Simple Interest Questions and Answers

Answer: B

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

The number of years equals the rate, so write both as \( R \).

\[ 432 = \frac{1200 \times R \times R}{100} = 12R^{2} \]

\[ R^{2} = 36 \quad\Longrightarrow\quad R = 6 \]

So the rate is 6% and the period 6 years. The data are complete, so "cannot be determined" is a trap; the negative root is discarded as a rate.

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: D

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

The rate is 8% per annum and the period two and a half years.

\[ \text{SI} = \frac{6500 \times 8 \times 2.5}{100} = \text{Rs. } 1300 \]

\[ \text{amount} = 6500 + 1300 = \text{Rs. } 7800 \]

The question asks what the sum will yield in total, so the principal is included; the interest alone is Rs. 1300.

Answer: C

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

The fall is \( 8 - 7\tfrac{3}{4} = 0.25 \) percentage points, over one year.

\[ \frac{P \times 0.25 \times 1}{100} = 61.50 \]

\[ 0.0025P = 61.50 \quad\Longrightarrow\quad P = \text{Rs. } 24{,}600 \]

A quarter of a percentage point is a small slice, so it takes a large capital to produce a Rs. 61.50 drop.

Answer: D

Let the sum be Rs. x, then it becomes Rs. 4x in eight years Rs. 3x is the interest on x for eight years.

R = (100 * 3x)/(x * 8) = 300/8 %

If the sum becomes ten times itself, then interest is 9x.

The required time period = (100 * 9x)/(x * 300/8) = (100 * 9x * 8)/(x * 300) = 24 years.

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: A

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

\[ 160 = \frac{800 \times R \times 4}{100} = 32R \]

\[ R = 5\% \]

The Rs. 160 is 20% of the principal, spread over four years - 5% a year.

Answer: D

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

\[ \text{SI} = 15500 - 12500 = \text{Rs. } 3000 \]

\[ 3000 = \frac{12500 \times R \times 4}{100} = 500R \]

\[ R = 6\% \]

The interest is 24% of the principal over four years, which is 6% a year.

Answer: D

If the interest earned is 25% more than the earlier interest then the rate of interest also should be 25% higher than the earlier rate. 

Let the earlier rate of interest be x%. 

Now it will be (x + 3)% 

% increase = (x + 3) - x/x * 100 = 25

=> x = 12

Answer: D

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

Write both amounts for the same \( P \) and \( T \).

\[ P + \frac{12PT}{100} = 3000, \qquad P + \frac{15PT}{100} = 3300 \]

Subtracting:

\[ \frac{3PT}{100} = 300 \quad\Longrightarrow\quad PT = 10{,}000 \]

Substituting back:

\[ P + 0.12 \times 10000 = 3000 \quad\Longrightarrow\quad P = \text{Rs. } 1800 \]

\[ T = \frac{10000}{1800} = 5.56 \approx 5.5 \text{ years} \]

Subtracting the two equations removes the principal and leaves the product \( PT \) - the key step.