Simple and Compound Interest — Practice

Question 1

The simple interest on ₹5000 at 8% per annum for 3 years is:
(a) ₹1000 (b) ₹1200 (c) ₹1250 (d) ₹1500

Accepted Answer

(b) ₹1200.

SI = 5000 × 8 × 3 / 100 = ₹1200 — the sum earns ₹400 every year.

Question 2

A sum of money doubles itself in 8 years at simple interest. The annual rate of interest is:
(a) 8% (b) 10% (c) 12.5% (d) 16%

Accepted Answer

(c) 12.5%.

Doubling means the interest equals the principal over 8 years, so R = 100/8 = 12.5% per annum.

Question 3

The compound interest on ₹10,000 at 10% per annum for 2 years is:
(a) ₹2000 (b) ₹2100 (c) ₹2200 (d) ₹1900

Accepted Answer

(b) ₹2100.

Multiplier: 1.1 × 1.1 = 1.21 → amount ₹12,100 → CI = ₹2100. The extra ₹100 over SI's ₹2000 is interest on the first year's ₹1000.

Question 4

At what rate of simple interest will ₹4500 amount to ₹5580 in 3 years?
(a) 6% (b) 7% (c) 8% (d) 9%

Accepted Answer

(c) 8%.

Interest = 5580 − 4500 = ₹1080 over 3 years = ₹360/year. Rate = 360/4500 × 100 = 8%.

Question 5

The difference between compound and simple interest on ₹8000 at 5% per annum for 2 years is:
(a) ₹10 (b) ₹20 (c) ₹25 (d) ₹40

Accepted Answer

(b) ₹20.

Two-year gap = P(R/100)² = 8000 × (0.05)² = 8000 × 0.0025 = ₹20 — interest on the first year's ₹400 interest.

Question 6

The compound interest on ₹8000 at 10% per annum for 1 year, compounded half-yearly, is:
(a) ₹800 (b) ₹810 (c) ₹820 (d) ₹840

Accepted Answer

(c) ₹820.

Convert: 5% per half-year, 2 periods. Amount = 8000 × 1.05² = 8000 × 1.1025 = ₹8820 → CI = ₹820, twenty rupees more than annual compounding's ₹800.

Question 7

The population of a town is 8000 and grows at 5% per annum. Its population after 2 years will be:
(a) 8800 (b) 8820 (c) 8850 (d) 9000

Accepted Answer

(b) 8820.

Population growth is CI in costume: 8000 × 1.05² = 8820. Option (a) is the SI-style answer (8000 + 2 × 400) that ignores growth on growth.

Question 8

The difference between CI and SI on ₹10,000 at 10% per annum for 3 years is:
(a) ₹300 (b) ₹310 (c) ₹320 (d) ₹331

Accepted Answer

(b) ₹310.

Three-year formula: P(R/100)²(3 + R/100) = 10000 × 0.01 × 3.1 = ₹310. Verify: CI = 13310 − 10000 = 3310; SI = 3000; gap = 310 ✓.

Question 9

A sum of money doubles itself in 4 years at compound interest. In how many years will it become 16 times itself?
(a) 8 (b) 12 (c) 16 (d) 64

Accepted Answer

(c) 16 years.

16 = 2⁴, i.e., four successive doublings at 4 years each: 2× at 4, 4× at 8, 8× at 12, 16× at 16 years. Multiplying 4 × 16 = 64 is the planted overreach.

Question 10

The simple interest on a sum is 1/9 of the principal, and the number of years equals the rate per cent per annum. The rate per annum is:
(a) 3% (b) 3⅓% (c) 5% (d) 9%

Accepted Answer

(b) 3⅓%.

PRT/100 = P/9 with T = R gives R² = 100/9, so R = 10/3 = 3⅓% (and T = 3⅓ years). Check: 3.33 × 3.33 / 100 ≈ 1/9 ✓.

Practice — Simple and Compound Interest

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