Simple Interest Questions and Answers

Answer: C

Under simple interest the growth is proportional to time, so work with the interest rather than the amount.

Becoming 7 times means the interest is \( 6P \) in 15 years:

\[ 6P = \frac{P \times R \times 15}{100} \quad\Longrightarrow\quad R = 40\% \]

Becoming 10 times needs interest of \( 9P \):

\[ 9P = \frac{P \times 40 \times T}{100} \quad\Longrightarrow\quad T = 22.5 \text{ years} \]

Equivalently the times are in the ratio \( 9 : 6 \), so \( 15 \times \tfrac{9}{6} = 22.5 \).

Answer: A

Apply the first year's rate, then compare with the final amount.

\[ 4500 \times 1.12 = \text{Rs. } 5040 \]

\[ 1 + \frac{R_{2}}{100} = \frac{5544}{5040} = 1.1 \]

\[ R_{2} = 10\% \]

With rates that differ by year the two growth factors multiply; they cannot be averaged.

Answer: C

\[ \text{SI} = \frac{P \times R \times T}{100} \]

Convert both periods to years: 15 months is \( \tfrac{5}{4} \) and 8 months is \( \tfrac{2}{3} \).

\[ \frac{P \times 7.5 \times \tfrac{5}{4}}{100} - \frac{P \times 12.5 \times \tfrac{2}{3}}{100} = 32.50 \]

\[ \frac{P(9.375 - 8.3333)}{100} = 32.50 \]

\[ \frac{1.04167P}{100} = 32.50 \quad\Longrightarrow\quad P = \text{Rs. } 3120 \]

The two \( R \times T \) products are close (9.375 against 8.333), so the small difference needs a large principal to reach Rs. 32.50.

Answer: B

Simple interest of Rs. 600 over two years means Rs. 300 in each year. Compound interest exceeds it by

\[ 645 - 600 = \text{Rs. } 45 \]

and that excess is one year's interest on the first year's Rs. 300:

\[ 300 \times \frac{R}{100} = 45 \quad\Longrightarrow\quad R = 15\% \]

The principal is never needed. This shortcut works for any two-year simple-versus-compound comparison.

Answer: C

\[ \text{SI} = \frac{P \times R \times T}{100} \]

B pays 10% and receives 11.5% on the same Rs. 3500, so only the 1.5% gap matters.

\[ \text{gain} = \frac{3500 \times 1.5 \times 3}{100} = \text{Rs. } 157.50 \]

Working out the two interests separately gives Rs. 1207.50 received against Rs. 1050 paid - the same Rs. 157.50, for more effort.

Answer: A

\[ \text{SI} = \frac{P \times R \times T}{100} \]

Find the rate first.

\[ \text{SI} = 9200 - 8000 = 1200 \quad\Longrightarrow\quad 1200 = \frac{8000 \times R \times 3}{100} = 240R \]

\[ R = 5\% \]

At \( 5 + 2 = 7\% \):

\[ \text{SI} = \frac{8000 \times 7 \times 3}{100} = 1680 \quad\Longrightarrow\quad \text{amount} = \text{Rs. } 9680 \]

The extra 2% over three years adds \( 8000 \times 0.06 = 480 \) - the same answer directly.

Answer: B

\[ \text{SI} = \frac{P \times R \times T}{100} \]

Equal interests give the ratio of the two parts.

\[ \frac{p_{1} \times 6.25 \times 8}{100} = \frac{p_{2} \times 7 \times 5}{100} \]

\[ 50\,p_{1} = 35\,p_{2} \quad\Longrightarrow\quad \frac{p_{1}}{p_{2}} = \frac{7}{10} \]

So the parts are 7 and 10 of 17 shares:

\[ p_{1} = \frac{7}{17} \times 20400 = \text{Rs. } 8400, \qquad p_{2} = \text{Rs. } 12{,}000 \]

The smaller part is Rs. 8400 - the one at the higher \( R \times T \) product, since it needs less capital for the same return.

Answer: B

\[ \text{SI} = \frac{P \times R \times T}{100} \]

Equal interests give the ratio of the parts.

\[ \frac{p_{1} \times 3 \times 8}{100} = \frac{p_{2} \times 5 \times 3}{100} \]

\[ 24\,p_{1} = 15\,p_{2} \quad\Longrightarrow\quad \frac{p_{1}}{p_{2}} = \frac{15}{24} = \frac{5}{8} \]

So the sum splits into 13 shares of \( 2665 \div 13 = 205 \):

\[ p_{2} = 8 \times 205 = \text{Rs. } 1640 \]

The first part is Rs. 1025, and both yield Rs. 246 of interest.

Answer: D

\[ \text{SI} = \frac{P \times R \times T}{100} \]

Tripling means the interest is twice the principal.

\[ 2P = \frac{P \times R \times 12}{100} \quad\Longrightarrow\quad 200 = 12R \]

\[ R = \frac{50}{3} = 16\tfrac{2}{3}\% \text{ p.a.} \]

That rate is not among the choices, so the answer is none of these. The multiple counts the amount, so the interest is one less than it - using 3 instead of 2 would give 25%.

Answer: D

Simple interest is proportional to time, so over eight years instead of five the interest rises by a fixed factor no matter what the rate is.

\[ \frac{\text{SI}_{8}}{\text{SI}_{5}} = \frac{8}{5} = 1.6 \]

That is a 60% increase for every rate and every sum, so the stated condition carries no information about \( R \).

The rate therefore cannot be determined. A condition that holds identically is not a constraint.