Compound Interest Questions and Answers

Answer: A

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

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

\[ 1 = P(0.04)^{2} = 0.0016P \quad\Longrightarrow\quad P = \text{Rs. } 625 \]

Checking: simple interest is Rs. 50 and compound interest Rs. 51. The Re. 1 gap is 4% of the Rs. 25 earned in the first year, which is exactly what the squared-rate formula encodes.

Answer: C

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

Find the rate from the simple-interest statement first.

\[ \frac{R \times 6}{100} = 0.60 \quad\Longrightarrow\quad R = 10\% \]

Now compound Rs. 12,000 for three years at that rate:

\[ \text{CI} = 12000\left[(1.1)^{3} - 1\right] = 12000(1.331 - 1) \]

\[ = 12000 \times 0.331 = \text{Rs. } 3972 \]

Simple interest for the same three years would be Rs. 3600; the extra Rs. 372 is the compounding.

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: 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: A

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

\[ \left(1 + \frac{R}{100}\right)^{2} = \frac{1348.32}{1200} = 1.1236 \]

\[ 1 + \frac{R}{100} = \sqrt{1.1236} = 1.06 \]

\[ R = 6\% \]

Averaging the total growth of 12.36% over two years to get 6.18% is close but wrong; the square root, not the halving, is what inverts two years of compounding.

Answer: B

The money more than doubles once \( (1.2)^{n} > 2 \). Test successive powers:

\[ (1.2)^{1} = 1.2, \qquad (1.2)^{2} = 1.44 \]

\[ (1.2)^{3} = 1.728, \qquad (1.2)^{4} = 2.0736 \]

The third year falls short and the fourth passes 2, so the answer is 4 years.

The question asks for complete years, so the answer must be a whole number even though the exact doubling time is about 3.8 years.

Answer: C

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

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

The interest is Rs. 820, of which Rs. 800 is simple interest and Rs. 20 is the second year's 5% on the first year's Rs. 400 - which is why Rs. 8800 is not the answer.

Answer: D

The effective rate is the single annual rate that produces the same growth as the nominal rate does with its compounding.

Half-yearly payment at a nominal 6% means 3% every six months:

\[ (1.03)^{2} = 1.0609 \]

\[ \text{effective rate} = 1.0609 - 1 = 6.09\% \]

The extra 0.09% over the nominal 6% is the second half-year's 3% earned on the first half-year's 3% - the whole point of the effective-rate idea.

Answer: C

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

Work out the compound interest first.

\[ \text{CI} = 4000\left[(1.1)^{2} - 1\right] = 4000(1.21 - 1) = \text{Rs. } 840 \]

The simple interest is half of that:

\[ \text{SI} = 420 \]

\[ 420 = \frac{P \times 8 \times 3}{100} = 0.24P \]

\[ P = \frac{420}{0.24} = \text{Rs. } 1750 \]

The two sums are unrelated; the only link is the numerical value of the interest, so the compound side must be finished before the simple side can begin.

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.