A invests a certain amount at 5\% simple interest per annum and B invests the remaining amount at 3\% simple interest per annum for 2 years. If the total amount invested by A and B together is \$4000 and both receive the same simple interest, what is the amount invested by A?

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**Solution:**

1. Let the amount invested by A be $x$ and by B be $4000 - x$.

2. The simple interest formula is:
   $$	ext{SI} = \frac{P 	imes R 	imes T}{100}$$

3. According to the problem, the simple interest received by A and B are equal:
   $$\frac{x 	imes 5 	imes 2}{100} = \frac{(4000 - x) 	imes 3 	imes 2}{100}$$

4. Simplify the equation:
   $$\frac{10x}{100} = \frac{6(4000 - x)}{100}$$

5. Multiply both sides by 100 to eliminate denominators:
   $$10x = 6(4000 - x)$$

6. Expand the right side:
   $$10x = 24000 - 6x$$

7. Bring like terms together:
   $$10x + 6x = 24000$$
   $$16x = 24000$$

8. Solve for $x$:
   $$x = \frac{24000}{16} = 1500$$

**Answer:** The amount invested by A is $\boxed{1500}$.

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