Simple Interest Questions and Answers

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.

Answer: C

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

\[ \text{SI} = 15600 - 12000 = \text{Rs. } 3600 \]

\[ 3600 = \frac{12000 \times 6 \times T}{100} = 720\,T \]

\[ T = 5 \text{ years} \]

The Rs. 15,600 is the amount, so the principal must be subtracted before the formula can be used.

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: B

Let the interest for one year be x.

As amount = Principal + Interest, we have 

P + 4x = 20720 --- (1)  ; P + 6x = 24080 --- (2)

Solving the equations (1) and (2), we can get 

P = Rs. 14000 and x = Rs. 1680

Interest for one year on Rs. 14000 is Rs. 1680

So, R = (100 * 1680)/(14000 * 1) = 12% p.a.

Answer: C

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

Find the original rate first.

\[ \text{SI} = 4600 - 4000 = 600 \quad\Longrightarrow\quad 600 = \frac{4000 \times R \times 5}{100} = 200R \]

\[ R = 3\% \]

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

\[ \text{SI} = \frac{4000 \times 6 \times 5}{100} = 1200 \quad\Longrightarrow\quad \text{amount} = \text{Rs. } 5200 \]

The extra 3% over 5 years adds \( 4000 \times 0.15 = 600 \), doubling the interest.