A man invested some amount at R% per annum on simple interest for eight years in scheme X. He also invested same amount at (R+4)% per annum on simple interest for same period of time in scheme Y. If interest received from scheme Y is Rs.1152 more than that from scheme X, then find the amount (in Rs.).

Correct Answer: The correct option is 3600 (or Rs. 3600).

Step-by-Step Explanation:

Step 1: Understand the given information

Let the principal amount invested in each scheme be $P$ (in Rs.).
- Scheme X: Rate of interest = $R\%$ per annum, Time period = $8$ years.
- Scheme Y: Rate of interest = $(R+4)\%$ per annum, Time period = $8$ years.
- The interest received from Scheme Y is Rs. 1152 more than that received from Scheme X.

Step 2: Formula for Simple Interest

The formula for Simple Interest ($SI$) is given by:

$SI=\frac{P\×R\×T}{100}$

Step 3: Calculate interest difference between Scheme Y and Scheme X

Interest from Scheme X ($S{I}_X$):

$S{I}_X=\frac{P\×R\×8}{100}$

Interest from Scheme Y ($S{I}_Y$):

$S{I}_Y=\frac{P\×(R+4)\×8}{100}$

Step 4: Set up the equation for difference in interest

Given that $S{I}_Y-S{I}_X=1152$:

$\frac{P\×(R+4)\×8}{100}-\frac{P\×R\×8}{100}=1152$

Factoring out common terms $\frac{8P}{100}$:

$\frac{8P}{100}\×\left[\right(R+4)-R]=1152$

$\frac{8P}{100}\×4=1152$

$\frac{32P}{100}=1152$

Step 5: Solve for principal amount ($P$)

$P=\frac{1152\×100}{32}$

$P=36\×100=3600$

Thus, the invested amount is Rs. 3600.