Simple Interest Questions and Answers

Answer: C

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

\[ \text{SI} = 3600 - 3000 = 600 \quad\Longrightarrow\quad 600 = \frac{3000 \times R \times 4}{100} = 120R \]

\[ R = 5\% \]

At 6%:

\[ \text{SI} = \frac{3000 \times 6 \times 4}{100} = 720 \quad\Longrightarrow\quad \text{amount} = \text{Rs. } 3720 \]

The data are complete, so "cannot be determined" is a trap: one extra percentage point over four years adds \( 3000 \times 0.04 = 120 \).

Answer: D

Let the sum be Rs. x, then it becomes Rs. 4x in eight years Rs. 3x is the interest on x for eight years.

R = (100 * 3x)/(x * 8) = 300/8 %

If the sum becomes ten times itself, then interest is 9x.

The required time period = (100 * 9x)/(x * 300/8) = (100 * 9x * 8)/(x * 300) = 24 years.

Answer: B

Compounded annually, doubling in two years means

\[ \left(1 + \frac{R}{100}\right)^{2} = 2 \]

\[ 1 + \frac{R}{100} = \sqrt{2} = 1.4142 \]

\[ R = 41.42 \approx 41.4\% \]

Under simple interest the answer would be 50%; compounding needs less, because the first year's interest itself earns in the second.

Answer: B

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

Let \( p \) be the part at 8%; the rest, \( 10000 - p \), earns 10%.

\[ 0.08p + 0.10(10000 - p) = 890 \]

\[ 0.08p + 1000 - 0.10p = 890 \quad\Longrightarrow\quad -0.02p = -110 \]

\[ p = \text{Rs. } 5500 \]

Checking: Rs. 440 from the 8% part and Rs. 450 from the Rs. 4500 at 10%.

Answer: D

Simple interest is proportional to time, so over eight years instead of five the interest rises by a fixed factor no matter what the rate is.

\[ \frac{\text{SI}_{8}}{\text{SI}_{5}} = \frac{8}{5} = 1.6 \]

That is a 60% increase for every rate and every sum, so the stated condition carries no information about \( R \).

The rate therefore cannot be determined. A condition that holds identically is not a constraint.

Answer: B

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

What Manoj pays Anwar on the borrowed Rs. 3450 at 6% for 3 years:

\[ \frac{3450 \times 6 \times 3}{100} = \text{Rs. } 621 \]

What he receives from Ramu on the whole sum \( S \) at 9% for 3 years:

\[ \frac{S \times 9 \times 3}{100} = 0.27S \]

His gain is the difference:

\[ 0.27S - 621 = 824.85 \quad\Longrightarrow\quad 0.27S = 1445.85 \]

\[ S = \text{Rs. } 5355 \]

So he added \( 5355 - 3450 = \text{Rs. } 1905 \) of his own money.

Answer: D

If the interest earned is 25% more than the earlier interest then the rate of interest also should be 25% higher than the earlier rate. 

Let the earlier rate of interest be x%. 

Now it will be (x + 3)% 

% increase = (x + 3) - x/x * 100 = 25

=> x = 12

Answer: D

Given that Rs.10,000 is invested in S.I for two years and Rs.8,000 in C.I for two years 

=>  C.I - S.I

=> 8,000 { [1 + r/100]2   - 1} = (10,000)2r /100

=> 8{ 1 + 2r/100 + r2 / (100)2 - 1} = r/5 

=> 16r/100 + 8r2/(100)2 = 20r/100

 => 4r/10 = 8r2/(100)2 

=> 8[r/100]2 - 4r/100 = 0 

=> r/100 {8r/100 -4} = 0 

=> r = 0% of 50% 

Since r!= 0%, r =50%

Answer: A

Apply the first year's rate, then compare with the final amount.

\[ 4500 \times 1.12 = \text{Rs. } 5040 \]

\[ 1 + \frac{R_{2}}{100} = \frac{5544}{5040} = 1.1 \]

\[ R_{2} = 10\% \]

With rates that differ by year the two growth factors multiply; they cannot be averaged.

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.