IBPS PO Quantitative Aptitude: Simple Interest questions with solutions

Q1. The simple interest on ₹1000 at 10% per annum for 2 years is?

1. ₹100
2. ₹200
3. ₹250
4. ₹150

Answer: ₹200

Simple interest = \(\frac{P \times R \times T}{100}\) = \(\frac{1000 \times 10 \times 2}{100}\) = 200. So the interest is ₹200.

Q2. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:

1. Rs. 650
2. Rs. 690
3. Rs. 698
4. Rs. 700

Answer: Rs. 698

The increase from 3 years to 4 years is Rs. 39, so the yearly simple interest is Rs. 39. In 3 years, interest is Rs. 117, hence principal = 815 - 117 = Rs. 698.

Q3. Mr. Thomas invested an amount of Rs. 13,900 divided into two different schemes A and B at simple interest rates of 14% p.a. and 11% p.a. respectively. If the total simple interest earned in 2 years is Rs. 3508, what was the amount invested in Scheme B?

1. Rs. 6400
2. Rs. 6500
3. Rs. 7200
4. Rs. 7500

Answer: Rs. 6400

If Scheme A gets $13900-x$ and Scheme B gets $x$, then total 2-year interest is $2\times 14\%\,(13900-x)+2\times 11\%\,x=3508$. Solving gives $x=6400$.

Q4. A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9% per annum in 5 years. What is the sum?

1. Rs. 4462.50
2. Rs. 8032.50
3. Rs. 8900
4. Rs. 8925

Answer: Rs. 8925

For simple interest, SI = (P × R × T) / 100. Here, P = (4016.25 × 100) / (9 × 5). This gives P = 8925.

Q5. How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum simple interest?

1. 3.5 years
2. 4 years
3. 4.5 years
4. 5 years

Answer: 4 years

Using SI = (P × R × T) / 100, we get 81 = (450 × 4.5 × T) / 100. Solving gives T = 4 years.

Q6. Reena took a loan of Rs. 1200 at simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?

1. 3.6
2. 6
3. 18
4. Cannot be determined

Answer: 6

If rate = r% and time = r years, then SI = (1200 × r × r) / 100 = 432. This gives r^2 = 36, so r = 6.

Q7. The banker's gain on a certain sum due 2 years hence at 10% per annum is Rs. 24. The present worth is:

1. Rs. 480
2. Rs. 520
3. Rs. 600
4. Rs. 960

Answer: Rs. 600

For a bill due after time t at rate r, banker's gain = banker's discount − true discount. Using the standard relation, the present worth can be obtained from the gain and the given time-rate data. Substituting the values gives the present worth as Rs. 600.

Q8. The banker's discount on a sum of money for 1 year is Rs. 558 and the true discount on the same sum for 2 years is Rs. 600. The rate percent is:

1. 10%
2. 13%
3. 12%
4. 15%

Answer: 12%

The banker's discount and true discount are connected through the rate and time. By forming equations for the given 1-year and 2-year cases and eliminating the sum, the rate is found to be 12%.

Q9. The banker's gain on a sum due 3 years hence at 12% per annum is Rs. 270. The banker's discount is:

1. Rs. 960
2. Rs. 840
3. Rs. 1020
4. Rs. 760

Answer: Rs. 1020

Banker's gain equals the difference between banker's discount and true discount. For a bill due in 3 years at 12%, the standard formula gives the banker's discount once the gain is known. Substituting the values yields Rs. 1020.

Q10. ₹16,000 is divided into two parts. One part is invested in a scheme that gives 15% simple interest annually for 2 years. The other part is invested in a scheme offering 10% simple interest annually for 2 years. If the total interest earned is ₹3,600, find the sum invested at 10% interest rate.

1. 14000
2. 12000
3. 10000
4. 4000

Answer: 4000

Let the amount invested at 10% be \(x\), so the amount at 15% is \(16000-x\). Interest for 2 years is \(0.2x\) and \(0.3(16000-x)\) respectively. Their sum equals 3600, which gives \(x=4000\).