Simple interest vs compound interest — compare for ₹10000 at 10% for 3 years

easy CBSE • NCERT Class 8 • 2 min read
Tags Comparing Quantities

Question

Compare the simple interest and compound interest on ₹10,000 at 10% per annum for 3 years (compounded annually). Find the difference.

(NCERT Class 8)

Solution — Step by Step

SI=P×R×T100=10000×10×3100=₹3000\text{SI} = \frac{P \times R \times T}{100} = \frac{10000 \times 10 \times 3}{100} = \mathbf{\text{₹}3000}

Amount after 3 years with SI: 10000+3000=₹13,00010000 + 3000 = \text{₹}13,000

Year 1: Interest on ₹10,000 = 10000×10100=₹1000\frac{10000 \times 10}{100} = \text{₹}1000. New principal = ₹11,000.

Year 2: Interest on ₹11,000 = 11000×10100=₹1100\frac{11000 \times 10}{100} = \text{₹}1100. New principal = ₹12,100.

Year 3: Interest on ₹12,100 = 12100×10100=₹1210\frac{12100 \times 10}{100} = \text{₹}1210. Final amount = ₹13,310.

CI=13310−10000=₹3310\text{CI} = 13310 - 10000 = \mathbf{\text{₹}3310}

CI−SI=3310−3000=₹310\text{CI} - \text{SI} = 3310 - 3000 = \mathbf{\text{₹}310}

Compound interest earns ₹310 more than simple interest over 3 years. This difference comes from “interest on interest” — in CI, each year’s interest is added to the principal, so you earn interest on the accumulated interest too.

Why This Works

In simple interest, the interest each year is always 10000×10100=₹1000\frac{10000 \times 10}{100} = \text{₹}1000 — fixed, because the principal doesn’t change. In compound interest, the principal grows each year, so the interest amount also grows.

The CI formula captures this directly:

A=P(1+R100)T=10000×(1.1)3=10000×1.331=₹13,310A = P\left(1 + \frac{R}{100}\right)^T = 10000 \times (1.1)^3 = 10000 \times 1.331 = \text{₹}13,310

Alternative Method — Using the CI Formula Directly

CI=P[(1+R100)T−1]=10000[(1.1)3−1]=10000×0.331=₹3310\text{CI} = P\left[\left(1 + \frac{R}{100}\right)^T - 1\right] = 10000\left[(1.1)^3 - 1\right] = 10000 \times 0.331 = \text{₹}3310

For 2 years, there’s a shortcut for the difference between CI and SI:

CI−SI=P(R100)2\text{CI} - \text{SI} = P\left(\frac{R}{100}\right)^2

For 2 years: 10000×(0.1)2=₹10010000 \times (0.1)^2 = \text{₹}100. For 3 years, the formula is more complex, so the year-by-year method is safer in exams.

Common Mistake

Students use the SI formula for CI problems or forget to update the principal each year. In SI, the principal stays ₹10,000 throughout. In CI, it grows: ₹10,000 → ₹11,000 → ₹12,100. If you compute 10% of ₹10,000 for all three years in a CI problem, you’ll get ₹3,000 — the SI answer, not CI.

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