Compound Interest Questions and Answers

Answer: C

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

For two years the difference is

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

\[ 2166 = 60000 \times \frac{R^{2}}{10000} = 6R^{2} \]

\[ R^{2} = 361 \quad\Longrightarrow\quad R = 19\% \]

The rate appears squared, so a perfect square is the signal that the arithmetic is on track - here \( 19^{2} = 361 \).

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

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

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

\[ 1 = P(0.04)^{2} = 0.0016P \]

\[ P = \frac{1}{0.0016} = \text{Rs. } 625 \]

Checking: SI = Rs. 50 and CI = \( 625(1.0816 - 1) = 51 \) - the Re. 1 gap is 4% of the first year's Rs. 25 of interest.

Answer: B

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

Over three years,

\[ \text{CI} - \text{SI} = P\left[(1.25)^{3} - 1 - 0.75\right] \]

\[ = P(1.953125 - 1.75) = 0.203125P \]

\[ 320 = 0.203125P \quad\Longrightarrow\quad P = \text{Rs. } 1575.38 \]

At a rate as high as 25% the gap is large relative to the principal - over 20% of it - which is why the answer is so much smaller than in the usual low-rate versions of this question.

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

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

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: D

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

The rate changes each year, so chain the three growth factors.

\[ P \times 1.05 \times 1.12 \times 1.08 = 6350.40 \]

\[ 1.05 \times 1.12 = 1.176, \qquad 1.176 \times 1.08 = 1.27008 \]

\[ P = \frac{6350.40}{1.27008} = \text{Rs. } 5000 \]

The order of the three rates makes no difference, since multiplication commutes - only their product matters.

Answer: D

Apply each year's change as a multiplier in turn; a decrease multiplies by less than 1.

\[ 50000 \times 1.25 \times 0.70 \times 1.40 \]

\[ 1.25 \times 0.70 = 0.875, \qquad 0.875 \times 1.40 = 1.225 \]

\[ 50000 \times 1.225 = 61{,}250 \]

Adding the percentages (\( +25 - 30 + 40 = +35\% \)) would give 67,500. Successive percentage changes never simply add, because each acts on a different base.

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%.