Compound Interest Questions and Answers

Answer: D

The population grows at 25% a year, so past figures are found by dividing, not subtracting.

\[ P_{-3} = \frac{100000}{(1.25)^{3}} = \frac{100000}{1.953125} = 51{,}200 \]

\[ P_{-2} = \frac{100000}{(1.25)^{2}} = \frac{100000}{1.5625} = 64{,}000 \]

\[ \text{difference} = 64000 - 51200 = 12{,}800 \]

Reducing by 25% a year instead of dividing by 1.25 would give 42,187 and 56,250 - a quite different pair. Growth of 25% is undone by a cut of 20%, not 25%.

Answer: D

The rate \( 6\tfrac{2}{3}\% \) is \( \tfrac{1}{15} \), so the yearly factor is \( \tfrac{16}{15} \). Over three years:

\[ \text{CI} - \text{SI} = P\left[\left(\frac{16}{15}\right)^{3} - 1 - \frac{3}{15}\right] \]

\[ \left(\frac{16}{15}\right)^{3} = \frac{4096}{3375} = 1.213629 \]

\[ 1.213629 - 1.2 = 0.013629 \]

\[ P = \frac{184}{0.013629} = \text{Rs. } 13{,}500 \]

Turning the awkward percentage into the fraction \( \tfrac{1}{15} \) first keeps the cube exact and avoids rounding drift.

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: 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: D

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

\[ A = 7500 \times (1.04)^{2} = 7500 \times 1.0816 = \text{Rs. } 8112 \]

The interest is Rs. 612, of which Rs. 600 is simple interest and Rs. 12 is the second year's 4% on the first year's Rs. 300 - which is why Rs. 8100 is offered as a near miss.

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: C

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

\[ \text{CI} = 25000\left[(1.12)^{3} - 1\right] \]

\[ (1.12)^{3} = 1.404928 \]

\[ \text{CI} = 25000(0.404928) = \text{Rs. } 10{,}123.20 \]

Simple interest would be \( 25000 \times 0.36 = 9000 \), so the compounding adds a little over Rs. 1100 - which is why the 9000-ish option is offered.

Answer: B

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

Reckoned half-yearly, the 10% becomes 5% per half-year over two periods.

\[ \text{CI} = 1200\left[(1.05)^{2} - 1\right] = 1200(1.1025 - 1) = \text{Rs. } 123 \]

\[ \text{SI} = 1200 \times 0.10 \times 1 = \text{Rs. } 120 \]

\[ \text{difference} = 123 - 120 = \text{Rs. } 3 \]

With annual compounding over a single year the two would be identical; it is the half-yearly rest that creates any gap at all.

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.

\[ \text{rate} = 2\% \text{ per half-year}, \qquad n = 1\tfrac{1}{2} \times 2 = 3 \]

\[ A = 5000 \times (1.02)^{3} = 5000 \times 1.061208 = \text{Rs. } 5306.04 \]

\[ \text{CI} = \text{Rs. } 306.04 \]

The one-and-a-half years become three whole half-year periods, which is exactly why half-yearly rests are convenient here.

Answer: D

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

The rate \( 6\tfrac{1}{4}\% \) is \( \tfrac{1}{16} \), so the yearly factor is \( \tfrac{17}{16} \).

\[ P \times \left(\frac{17}{16}\right)^{3} = 4913 \]

Since \( 4913 = 17^{3} \):

\[ P = 4913 \times \frac{16^{3}}{17^{3}} = \frac{17^{3} \times 4096}{17^{3}} = \text{Rs. } 4096 \]

Recognising 4913 as \( 17^{3} \) and 4096 as \( 16^{3} \) turns the whole problem into a single cancellation.