Compound Interest Questions and Answers

Answer: C

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

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: 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.

\[ \text{rate} = 5\% \text{ per half-year}, \qquad n = 2 \]

\[ A = 15000 \times (1.05)^{2} = 15000 \times 1.1025 = \text{Rs. } 16{,}537.50 \]

Annual compounding at 10% would give Rs. 16,500; the extra Rs. 37.50 is the second half-year's 5% on the first half-year's Rs. 750.

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.

Answer: A

C.I. when interest is compounded yearly 

= [5000 * (1 + 4/100) * (1 + (1/2 * 4)/100] 

= 5000 * 26/25 * 51/50 = Rs. 5304 

C.I. when interest is compounded half-yearly 

= [5000 * (1 + 2/100)²] 

= (5000 * 51/50 * 51/50 * 51/50) = Rs. 5306.04 Difference = (5306.04 - 5304) = Rs. 2.04

Answer: C

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

Quarterly compounding quarters the rate, and nine months is three quarters.

\[ \text{rate} = \frac{16}{4} = 4\% \text{ per quarter}, \qquad n = 3 \]

\[ A = 15625 \times (1.04)^{3} = 15625 \times 1.124864 = \text{Rs. } 17{,}576 \]

\[ \text{CI} = 17576 - 15625 = \text{Rs. } 1951 \]

The principal is \( 25^{3} \) and the amount \( 26^{3} \), since the factor is \( \left(\tfrac{26}{25}\right)^{3} \) - which is why both come out as whole numbers.

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.

\[ 50 = \frac{P \times 5 \times 2}{100} = 0.1P \quad\Longrightarrow\quad P = \text{Rs. } 500 \]

\[ \text{CI} = 500\left[(1.05)^{2} - 1\right] = 500(1.1025 - 1) \]

\[ = \text{Rs. } 51.25 \]

The Rs. 1.25 excess is 5% of the first year's Rs. 25 of interest - the only difference compounding makes over two years.

Answer: D

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

\[ \text{SI} = 1000 \times 0.10 \times 4 = \text{Rs. } 400 \]

\[ \text{CI} = 1000\left[(1.1)^{4} - 1\right] = 1000(1.4641 - 1) = \text{Rs. } 464.10 \]

\[ \text{difference} = 464.10 - 400 = \text{Rs. } 64.10 \]

The gap widens sharply with time: over three years it is only Rs. 31, and over two years Rs. 10. That growth is the compounding acting on ever more accumulated interest.

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: 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: 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.