Quantitative Aptitude Quiz: SI & CI Concepts (Set 1)

Question 1

1. The difference between simple interest and compound interest on a certain sum of money for 3 years at 10% per annum is ₹93. Find the principal sum.

A) ₹2,500

B) ₹2,800

C) ₹3,000

D) ₹3,500

Answer

Correct Answer: C

Solution: For 3 years, the difference (D) between CI and SI is given by: $$D = P \left(\frac{R}{100}\right)^2 \left(3 + \frac{R}{100}\right)$$ Given D = 93 and R = 10: $$93 = P \left(\frac{10}{100}\right)^2 \left(3 + \frac{10}{100}\right)$$ $$93 = P \left(\frac{1}{100}\right) \left(3.1\right)$$ $$93 = P \times 0.031 ⇒ P = \frac{93}{0.031} = 3000$$ The principal is ₹3,000.

Question 2

2. What equal annual installment will discharge a debt of ₹6,450 due in 4 years at 5% per annum simple interest?

A) ₹1,400

B) ₹1,500

C) ₹1,550

D) ₹1,600

Answer

Correct Answer: B

Solution: Let the annual installment be x. The sum of the present value of installments plus their interests should equal the debt. $$x + \left(x + \frac{x \times 5 \times 1}{100}\right) + \left(x + \frac{x \times 5 \times 2}{100}\right) + \left(x + \frac{x \times 5 \times 3}{100}\right) = 6450$$ $x + 1.05x + 1.10x + 1.15x = 6450$ $$4.30x = 6450 ⇒ x = \frac{6450}{4.3} = 1500$$ The installment is ₹1,500.

Question 3

3. A man borrows ₹21,000 at 10% compound interest. If he pays it back in two equal annual installments, what is the value of each installment?

A) ₹12,100

B) ₹12,000

Question 4

4. At what rate percent per annum will a sum of money double itself in 8 years at simple interest?

A) 10%

B) 12%

C) 12.25%

D) 12.5%

Answer

Correct Answer: D

Solution: Let Principal = P. Since the sum doubles, Amount = 2P. SI = Amount - Principal = 2P - P = P. $$SI = \frac{P \times R \times T}{100} ⇒ P = \frac{P \times R \times 8}{100}$$ $$1 = \frac{8R}{100} ⇒ R = \frac{100}{8} = 12.5%$$

Question 5

5. If the compound interest on a certain sum for 2 years at 4% p.a. is ₹510, what would be the simple interest on the same sum for the same period and same rate?

A) ₹480

B) ₹500

C) ₹450

D) ₹490

Answer

Correct Answer: B

Solution: Relationship: $CI = SI \left(1 + \frac{R}{200}\right)$ for 2 years. $$510 = SI \left(1 + \frac{4}{200}\right)$$ $$510 = SI \left(1 + 0.02\right)$$ $$510 = 1.02 \times SI ⇒ SI = \frac{510}{1.02} = 500$$ The simple interest is ₹500.

Question 6

6. A sum of money becomes 1.331 times of itself in 3 years at compound interest. What is the annual rate of interest?

A) 8%

B) 11%

C) 10%

D) 15%

Answer

Correct Answer: C

Solution: $$A = P(1 + r/100)^n$$ $$1.331P = P(1 + r/100)^3$$ $$1.331 = (1 + r/100)^3$$ Take the cube root of both sides: $$\sqrt[3]{1.331} = 1 + r/100$$ $1.1 = 1 + r/100 ⇒ 0.1 = r/100 ⇒ r = 10%$

Question 7

7. A sum was put at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate, it would have fetched ₹360 more. Find the sum.

A) ₹6,000

B) ₹5,000

C) ₹4,500

D) ₹7,200

Answer

Correct Answer: A

Solution: The extra interest is due to the increase in rate over 3 years. Total extra percentage = $3 years \times 2% = 6%$. According to the problem, 6% of Principal (P) = 360. $$\frac{6}{100} \times P = 360$$ $$P = \frac{360 \times 100}{6} = 6000$$ The sum is ₹6,000.

Question 8

8. If the simple interest on a sum of money for 2 years at 5% per annum is ₹50, what is the compound interest on the same sum at the same rate and for the same time?

A) ₹50.50

B) ₹51.25

C) ₹52.00

D) ₹55.00

Answer

Correct Answer: B

Solution: SI = 50, T = 2, R = 5. $P = \frac{SI \times 100}{R \times T} = \frac{50 \times 100}{5 \times 2} = 500$. CI for 2 years: $A = 500(1 + 5/100)^2 = 500(1.05)^2 = 500 \times 1.1025 = 551.25$. CI = 551.25 - 500 = 51.25.

Question 9

9. The population of a town increases by 5% annually. If the current population is 92,610, what was it 3 years ago?

A) 75,000

B) 85,000

C) 80,000

D) 90,000

Answer

Correct Answer: C

Solution: Population growth follows CI formula: $P_{now} = P_{ago} (1 + r/100)^n$ $$92610 = P_{ago} (1 + 5/100)^3$$ $$92610 = P_{ago} (1.05)^3$$ $$92610 = P_{ago} \times 1.157625$$ $$P_{ago} = \frac{92610}{1.157625} = 80000$$

Question 10

10. A sum of money at compound interest amounts to ₹650 at the end of the first year and ₹676 at the end of the second year. Find the sum.

A) ₹600

B) ₹620

C) ₹630

D) ₹625

Answer

Correct Answer: D

Solution: Interest on 650 for one year = 676 - 650 = 26. Rate $R = \frac{26}{650} \times 100 = 4%$. For the first year: A = P(1 + R/100) 650 = P(1 + 4/100) ⇒ 650 = 1.04P $P = \frac{650}{1.04} = 625$. The sum is ₹625.

Accepted Answer

Correct Answer: A

Solution: For CI installments, $P = \frac{x}{(1 + r/100)^1} + \frac{x}{(1 + r/100)^2}$ $$21000 = \frac{x}{1.1} + \frac{x}{1.21}$$ Multiply by 1.21: $$21000 \times 1.21 = 1.1x + x$$ $$25410 = 2.1x ⇒ x = \frac{25410}{2.1} = 12100$$ Each installment is ₹12,100.