Class 8 Compound Interest worksheet from Edugain.com featuring math problems on population increase, compound interest, and depreciation.

I'll solve each problem step by step, working through the calculations carefully.

Problem (1): Population increases by 1% every year. Current population is 900,000. Find population after 2 years.

Let me calculate this:
- Initial population = 900,000
- Rate = 1% per year
- Time = 2 years

Using compound interest formula: A = P(1 + r/100)^n
A = 900,000 × (1 + 1/100)²
A = 900,000 × (1.01)²
A = 900,000 × 1.0201
A = 918,090

Problem (2): Compound interest on Rs.4,000 for 2 years at 20% per annum compounded annually.

P = 4,000, R = 20%, T = 2 years

Amount = P(1 + R/100)^T
Amount = 4,000 × (1 + 20/100)²
Amount = 4,000 × (1.2)²
Amount = 4,000 × 1.44
Amount = 5,760

Compound Interest = Amount - Principal
CI = 5,760 - 4,000 = 1,760

Problem (3): Compound interest on Rs.30,000 for 2 years and 3 months at 40% per annum compounded annually.

For 2 years and 3 months, I need to calculate:
- First 2 years with compound interest
- Then 3 months (which is 1/4 of a year) with simple interest on the amount

Step 1: Amount after 2 years
A₁ = 30,000 × (1 + 40/100)²
A₁ = 30,000 × (1.4)²
A₁ = 30,000 × 1.96
A₁ = 58,800

Step 2: Simple interest for 3 months on Rs.58,800
SI = (P × R × T)/100
SI = (58,800 × 40 × 1/4)/100
SI = (58,800 × 10)/100
SI = 5,880

Final Amount = 58,800 + 5,880 = 64,680
Compound Interest = 64,680 - 30,000 = 34,680

Problem (4): Workers increase by 30% every year. Currently 7,000 workers. Find workers after 2 years.

Initial workers = 7,000
Rate = 30%
Time = 2 years

Workers after 2 years = 7,000 × (1 + 30/100)²
= 7,000 × (1.3)²
= 7,000 × 1.69
= 11,830 workers

Problem (5): Vinayak borrowed Rs.80,000 at 10% per annum compounded annually. Amount payable after 3 years?

P = 80,000, R = 10%, T = 3 years

Amount = 80,000 × (1 + 10/100)³
Amount = 80,000 × (1.1)³
Amount = 80,000 × 1.331
Amount = 106,480

Answer: d. Rs. 106,480

Problem (6): At 10% per annum compounded annually, how long for Rs.40,000 to compound to Rs.53,240?

P = 40,000, A = 53,240, R = 10%, T = ?

A = P(1 + R/100)^T
53,240 = 40,000 × (1.1)^T
53,240/40,000 = (1.1)^T
1.331 = (1.1)^T

Now I need to find T such that 1.1^T = 1.331
1.1¹ = 1.1
1.1² = 1.21
1.1³ = 1.331 ✓

So T = 3 years

Answer: b. Three years

Problem (7): Sara bought scooter at Rs.40,000. Value depreciates at 10% per annum. Value after 3 years?

This is depreciation, so we use: V = P(1 - R/100)^T

V = 40,000 × (1 - 10/100)³
V = 40,000 × (0.9)³
V = 40,000 × 0.729
V = 29,160

Problem (8): Compound interest on Rs.6,000 for one year at 6% per annum compounded half yearly.

When compounded half-yearly:
- Rate per half-year = 6%/2 = 3%
- Number of periods in 1 year = 2

Amount = 6,000 × (1 + 3/100)²
Amount = 6,000 × (1.03)²
Amount = 6,000 × 1.0609
Amount = 6,365.40

Compound Interest = 6,365.40 - 6,000 = 365.40

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Final Answer:
(1) 918,090
(2) Rs. 1,760
(3) Rs. 34,680
(4) 11,830 workers
(5) d. Rs. 106,480
(6) b. Three years
(7) Rs. 29,160
(8) Rs. 365.40