Simple Interest Questions and Answers

Answer: A

Doubling means the interest equals the principal, so

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

Trebling needs interest of \( 2P \):

\[ 2P = \frac{P \times 10 \times T}{100} \quad\Longrightarrow\quad T = 20 \text{ years} \]

Under simple interest the interest accumulates linearly, so twice the interest takes exactly twice the time - under compound interest it would take less.

Answer: D

Let \( p \) be the part at 6%; the rest, \( 1500 - p \), earns 5%.

\[ 0.06p + 0.05(1500 - p) = 85 \]

\[ 0.06p + 75 - 0.05p = 85 \quad\Longrightarrow\quad 0.01p = 10 \]

\[ p = \text{Rs. } 1000 \text{ at } 6\% \]

\[ \text{lent at } 5\% = 1500 - 1000 = \text{Rs. } 500 \]

The question asks for the 5% part, so the Rs. 1000 found first is not the answer.

Answer: A

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

The time equals the rate, so write both as \( R \).

\[ \frac{4}{9}P = \frac{P \times R \times R}{100} \]

The \( P \) cancels:

\[ \frac{4}{9} = \frac{R^{2}}{100} \quad\Longrightarrow\quad R^{2} = \frac{400}{9} \]

\[ R = \frac{20}{3} = 6\tfrac{2}{3} \]

So the rate is \( 6\tfrac{2}{3}\% \) and the time \( 6\tfrac{2}{3} \) years.

Answer: A

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

Work out the target interest first.

\[ \frac{5000 \times 4 \times 5}{100} = \text{Rs. } 1000 \]

Now find how long Rs. 4000 at 3% takes to earn the same:

\[ 1000 = \frac{4000 \times 3 \times T}{100} = 120\,T \]

\[ T = \frac{1000}{120} = 8\tfrac{1}{3} \text{ years} \]

The smaller principal at the lower rate needs longer, so the answer must exceed the 5 years - a quick sanity check.

Answer: B

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

Doubling means the interest equals the principal.

\[ P = \frac{P \times 5 \times T}{100} \quad\Longrightarrow\quad 100 = 5T \]

\[ T = 20 \text{ years} \]

The general rule for doubling under simple interest is \( T = \tfrac{100}{R} \).

Answer: A

100 ---- 200 ---- 5

             200 ---- 5                                         

             100 ---- 2 1/2

             ------------------

              600 ---- 12 ½ years

Answer: A

Doubling in 8 years fixes the rate:

\[ P = \frac{P \times R \times 8}{100} \quad\Longrightarrow\quad R = 12.5\% \]

After 10 years:

\[ \text{amount} = P\left(1 + \frac{12.5 \times 10}{100}\right) = P(1 + 1.25) = 2.25P \]

So it becomes \( 2\tfrac{1}{4} \) times itself. Under simple interest the multiple grows linearly with time, so ten years gives \( \tfrac{10}{8} \) of the interest, not of the amount.

Answer: C

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

Equal interests over the same period give

\[ \frac{p_{1} \times 9 \times T}{100} = \frac{p_{2} \times 12 \times T}{100} \]

\[ 9\,p_{1} = 12\,p_{2} \quad\Longrightarrow\quad \frac{p_{1}}{p_{2}} = \frac{12}{9} = \frac{4}{3} \]

The sums are in the inverse ratio of the rates: the lower rate needs the larger capital to earn as much.

Answer: B

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

Both banks take the same sum at the same rate, so the difference comes from the 1.5 extra years.

\[ \frac{P \times 15 \times 5}{100} - \frac{P \times 15 \times 3.5}{100} = 144 \]

\[ \frac{P \times 15 \times 1.5}{100} = 0.225P = 144 \]

\[ P = \frac{144}{0.225} = \text{Rs. } 640 \]

Checking: Rs. 480 over 5 years against Rs. 336 over 3.5 - a difference of exactly Rs. 144.

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.