Simple Interest Questions and Answers

Answer: B

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

The two investments share the rate and the four years, so the Rs. 130 gap comes entirely from the Rs. 250 difference in principal.

\[ \frac{(2250 - 2000) \times R \times 4}{100} = 130 \]

\[ \frac{250 \times 4R}{100} = 10R = 130 \quad\Longrightarrow\quad R = 13\% \]

Working out each interest separately gives the same answer but needs two calculations instead of one.

Answer: D

Let \( p \) be the part at 5%; the rest, \( 2500 - p \), earns 6%.

\[ 0.05p + 0.06(2500 - p) = 140 \]

\[ 0.05p + 150 - 0.06p = 140 \quad\Longrightarrow\quad -0.01p = -10 \]

\[ p = \text{Rs. } 1000 \]

Checking: Rs. 50 from the 5% part and Rs. 90 from the Rs. 1500 at 6%.

Answer: A

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

Find the original rate first.

\[ \text{SI} = 2600 - 2000 = 600 \quad\Longrightarrow\quad 600 = \frac{2000 \times R \times 5}{100} = 100R \]

\[ R = 6\% \]

At \( 6 + 3 = 9\% \):

\[ \text{SI} = \frac{2000 \times 9 \times 5}{100} = 900 \quad\Longrightarrow\quad \text{amount} = \text{Rs. } 2900 \]

The extra 3% over 5 years adds \( 2000 \times 0.15 = 300 \) - the same answer in one step.

Answer: C

Present worth is the sum that would grow to Rs. 132 over the two years.

\[ \text{PW}\left(1 + \frac{5 \times 2}{100}\right) = 132 \]

\[ 1.1 \times \text{PW} = 132 \quad\Longrightarrow\quad \text{PW} = \text{Rs. } 120 \]

Taking 10% off Rs. 132 would give Rs. 118.80 - the second option, and wrong, because the interest accrues on the present worth, not on the future amount.

Answer: D

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

The Rs. 8470 is the amount, which is the principal plus nine years of interest.

\[ 8470 = P\left(1 + \frac{6 \times 9}{100}\right) = P(1.54) \]

\[ P = \frac{8470}{1.54} = \text{Rs. } 5500 \]

Checking: the interest is \( 5500 \times 0.54 = 2970 \), and \( 5500 + 2970 = 8470 \).

Answer: A

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

Becoming \( \tfrac{7}{6} \) of itself means the interest is \( \tfrac{1}{6} \) of the principal.

\[ \frac{P}{6} = \frac{P \times R \times 3}{100} \]

\[ \frac{1}{6} = \frac{3R}{100} \quad\Longrightarrow\quad R = \frac{100}{18} = \frac{50}{9} = 5\tfrac{5}{9}\% \]

The multiple applies to the amount, so subtracting 1 gives the interest as a fraction of the sum.

Answer: D

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

\[ 4016.25 = \frac{P \times 9 \times 5}{100} = 0.45P \]

\[ P = \frac{4016.25}{0.45} = \text{Rs. } 8925 \]

Five years at 9% comes to 45% of the sum, so the principal is a little over twice the interest.

Answer: B

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

\[ 70 = \frac{P \times 3.5 \times 4}{100} = 0.14P \]

\[ P = \frac{70}{0.14} = \text{Rs. } 500 \]

Over four years at 3.5% the interest comes to 14% of the sum.

Answer: C

Let \( p \) be the part at 3%; the rest, \( 4000 - p \), earns 5%.

\[ 0.03p + 0.05(4000 - p) = 144 \]

\[ 0.03p + 200 - 0.05p = 144 \quad\Longrightarrow\quad -0.02p = -56 \]

\[ p = \text{Rs. } 2800 \]

Checking: Rs. 84 from the 3% part and Rs. 60 from the Rs. 1200 at 5%.

Answer: C

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

\[ \text{SI} = 900 - 750 = \text{Rs. } 150 \]

\[ 150 = \frac{750 \times R \times 5}{100} = 37.5R \]

\[ R = 4\% \]

The Rs. 900 is the amount, so the principal must be subtracted first.