Class 8 Compound Interest — practice questions with answers

Comparing Quantities · 5 sample questions from the school syllabus. Try each one, then open the answer and the worked solution.

1. Which formula gives the amount A when principal P is compounded annually at rate R% per annum for n years?

A $A = P (1 + \frac{R}{100})^{n}$
B $A = P (1 + \frac{R n}{100})$
C $A = P + \frac{P \cdot R \cdot n}{100}$
D $A = P (1 - \frac{R}{100})^{n}$

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Answer: A)

A = P (1 + \frac{R}{100})^{n}

For compound interest the amount is

A = P (1 + \frac{R}{100})^{n}

. The simple-interest form is

A = P (1 + \frac{R n}{100})

2. A scooter costing Rs 50000 depreciates at 10% per annum. Its value after 1 year is:

ARs 45000
BRs 55000
CRs 40000
DRs 49000

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Answer: A) Rs 45000

Value =

50000 (1 - \frac{10}{100}) = 50000 \times 0.9 =

Rs 45000.

3. The amount on Rs 10000 at 8% per annum compounded half-yearly for 1 year (i.e. 4% per half-year, 2 periods) is:

ARs 10816
BRs 10800
CRs 10824
DRs 10400

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Answer: A) Rs 10816

A = 10000(1.04)² = 10000 * 1.0816 = Rs 10816.

4. At what rate per annum (compounded annually) will Rs 2000 amount to Rs 2205 in 2 years?

A5%
B10%
C4%
D6%

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Answer: A) 5%

(1 + \frac{R}{100})^{2} = \frac{2205}{2000} = 1.1025 = (1.05)^{2}

, so R = 5%.

5. A startup in Bengaluru has 6250 users, growing at 20% per year (compounded annually).
After how many full years will the user base first exceed 10000?

A3 years
B2 years
C4 years
D5 years

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Answer: A) 3 years

Year 1: 6250*1.2 = 7500. Year 2: 9000. Year 3: 10800 > 10000. So it first exceeds 10000 after 3 years.

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