Question
A man invests a total of Rs 10,000 in two schemes:
- I Scheme A: Simple interest at 10% p.a. for 2 years. I
- I Scheme B: Compound interest at 20% p.a., compounded annually, for 2 years. The interest earned from Scheme B is Rs 560 more than the interest earned from Scheme A. How much money (in rupees) did he invest in Scheme B?
Solution
Let amount invested in Scheme A = x Then amount in Scheme B = 10,000 − x Interest from Scheme A (SI at 10% for 2 years): I_A = x × 10 × 2 / 100 = 0.20x Interest from Scheme B (CI at 20% for 2 years): Amount factor = 1.2² = 1.44 I_B = (10,000 − x) × (1.44 − 1) = (10,000 − x) × 0.44 Given: I_B is Rs 560 more than I_A: (10,000 − x) × 0.44 = 0.20x + 560 4,400 − 0.44x = 0.20x + 560 4,400 − 560 = 0.20x + 0.44x 3,840 = 0.64x x = 3,840 / 0.64 = 6,000 So amount in Scheme A = Rs 6,000 Amount in Scheme B = 10,000 − 6,000 = Rs 4,000
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