Simple Interest Questions and Answers

Answer: D

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

The rate \( 15\tfrac{2}{5}\% \) is 15.4%, and nine months is \( \tfrac{3}{4} \) of a year.

\[ \text{SI} = \frac{78000 \times 15.4 \times 0.75}{100} \]

\[ = 780 \times 15.4 \times 0.75 = 780 \times 11.55 = \text{Rs. } 9009 \]

Leaving the period as 9 rather than \( \tfrac{9}{12} \) would inflate the answer twelvefold.

Answer: B

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

Doubling means the interest equals the principal.

\[ P = \frac{P \times 5 \times T}{100} \quad\Longrightarrow\quad 100 = 5T \]

\[ T = 20 \text{ years} \]

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

Answer: A

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

The two schemes share the sum and the two years, so only the 6% gap in rate matters.

\[ \frac{P \times (18 - 12) \times 2}{100} = 840 \]

\[ 0.12P = 840 \quad\Longrightarrow\quad P = \text{Rs. } 7000 \]

Checking: Rs. 2520 at 18% against Rs. 1680 at 12% - a difference of exactly Rs. 840.

Answer: A

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

Work out the target interest first.

\[ \frac{5000 \times 4 \times 5}{100} = \text{Rs. } 1000 \]

Now find how long Rs. 4000 at 3% takes to earn the same:

\[ 1000 = \frac{4000 \times 3 \times T}{100} = 120\,T \]

\[ T = \frac{1000}{120} = 8\tfrac{1}{3} \text{ years} \]

The smaller principal at the lower rate needs longer, so the answer must exceed the 5 years - a quick sanity check.

Answer: D

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

\[ 840 = P\left(1 + \frac{3 \times 10}{100}\right) = 1.3P \]

\[ P = \frac{840}{1.3} = \text{Rs. } 646.15 \]

That value appears nowhere on the list, so the answer is none. The Rs. 525 offered would amount to only Rs. 682.50 over the same ten years.

Answer: A

Doubling means the interest equals the principal, so

\[ P = \frac{P \times R \times 10}{100} \quad\Longrightarrow\quad R = 10\% \]

Trebling needs interest of \( 2P \):

\[ 2P = \frac{P \times 10 \times T}{100} \quad\Longrightarrow\quad T = 20 \text{ years} \]

Under simple interest the interest accumulates linearly, so twice the interest takes exactly twice the time - under compound interest it would take less.

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: B

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

\[ 9600 = \frac{P \times 16 \times 5}{100} = 0.8P \]

\[ P = \frac{9600}{0.8} = \text{Rs. } 12{,}000 \]

Over five years at 16% the interest comes to 80% of the loan, which is why the principal is only a little above the interest itself.

Answer: B

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

\[ 81 = \frac{450 \times 4.5 \times T}{100} = 20.25\,T \]

\[ T = \frac{81}{20.25} = 4 \text{ years} \]

The interest is 18% of the principal, and at 4.5% a year that takes exactly four years.

Answer: D

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

Carry everything forward to the date of the last installment, 19 years after purchase.

The debt grows for 19 years:

\[ 20000\left(1 + \frac{6 \times 19}{100}\right) = 20000 \times 2.14 = \text{Rs. } 42{,}800 \]

The first 19 installments of Rs. 1000 earn interest for 19, 18, …, 1 years:

\[ 1000\left[19 + \frac{6}{100}(19 + 18 + \cdots + 1)\right] = 1000\left[19 + 0.06 \times 190\right] \]

\[ = 1000(19 + 11.4) = \text{Rs. } 30{,}400 \]

\[ \text{last installment} = 42800 - 30400 = \text{Rs. } 12{,}400 \]

That figure is not among the choices, so the answer is none of these.