Compound Interest Questions and Answers

Answer: C

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

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: 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

Over two years the gap between compound and simple interest is

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

\[ 15 = P(0.10)^{2} = 0.01P \]

\[ P = \text{Rs. } 1500 \]

Checking: SI = Rs. 300 and CI = \( 1500(1.21 - 1) = 315 \) - a Rs. 15 gap, which is 10% of the first year's Rs. 150 of interest.

Answer: D

Let the principal be Rs.P 

S.I at 5% p.a in 8 years on Rs.P = Rs.500

(P)(8)(5)/100 = 500

P = 1250

C.I on Rs.2P i.e., Rs.2500 at 5% p.a in two years

=2500{ [1 + 5/100]² - 1} = 2500{ 21² - 20² /20²} 

= 2500/400(441 - 400)

= 25/4(41) = 1025/4 = Rs.256.25

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

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

Recover the principal from the simple interest.

\[ 80 = \frac{P \times 4 \times 2}{100} = 0.08P \quad\Longrightarrow\quad P = \text{Rs. } 1000 \]

\[ \text{CI} = 1000\left[(1.04)^{2} - 1\right] = 1000(1.0816 - 1) = \text{Rs. } 81.60 \]

The Rs. 1.60 excess is 4% of the first year's Rs. 40 of interest - the whole of what compounding adds over two years.

Answer: B

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

P is 13 and so waits 12 years; Q is 14 and waits 11. Both end with the same amount:

\[ P_{\text{share}}(1.04)^{12} = Q_{\text{share}}(1.04)^{11} \]

\[ Q_{\text{share}} = P_{\text{share}} \times 1.04 \]

The younger boy's money compounds for an extra year, so he needs less of it. Their shares are therefore in the ratio \( 25 : 26 \):

\[ P_{\text{share}} + 1.04\,P_{\text{share}} = 25500 \quad\Longrightarrow\quad 2.04\,P_{\text{share}} = 25500 \]

\[ P_{\text{share}} = 12{,}500, \qquad Q_{\text{share}} = \text{Rs. } 13{,}000 \]

Checking: \( 12500 \times 1.04 = 13000 \), so after Q's extra year the two balances coincide.

Answer: D

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

\[ A = 8000 \times (1.05)^{2} = 8000 \times 1.1025 = \text{Rs. } 8820 \]

Rs. 8820 is not among the choices, so the answer is none of these.

The Rs. 8800 on offer is what simple interest would give (Rs. 800 of interest); the extra Rs. 20 is the second year's 5% on the first year's Rs. 400.

Answer: D

Compound interest needs both the principal and the rate, but the question supplies only the final amount.

\[ P + \frac{P \times R \times 2}{100} = 1344 \]

That is one equation in two unknowns, so infinitely many pairs fit. For example:

• \( P = 1200 \) at 6% gives 1344, and CI would be Rs. 148.32
• \( P = 1120 \) at 10% also gives 1344, and CI would be Rs. 235.20

Since the two produce different answers, the data are insufficient. Any of these questions that fixes only the amount, and not the principal or rate separately, is underdetermined.

Answer: B

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

Recover the principal from the compound interest.

\[ 525 = P\left[(1.1)^{2} - 1\right] = 0.21P \quad\Longrightarrow\quad P = \text{Rs. } 2500 \]

Double the time is 4 years and half the rate is 5%:

\[ \text{SI} = \frac{2500 \times 5 \times 4}{100} = \text{Rs. } 500 \]

Doubling the time while halving the rate leaves simple interest unchanged, since \( Rn \) is the same - the answer would be identical for 2 years at 10%.