Simple Interest Questions and Answers

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.

Answer: B

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

What Manoj pays Anwar on the borrowed Rs. 3450 at 6% for 3 years:

\[ \frac{3450 \times 6 \times 3}{100} = \text{Rs. } 621 \]

What he receives from Ramu on the whole sum \( S \) at 9% for 3 years:

\[ \frac{S \times 9 \times 3}{100} = 0.27S \]

His gain is the difference:

\[ 0.27S - 621 = 824.85 \quad\Longrightarrow\quad 0.27S = 1445.85 \]

\[ S = \text{Rs. } 5355 \]

So he added \( 5355 - 3450 = \text{Rs. } 1905 \) of his own money.

Answer: D

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

The interest falls Rs. 340 short of the principal.

\[ \frac{P \times 4 \times 8}{100} = P - 340 \]

\[ 0.32P = P - 340 \quad\Longrightarrow\quad 0.68P = 340 \]

\[ P = \text{Rs. } 500 \]

Checking: the interest is Rs. 160, which is indeed Rs. 340 less than Rs. 500.

Answer: C

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

The rate changes with the period, so add the three slabs. Beyond five years leaves \( 9 - 5 = 4 \) years at 14%.

\[ P\left(\frac{6 \times 2}{100} + \frac{9 \times 3}{100} + \frac{14 \times 4}{100}\right) = 11400 \]

\[ P(0.12 + 0.27 + 0.56) = 0.95P = 11400 \]

\[ P = \text{Rs. } 12{,}000 \]

The final slab runs for four years, not nine - "beyond five years" means the remainder of the nine.

Answer: B

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

Under simple interest the yearly addition is constant, so the gap between the two amounts covers two years.

\[ 1875 - 1725 = 150 \text{ over 2 years} \quad\Longrightarrow\quad 75 \text{ per year} \]

\[ P = 1725 - 3 \times 75 = \text{Rs. } 1500 \]

\[ R = \frac{75 \times 100}{1500} = 5\% \]

Working back three years from the first amount recovers the principal.

Answer: A

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

The period \( 2\tfrac{2}{5} \) years is 2.4 years, and Rs. 1120 is the amount.

\[ 1120 = P\left(1 + \frac{5 \times 2.4}{100}\right) = P(1.12) \]

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

The interest is Rs. 120, which is 12% of the principal - 5% a year for 2.4 years.