Simple Interest Questions and Answers

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.

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: B

"6 paisa per month" means 6 paise on each rupee each month, or 6% per month.

\[ \text{SI} = 500 \times \frac{6}{100} \times 9 \]

\[ = 500 \times 0.06 \times 9 = \text{Rs. } 270 \]

The rate is already monthly, so the nine months are used directly - no conversion to years is needed.

Answer: B

Compounded annually, doubling in two years means

\[ \left(1 + \frac{R}{100}\right)^{2} = 2 \]

\[ 1 + \frac{R}{100} = \sqrt{2} = 1.4142 \]

\[ R = 41.42 \approx 41.4\% \]

Under simple interest the answer would be 50%; compounding needs less, because the first year's interest itself earns in the second.

Answer: C

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

\[ 5400 = \frac{P \times 12 \times 3}{100} = 0.36P \]

\[ P = \frac{5400}{0.36} = \text{Rs. } 15{,}000 \]

Three years at 12% comes to 36% of the principal, so the principal is a little under three times the interest.

Answer: A

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

Beyond eight years leaves \( 11 - 8 = 3 \) years at 13%, so the three slabs are:

\[ P\left(\frac{6 \times 3 + 9 \times 5 + 13 \times 3}{100}\right) = 8160 \]

\[ P\left(\frac{18 + 45 + 39}{100}\right) = 1.02P = 8160 \]

\[ P = \text{Rs. } 8000 \]

The final slab covers only the last three years, not all eleven.

Answer: B

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

Let \( p \) be the part at 8%; the rest, \( 10000 - p \), earns 10%.

\[ 0.08p + 0.10(10000 - p) = 890 \]

\[ 0.08p + 1000 - 0.10p = 890 \quad\Longrightarrow\quad -0.02p = -110 \]

\[ p = \text{Rs. } 5500 \]

Checking: Rs. 440 from the 8% part and Rs. 450 from the Rs. 4500 at 10%.

Answer: C

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

Write the interest as a fraction of the principal.

\[ \frac{9}{25}P = \frac{P \times R \times 4}{100} \]

The \( P \) cancels:

\[ \frac{9}{25} = \frac{4R}{100} \quad\Longrightarrow\quad R = \frac{9 \times 100}{25 \times 4} = 9\% \]

Because the interest is quoted as a fraction of the principal, no actual sum is needed.