Compound Interest Questions and Answers

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.

Answer: D

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

\[ (1.05)^{n} = \frac{9261}{8000} = 1.157625 \]

Testing the small powers:

\[ (1.05)^{2} = 1.1025, \qquad (1.05)^{3} = 1.157625 \]

\[ n = 3 \text{ years} \]

The numbers 8000 and 9261 are \( 20^{3} \) and \( 21^{3} \), so the ratio is \( \left(\tfrac{21}{20}\right)^{3} \) - the cube is visible without any multiplication.

Answer: C

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

Recover the principal from the compound interest first.

\[ \text{CI} = P\left[(1.03)^{2} - 1\right] = P(0.0609) = 101.50 \]

\[ P = \frac{101.50}{0.0609} = \text{Rs. } 1666.67 \]

\[ \text{SI} = 1666.67 \times 0.03 \times 2 = \text{Rs. } 100 \]

The Rs. 1.50 gap between the two is interest on the first year's Rs. 50 of interest - which is the whole of what compounding adds over two years.

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: B

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

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

\[ 1 = P(0.04)^{2} = 0.0016P \]

\[ P = \frac{1}{0.0016} = \text{Rs. } 625 \]

Checking: SI = Rs. 50 and CI = \( 625(1.0816 - 1) = 51 \) - the Re. 1 gap is 4% of the first year's Rs. 25 of interest.

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.