Multiple choice

Ravi invests ₹75,000 in three schemes: X, Y, and Z. The money invested in Scheme Z is ₹600 more than the money in Scheme X. • Scheme X: 7% interest, compounded annually • Scheme Y: 9% simple interest • Scheme Z: 4% interest, compounded half-yearly At the end of one year, and the amount of Scheme X is ₹223.8 more than the amount of Scheme Z. Find the money invested in Scheme Y.

  1. 15120 rupees

  2. 16500 rupees

  3. 17100 rupees

  4. 18550 rupees
Reveal answer Fill a bubble to check yourself
C Correct answer
AI explanation

Let the investment in Scheme X be x. The investment in Scheme Z is x + 600. Using the compound interest formula for Scheme X, the amount is 1.07x. Using the compound interest formula for Scheme Z with a 2% half-yearly rate over two half-year periods, the amount is (x + 600)(1.02)^2 = 1.0404(x + 600). The problem states that the Scheme X amount is 223.8 more than Scheme Z, giving the equation 1.07x = 1.0404(x + 600) + 223.8. Expanding and solving gives 1.07x = 1.0404x + 624.24 + 223.8, so 0.0296x = 848.04 and x = 28650. The investment in Scheme Y is the remaining amount: 75000 - (28650 + 29250) = 17100 rupees.

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