Simple Interest Questions and Answers

Answer: C

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

Two years and six months is 2.5 years.

\[ \text{SI} = \frac{8200 \times 17.5 \times 2.5}{100} \]

\[ = 82 \times 17.5 \times 2.5 = 82 \times 43.75 = \text{Rs. } 3587.50 \]

The six months must be converted to 0.5 of a year, not left as a separate term.

Answer: C

Rs. 600 over ten years means Rs. 60 a year on the original principal.

First five years at the original principal:

\[ 5 \times 60 = \text{Rs. } 300 \]

Next five years with the principal trebled, so three times the yearly interest:

\[ 5 \times 180 = \text{Rs. } 900 \]

\[ \text{total} = 300 + 900 = \text{Rs. } 1200 \]

Trebling the whole Rs. 600 would give Rs. 1800; only the second half of the period is affected.

Answer: A

Let the second amount be Rs. x. Then,

(12000 * 10 * 1)/100 + (x * 20 * 1)/100 = [(12000 + x) * 14 * 1] / 100

120000 + 20x = 16800 + 14x

x = 8000

Total investment = 12000 + 8000 = Rs. 20,000.

Answer: B

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

Doubling means the interest equals the principal.

\[ P = \frac{P \times 3 \times T}{100} \quad\Longrightarrow\quad 100 = 3T \]

\[ T = \frac{100}{3} = 33\tfrac{1}{3} \text{ years} \]

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

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: B

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

The extra income comes from the \( 12\tfrac{1}{2} - 10 = 2.5 \) percentage points, over one year.

\[ \frac{P \times 2.5 \times 1}{100} = 1250 \]

\[ 0.025P = 1250 \quad\Longrightarrow\quad P = \text{Rs. } 50{,}000 \]

Checking: Rs. 5000 a year at 10% against Rs. 6250 at 12.5%.

Answer: A

P= Rs 900

Rate = 4.5*

SI= Rs 81

=> SI= Time*P*rate

calculating for rate,

=> Time= SI/rate*P

= 81/4.5%*900

=> 2 years

Thus, it will take 2 years to yield Rs 81.

Answer: C

Simple interest of Rs. 800 over two years is Rs. 400 a year. Compound interest exceeds it by

\[ 840 - 800 = \text{Rs. } 40 \]

which is one year's interest on that first Rs. 400:

\[ 400 \times \frac{R}{100} = 40 \quad\Longrightarrow\quad R = 10\% \text{ p.a.} \]

The sum works out at Rs. 4000, and \( 4000(1.21 - 1) = 840 \) confirms it - but the principal is not needed to answer.

Answer: B

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

The number of years equals the rate, so write both as \( R \).

\[ 432 = \frac{1200 \times R \times R}{100} = 12R^{2} \]

\[ R^{2} = 36 \quad\Longrightarrow\quad R = 6 \]

So the rate is 6% and the period 6 years. The data are complete, so "cannot be determined" is a trap; the negative root is discarded as a rate.

Answer: A

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

The time equals the rate, so write both as \( R \).

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

The \( P \) cancels:

\[ \frac{4}{9} = \frac{R^{2}}{100} \quad\Longrightarrow\quad R^{2} = \frac{400}{9} \]

\[ R = \frac{20}{3} = 6\tfrac{2}{3} \]

So the rate is \( 6\tfrac{2}{3}\% \) and the time \( 6\tfrac{2}{3} \) years.