Simple Interest Questions and Answers

Answer: A

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

Doubling means the interest equals the principal.

\[ P = \frac{P \times R \times 5}{100} \quad\Longrightarrow\quad R = 20\% \]

\[ \text{Raj's interest} = \frac{6250 \times 20 \times 3}{100} = \text{Rs. } 3750 \]

Aman's Rs. 8000 never enters Raj's calculation; it serves only to fix the rate, and even that cancels out of the doubling condition.

Answer: A

Doubling in 8 years fixes the rate:

\[ P = \frac{P \times R \times 8}{100} \quad\Longrightarrow\quad R = 12.5\% \]

After 10 years:

\[ \text{amount} = P\left(1 + \frac{12.5 \times 10}{100}\right) = P(1 + 1.25) = 2.25P \]

So it becomes \( 2\tfrac{1}{4} \) times itself. Under simple interest the multiple grows linearly with time, so ten years gives \( \tfrac{10}{8} \) of the interest, not of the amount.

Answer: B

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

Both the rate and the time are \( x \), and the interest is also \( x \).

\[ x = \frac{P \times x \times x}{100} = \frac{P x^{2}}{100} \]

Dividing both sides by \( x \):

\[ 1 = \frac{Px}{100} \quad\Longrightarrow\quad P = \frac{100}{x} \]

Only one power of \( x \) survives the cancellation, which is what separates \( \tfrac{100}{x} \) from \( \tfrac{100}{x^{2}} \).

Answer: A

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

\[ 16500 = \frac{5000 \times 15 \times T}{100} = 750\,T \]

\[ T = \frac{16500}{750} = 22 \text{ years} \]

The interest is more than three times the principal, so a long period is expected - at 15% a year it takes over six years just to match the sum.

Answer: A

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

The two deposits share one rate \( r \), so add their interests.

\[ \frac{500 \times r \times 2}{100} + \frac{300 \times r \times 4}{100} = 220 \]

\[ 10r + 12r = 220 \quad\Longrightarrow\quad 22r = 220 \]

\[ r = 10\% \]

Checking: Rs. 100 from SBI and Rs. 120 from ICICI.

Answer: C

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

With the sum and the rate the same, simple interest is proportional to time.

\[ \frac{\text{SI}_{6}}{\text{SI}_{9}} = \frac{6}{9} = \frac{2}{3} \]

So the ratio is \( 2 : 3 \). Neither the sum nor the rate is needed - both cancel - so "data inadequate" is a trap.

Answer: A

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

Set the two interests equal and see what it forces.

\[ \frac{p_{1} \times 15 \times 4}{100} = \frac{p_{2} \times 20 \times 3}{100} \]

\[ 0.6\,p_{1} = 0.6\,p_{2} \quad\Longrightarrow\quad p_{1} = p_{2} \]

\[ \text{difference} = \text{Rs. } 0 \]

Both products \( R \times T \) come to 60, so the two schemes are identical and the salary splits exactly in half - Rs. 4300 each.

Answer: B

Simple Interest =Principal * Rate * Time

Principal = 2600 Rs

Interest Rate = 6.67%

=> SI = 2600 * (20/3)*Time

SI of the number will be a whole number for the first time when time is three as denominator of rate is 3.

Hence time should be set to 3 years.

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: C

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

Equal interests over the same period give

\[ \frac{p_{1} \times 9 \times T}{100} = \frac{p_{2} \times 12 \times T}{100} \]

\[ 9\,p_{1} = 12\,p_{2} \quad\Longrightarrow\quad \frac{p_{1}}{p_{2}} = \frac{12}{9} = \frac{4}{3} \]

The sums are in the inverse ratio of the rates: the lower rate needs the larger capital to earn as much.