Simple Interest Questions and Answers

Answer: C

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

B pays 10% and receives 11.5% on the same Rs. 3500, so only the 1.5% gap matters.

\[ \text{gain} = \frac{3500 \times 1.5 \times 3}{100} = \text{Rs. } 157.50 \]

Working out the two interests separately gives Rs. 1207.50 received against Rs. 1050 paid - the same Rs. 157.50, for more effort.

Answer: D

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

The interest falls Rs. 340 short of the principal.

\[ \frac{P \times 4 \times 8}{100} = P - 340 \]

\[ 0.32P = P - 340 \quad\Longrightarrow\quad 0.68P = 340 \]

\[ P = \text{Rs. } 500 \]

Checking: the interest is Rs. 160, which is indeed Rs. 340 less than Rs. 500.

Answer: C

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

Find the rate from the four-year figure.

\[ \text{SI} = 540 - 450 = 90 \quad\Longrightarrow\quad 90 = \frac{450 \times R \times 4}{100} = 18R \]

\[ R = 5\% \]

\[ \text{SI over 6 years} = \frac{450 \times 5 \times 6}{100} = 135 \quad\Longrightarrow\quad \text{amount} = \text{Rs. } 585 \]

Since simple interest is linear in time, the shortcut is \( 90 \times \tfrac{6}{4} = 135 \).

Answer: D

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

\[ \text{SI} = 920 - 800 = 120 \quad\Longrightarrow\quad 120 = \frac{800 \times R \times 3}{100} = 24R \]

\[ R = 5\% \]

At \( 5 + 3 = 8\% \):

\[ \text{SI} = \frac{800 \times 8 \times 3}{100} = 192 \quad\Longrightarrow\quad \text{amount} = \text{Rs. } 992 \]

The extra 3% over three years adds \( 800 \times 0.09 = 72 \) - the same answer in one step.

Answer: D

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

\[ \text{SI} = \frac{18000 \times 15 \times 4}{100} = \text{Rs. } 10{,}800 \]

\[ \text{amount} = 18000 + 10800 = \text{Rs. } 28{,}800 \]

The question asks what Kiran gets, so the principal is included; the interest alone is Rs. 10,800.

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: 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: 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: D

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

Tripling means the interest is twice the principal.

\[ 2P = \frac{P \times R \times 12}{100} \quad\Longrightarrow\quad 200 = 12R \]

\[ R = \frac{50}{3} = 16\tfrac{2}{3}\% \text{ p.a.} \]

That rate is not among the choices, so the answer is none of these. The multiple counts the amount, so the interest is one less than it - using 3 instead of 2 would give 25%.

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.