Example 1 · level 1
Kapoor Electronics invests ₹1,80,000 at 10% p.a. compounded annually for 2 years. Find the compound interest (to the nearest rupee).
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- i = 10% per period, n = 2 periods.
- A = 1,80,000 × 1.1^2 = ₹2,17,800.
- CI = ₹2,17,800 − 1,80,000 = ₹37,800.
Answer: ₹37,800