Competitive Exams : Quantitative Aptitude - SI/CI (Set 2) | Concept of Maths

Question 1

1. The difference between Simple Interest and Compound Interest (compounded annually) on a certain sum of money for 2 years at 15% per annum is ₹144. Find the principal sum.

A) ₹6,000

B) ₹6,200

C) ₹6,400

D) ₹6,600

Answer

Correct Answer: C

Solution: For 2 years, the difference between CI and SI is given by the formula: $$D = P\left(\frac{r}{100}\right)^2$$ Given D = 144, r = 15. $144 = P \times \left(\frac{15}{100}\right)^2 = P \times \left(\frac{3}{20}\right)^2$ $144 = P \times \frac{9}{400}$ $P = \frac{144 \times 400}{9} = 16 \times 400 = 6,400$. Correct Answer: ₹6,400

Question 2

2. A man borrows ₹2,550 and agrees to pay it back in two equal annual installments at 4% per annum compound interest. Find the value of each installment.

A) ₹1,324

B) ₹1,352

C) ₹1,275

D) ₹1,400

Answer

Correct Answer: B

Solution: Let the installment be x. Formula for loan with installments: $P = \frac{x}{(1 + r/100)^1} + \frac{x}{(1 + r/100)^2}$ $2550 = \frac{x}{1.04} + \frac{x}{(1.04)^2}$ $2550 = \frac{x}{26/25} + \frac{x}{(26/25)^2} = \frac{25x}{26} + \frac{625x}{676}$ $2550 = \frac{650x + 625x}{676} = \frac{1275x}{676}$ $x = \frac{2550 \times 676}{1275} = 2 \times 676 = 1,352$. Correct Answer: ₹1,352

Question 3

3. A sum of money triples itself in 4 years at compound interest. In how many years will it become 27 times itself at the same rate?

Question 4

4. Find the difference between the compound interest on ₹10,000 for 1 year at 10% per annum when compounded annually and when compounded half-yearly.

A) ₹25

B) ₹50

C) ₹0

D) ₹100

Answer

Correct Answer: A

Solution: Case 1 (Annual): $CI = 10,000 \times 10% = ₹1,000$. Case 2 (Half-yearly): Rate = 10/2 = 5%, Periods n = 2. $A = 10,000(1 + 5/100)^2 = 10,000(1.05)^2 = 10,000 \times 1.1025 = 11,025$. CI = 11,025 - 10,000 = ₹1,025. Difference = 1,025 - 1,000 = ₹25. Correct Answer: ₹25

Question 5

5. A sum of money invested at compound interest amounts to ₹4,500 in 2 years and ₹6,750 in 4 years. What is the principal sum?

A) ₹2,500

B) ₹3,000

C) ₹3,200

D) ₹3,500

Answer

Correct Answer: B

Solution: Let Principal be P and multiplier be M = (1+r/100). $P \times M^2 = 4500$ --- (i) $P \times M^4 = 6750$ --- (ii) Divide (ii) by (i): $M^2 = 6750 / 4500 = 1.5$. Substitute $M^2$ in (i): $P \times 1.5 = 4500 \Rightarrow P = 4500 / 1.5 = 3,000$. Correct Answer: ₹3,000

Question 6

6. Calculate the compound interest on ₹12,000 for 9 months at 20% per annum, interest being compounded quarterly.

A) ₹1,800

B) ₹1,902

C) ₹1,891.50

D) ₹2,010.50

Answer

Correct Answer: C

Solution: Principal P = 12,000. Rate = 20% p.a. = 5% per quarter. Time = 9 months = 3 quarters. $A = P(1 + r/100)^n = 12,000(1 + 5/100)^3 = 12,000(1.05)^3$ $A = 12,000 \times 1.157625 = 13,891.50$. CI = 13,891.50 - 12,000 = 1,891.50. Correct Answer: ₹1,891.50

Question 7

7. If the simple interest on a sum of money for 2 years is ₹480 and the compound interest for the same period at the same rate is ₹492, what is the rate of interest per annum?

A) 5%

B) 6%

C) 8%

D) 10%

Answer

Correct Answer: A

Solution: SI for 2 years = 480 \Rightarrow SI for 1 year = 240. The difference between CI and SI for 2 years is the interest earned on the first year's interest. Difference = 492 - 480 = ₹12. This ₹12 is the interest on ₹240 (the first year's SI). Rate $r = (12 / 240) \times 100 = (1/20) \times 100 = 5%$. Correct Answer: 5%

Question 8

8. At what rate percent per annum will ₹2,000 amount to ₹2,420 in 2 years, interest being compounded annually?

A) 8%

B) 9%

C) 12%

D) 10%

Answer

Correct Answer: D

Solution: $A = P(1 + r/100)^n$ $2420 = 2000(1 + r/100)^2$ $1.21 = (1 + r/100)^2$ Taking square root: 1.1 = 1 + r/100 r/100 = 0.1 \Rightarrow r = 10%. Correct Answer: 10%

Question 9

9. A sum of ₹12,600 is divided into two parts such that the simple interest on the first part for 3 years at 2% p.a. is equal to the simple interest on the second part for 4 years at 3% p.a. Find the first part.

A) ₹4,000

B) ₹8,400

C) ₹4,200

D) ₹7,500

Answer

Correct Answer: B

Solution: Let parts be $P_1$ and $P_2$. $SI_1 = P_1 \times 3 \times 2 / 100 = 6P_1/100$ $SI_2 = P_2 \times 4 \times 3 / 100 = 12P_2/100$ Given $SI_1 = SI_2 \Rightarrow 6P_1 = 12P_2 \Rightarrow P_1 = 2P_2$. Also $P_1 + P_2 = 12600 \Rightarrow 2P_2 + P_2 = 12600 \Rightarrow 3P_2 = 12600$ $P_2 = 4200$. $P_1 = 2 \times 4200 = 8400$. Correct Answer: ₹8,400

Question 10

10. If the simple interest on a sum of money for 3 years at 4% per annum is ₹600, what would be the compound interest on the same sum for 2 years at 5% per annum?

A) ₹512.50

B) ₹500.00

C) ₹525.25

D) ₹550.00

Answer

Correct Answer: A

Solution: Step 1: Find Principal from SI. $SI = (P \times R \times T) / 100 \Rightarrow 600 = (P \times 4 \times 3) / 100$ $600 = 12P / 100 \Rightarrow P = (600 \times 100) / 12 = 5,000$. Step 2: Find CI for 2 years at 5%. $A = 5000(1 + 5/100)^2 = 5000(1.05)^2 = 5000 \times 1.1025 = 5512.50$. CI = 5512.50 - 5000 = 512.50. Correct Answer: ₹512.50

Accepted Answer

Correct Answer: D

Solution: In CI, if a sum becomes n times in t years, it becomes $n^k$ times in kt years. Here, n=3 and t=4 years. We want it to become 27 times. $27 = 3^3$. So, k=3. Time taken $= k \times t = 3 \times 4 = 12$ years. Correct Answer: 12 years