Compound Interest Questions and Answers

Answer: C

Going backwards in time means dividing by the growth factor.

\[ P_{-4} = \frac{50000}{(1.1)^{4}} = \frac{50000}{1.4641} \]

\[ = 34{,}150.67 \approx 34{,}151 \]

Subtracting 10% four times instead would give \( 50000 \times 0.9^{4} = 32{,}805 \) - a different figure, because a 10% rise is undone by a fall of \( \tfrac{1}{11} \), not a fall of 10%.

Answer: A

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

Quarterly compounding splits the annual rate into four, and six months is two such quarters.

\[ \text{rate per quarter} = \frac{25}{4} = 6.25\%, \qquad n = 2 \]

\[ A = 5500 \times (1.0625)^{2} = 5500 \times 1.12890625 = 6208.98 \]

\[ \text{interest} = 6208.98 - 5500 = \text{Rs. } 708.98 \]

Simple interest for the same six months would be \( 5500 \times 0.25 \times 0.5 = 687.50 \); the extra Rs. 21.48 is the second quarter's interest on the first quarter's interest.

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: 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: B

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

The ratio of consecutive amounts gives the rate directly.

\[ 1 + \frac{R}{100} = \frac{1000}{900} = \frac{10}{9} = 1.1111 \quad\Longrightarrow\quad R = 11.11\% \]

Now work back three years from the Rs. 900:

\[ P = \frac{900}{(10/9)^{3}} = 900 \times \frac{729}{1000} = \text{Rs. } 656.10 \]

The rate \( \tfrac{1}{9} \) is what makes the cube \( \tfrac{729}{1000} \) exact, so the principal comes out to the paisa.

Answer: B

Compound growth multiplies, so two doubling periods give a quadrupling.

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

So it takes 6 years. The rate itself is never needed.

Under simple interest the answer would instead be 9 years, since there the interest earned each year stays constant.

Answer: A

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

\[ \text{SI} = 1000 \times 0.10 \times 3 = \text{Rs. } 300 \]

\[ \text{CI} = 1000\left[(1.1)^{3} - 1\right] = 1000(1.331 - 1) = \text{Rs. } 331 \]

\[ \text{difference} = 331 - 300 = \text{Rs. } 31 \]

Using the three-year formula \( P\left[\left(1+\tfrac{R}{100}\right)^{3} - 1 - \tfrac{3R}{100}\right] = 1000(1.331 - 1.3) \) gives the same Rs. 31 in one step.

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: C

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

The same growth factor applies to each three-year stretch, so compare the two amounts.

\[ \frac{20070}{13380} = 1.5 \]

That 1.5 is the three-year multiplier, and it also carries the sum from the start to year 3:

\[ P \times 1.5 = 13380 \quad\Longrightarrow\quad P = \text{Rs. } 8920 \]

The annual rate works out at \( \sqrt[3]{1.5} - 1 = 14.47\% \), but it is never needed - the three-year factor does all the work.

Answer: B

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

Half-yearly compounding at 5% per annum means 2.5% per half-year. The January deposit earns for two half-years and the July deposit for one.

\[ A = 1600(1.025)^{2} + 1600(1.025) \]

\[ = 1600(1.050625) + 1640 = 1681 + 1640 = \text{Rs. } 3321 \]

\[ \text{interest} = 3321 - 3200 = \text{Rs. } 121 \]

Each deposit must be tracked from its own date; treating the full Rs. 3200 as present all year would give Rs. 162.