A man invested a total of Rs. 47,000. He invested part of this amount at simple interest at 12% per annum for 2 years, and the remaining amount at compound interest at 15% per annum for 2 years. If the compound interest earned is Rs. 532.5 more than the simple interest, what is the amount invested at compound interest?

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**Solution:**

1. Let the amount invested at simple interest be *x* rupees.

2. Then, the amount invested at compound interest will be *(47000 - x)* rupees.

3. Calculate the simple interest (SI) for 2 years at 12% p.a.:
   $$SI = \frac{x \times 12 \times 2}{100} = \frac{24x}{100} = 0.24x$$

4. Calculate the compound interest (CI) for 2 years at 15% p.a.:
   Compound amount = $ $(47000 - x) \times \left(1 + \frac{15}{100}\right)^2 = (47000 - x) \times (1.15)^2 = (47000 - x) \times 1.3225$ $

Compound interest = Compound amount - Principal =
   $$1.3225(47000 - x) - (47000 - x) = 0.3225(47000 - x)$$

5. According to the problem, the compound interest is Rs. 532.5 more than the simple interest:
   $$0.3225(47000 - x) - 0.24x = 532.5$$

6. Simplify the equation:
   $$0.3225 \times 47000 - 0.3225x - 0.24x = 532.5$$
   $$15157.5 - 0.5625x = 532.5$$

7. Rearranging:
   $$15157.5 - 532.5 = 0.5625x$$
   $$14625 = 0.5625x$$

8. Solve for *x*:
   $$x = \frac{14625}{0.5625} = 26000$$

9. Amount invested at compound interest =
   $$47000 - 26000 = 21000$$

**Answer:** Rs. 21,000

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