Quantitative Aptitude

Simple and Compound Interest

3,394 Questions
All Simple interest and its inverse forms 768 Compound interest with annual compounding 742 Simple-interest growth and split sums 626 Difference between compound and simple interest 479 Half-yearly and quarterly compounding 252 Changing rates and fractional years 201 Installments (CI based) 190 Ratio of principal to interest 118 Population growth and depreciation 16

Simple and Compound Interest Questions

Multiple choice

Amit has Rs. 18,000 that he wants to divide into four parts: Fund A, Fund B, Fund C, and Fund D. He wants to ensure that the simple interest earned on each part is equal. If he invests Fund A for 2 years at a rate of 5% per annum, Fund B for 3 years at a rate of 7% per annum, Fund C for 4 years at a rate of 6% per annum, and Fund D for 5 years at a rate of 8% per annum, how much does he invest in Fund A, Fund B, Fund C, and Fund D, respectively (in Rs.)?

  1. 7,400; 4,000; 3,500 and 3,100

  2. 7,400; 5,000; 3,500 and 2,100

  3. 8,400; 4,000; 3,500 and 2,100

  4. 8,400; 5,000; 3,500 and 1,100
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Simple interest is P * R * T / 100. Since interest is equal, P1*0.05*2 = P2*0.07*3 = P3*0.06*4 = P4*0.08*5. This gives ratios 0.1P1 = 0.21P2 = 0.24P3 = 0.4P4. Solving for P1+P2+P3+P4 = 18000 yields 8400, 4000, 3500, and 2100.

Multiple choice

Ingrid invested $20,000 in a heritage preservation bond offered by Aurora Capital, which paid simple interest at a rate of 6% per year. After a few years, she rolled over the entire maturity amount into a long-term sustainability fund at Solstice Group, where it earned simple interest at a rate of 7.5% per year for 8 years. The total interest earned from both investments was in the ratio 5 : 8. For how many years did Ingrid keep her funds with Aurora Capital?

  1. 4 years

  2. 6 years

  3. 8 years

  4. 10 years
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let t be the years at Aurora. Interest 1 = 20,000 * 0.06 * t = 1200t. Maturity amount = 20,000 + 1200t. Interest 2 = (20,000 + 1200t) * 0.075 * 8 = (20,000 + 1200t) * 0.6 = 12,000 + 720t. Ratio 1200t / (12,000 + 720t) = 5/8. Cross-multiplying: 9600t = 60,000 + 3600t, so 6000t = 60,000, t = 10.

Multiple choice

Ramesh invests Rs. 20,000 for 8 years in a savings scheme offering 10% interest per annum, compounded annually. Separately, Sunil invests a certain sum, in the same scheme for 4 years. After 4 years, Sunil reinvests the amount he receives at the end of 4 years for another 4 years at 2.5% simple interest. If both Ramesh and Sunil receive the same amount at the end of 8 years, what was the initial amount, in rupees, deposited by Sunil?

  1. 22,820

  2. 23,600

  3. 24,850

  4. 26,620
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Ramesh: 20000(1.1)^8 = 42871.77. Sunil: P(1.1)^4 * (1 + 0.025*4) = P(1.4641)(1.1) = P(1.61051). Setting P(1.61051) = 42871.77 gives P = 26620.

Multiple choice

A woman invests 70% of her monthly income in a savings account that gives her an annual interest rate of 5%. She decides to invest the remaining 30% in a different account that offers an annual interest rate of 8%. If her monthly income increases by 15% and she keeps the same investment distribution, what is the percentage increase in her total annual interest earned?

  1. 12%

  2. 16%

  3. 15%

  4. 20%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let income be 100. Interest = 0.7*100*0.05 + 0.3*100*0.08 = 3.5 + 2.4 = 5.9. New income = 115. New interest = 0.7*115*0.05 + 0.3*115*0.08 = 4.025 + 2.76 = 6.785. Increase = 6.785 - 5.9 = 0.885. Percentage increase = (0.885/5.9) * 100 = 15%.

Multiple choice

Neha borrows Rs. 1.5 lakh from a bank at an interest rate of 10% per annum, compounded quarterly. She repays Rs. 15,000 at the end of the first year and plans to close the loan by paying the outstanding amount at the end of the second year. What is the total interest paid by Neha over the two years to the nearest rupees?

  1. 31,823

  2. 34,123

  3. 31,203

  4. 32,123
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Quarterly rate = 10%/4 = 2.5%. Amount after 1 year (4 quarters) = 150000 * (1.025)^4 = 165566. After paying 15000, balance = 150566. Amount after 2nd year = 150566 * (1.025)^4 = 166203. Total paid = 15000 + 166203 = 181203. Total interest = 181203 - 150000 = 31203.

Multiple choice

Ravi invests ₹75,000 in three schemes: X, Y, and Z. The money invested in Scheme Z is ₹600 more than the money in Scheme X. • Scheme X: 7% interest, compounded annually • Scheme Y: 9% simple interest • Scheme Z: 4% interest, compounded half-yearly At the end of one year, and the amount of Scheme X is ₹223.8 more than the amount of Scheme Z. Find the money invested in Scheme Y.

  1. 15120 rupees

  2. 16500 rupees

  3. 17100 rupees

  4. 18550 rupees
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice

The simple interest received on Rs. 6800 at 14.5% simple interest for 3 years is x. If Rs. (2x + 1114) is invested at 30% per annum compounded annually for 2 years, then find the compound interest received after 2 years y. Find the value of x + y ?

  1. 8452.5

  2. 9564.8

  3. 7464.3

  4. 7808.7
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

x = (6800 * 14.5 * 3) / 100 = 2958. Investment = 2x + 1114 = 2(2958) + 1114 = 7030. Compound interest on 7030 at 30% for 2 years = 7030 * (1.3^2 - 1) = 7030 * 0.69 = 4850.7. x + y = 2958 + 4850.7 = 7808.7.

Multiple choice

A invests Rs. X at rate of 20% p.a. compound interest for 2 years. At the end of 2 years, he received Rs. 1698.4 as interest. If he invests Rs. 'X + 500' at the rate of 23% simple interest for 3.5 years, then find simple interest received by him.

  1. 3964.8

  2. 3509.8

  3. 3254.8

  4. 3658.8
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

CI for 2 years at 20% is 44% of X. 0.44X = 1698.4, so X = 3860. New principal = 3860 + 500 = 4360. SI = (P*R*T)/100 = (4360 * 23 * 3.5) / 100 = 43.6 * 80.5 = 3509.8.