Simple Interest Questions and Answers

Answer: C

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

Under simple interest each year adds the same amount, so the gap between the two figures is one year's interest.

\[ 854 - 815 = \text{Rs. } 39 \text{ per year} \]

\[ P = 815 - 3 \times 39 = 815 - 117 = \text{Rs. } 698 \]

Working back three years from the three-year amount recovers the principal.

Answer: B

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

The five years between the two figures account for the whole increase.

\[ \frac{1020 - 720}{5} = \frac{300}{5} = \text{Rs. } 60 \text{ per year} \]

\[ P = 720 - 2 \times 60 = \text{Rs. } 600 \]

"A further period of 5 years" means seven years in all, not five - but only the gap of five is needed.

Answer: C

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

The three years between the two amounts give the yearly interest.

\[ \frac{12005 - 9800}{3} = \frac{2205}{3} = \text{Rs. } 735 \text{ per year} \]

\[ P = 9800 - 5 \times 735 = 9800 - 3675 = \text{Rs. } 6125 \]

\[ R = \frac{735 \times 100}{6125} = 12\% \]

Answer: B

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

Doubling means the interest equals the principal.

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

\[ R = \frac{25}{3} = 8\tfrac{1}{3}\% \]

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

Answer: A

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

\[ \frac{2}{5}P = \frac{P \times R \times 10}{100} \]

The \( P \) cancels:

\[ \frac{2}{5} = \frac{10R}{100} = \frac{R}{10} \quad\Longrightarrow\quad R = 4\% \]

Checking: 4% for 10 years gives 40% of the sum, which is \( \tfrac{2}{5} \).

Answer: A

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

\[ 0.125P = \frac{P \times 10 \times T}{100} = 0.1PT \]

\[ T = \frac{0.125}{0.1} = 1.25 = 1\tfrac{1}{4} \text{ years} \]

At 10% a year, interest of 12.5% of the sum takes a year and a quarter.

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: B

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

Work out B's share of the interest first.

\[ \frac{2500 \times 7 \times 4}{100} = \text{Rs. } 700 \]

\[ \text{from C} = 1120 - 700 = \text{Rs. } 420 \]

\[ 420 = \frac{C \times 7 \times 4}{100} = 0.28C \quad\Longrightarrow\quad C = \text{Rs. } 1500 \]

Both loans run at the same rate for the same time, so their interests are in proportion to the sums.

Answer: D

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

Both loans share one rate \( R \), so add the two interests.

\[ \frac{5000 \times R \times 2}{100} + \frac{3000 \times R \times 4}{100} = 2200 \]

\[ 100R + 120R = 2200 \quad\Longrightarrow\quad 220R = 2200 \]

\[ R = 10\% \]

Checking: Rs. 1000 from B and Rs. 1200 from C.

Answer: C

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

The sum and the period are common, so only the gap in rates produces the difference.

\[ \frac{1500 \times \Delta R \times 3}{100} = 13.50 \]

\[ 45\,\Delta R = 13.50 \quad\Longrightarrow\quad \Delta R = 0.3\% \]

Neither rate needs to be known individually - only their difference enters the calculation.