a quarter, not a full year's 30%.