A man invested an equal amount of money in two different schemes, X and Y, both for 8 years. Scheme X offers simple interest at a rate of $R\%$ per annum, while scheme Y offers simple interest at a rate of $(R + 4)\%$ per annum. The interest earned from scheme Y exceeds that from scheme X by Rs. 1152. What is the amount (in Rs.) invested by the man?

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**Solution:**

1. Let the amount invested be $P$ rupees.

2. Interest from scheme X for 8 years at $R\%$ per annum (simple interest) is:
   $$I_X = \frac{P 	imes R 	imes 8}{100} = \frac{8PR}{100}$$

3. Interest from scheme Y for 8 years at $(R + 4)\%$ per annum is:
   $$I_Y = \frac{P 	imes (R + 4) 	imes 8}{100} = \frac{8P(R + 4)}{100}$$

4. Given that the interest from scheme Y exceeds that from scheme X by Rs. 1152:
   $$I_Y - I_X = 1152$$

5. Substitute the values:
   $$\frac{8P(R + 4)}{100} - \frac{8PR}{100} = 1152$$

6. Simplify the left side:
   $$\frac{8P(R + 4 - R)}{100} = 1152$$
   $$\frac{8P 	imes 4}{100} = 1152$$

7. Calculate:
   $$\frac{32P}{100} = 1152$$

8. Multiply both sides by 100:
   $$32P = 1152 	imes 100 = 115200$$

9. Divide both sides by 32:
   $$P = \frac{115200}{32} = 3600$$

**Answer:** Rs. 3600

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