A man invested Rs X and Rs x + 400 in two different schemes A & B respectively. The rate of interest offered by schemes A & B is 10% p.a. and 12% p.a. respectively. If the sum of interest received by man from scheme A after four years and from scheme B after two year is Rs 640, then find X.

The correct answer is 850.

Let's solve the problem step-by-step.

First, let's understand the details of the investments in the two schemes:
1. In Scheme A:
- Principal invested, $P_A=\; X$
- Rate of interest, $R_A=\; 10\%$ per annum simple interest
- Time period, $T_A=\; 4$ years

2. In Scheme B:
- Principal invested, $P_B=\; X\; +\; 400$
- Rate of interest, $R_B=\; 12\%$ per annum simple interest
- Time period, $T_B=\; 2$ years

The formula for Simple Interest is:
$\text{Simple Interest (SI)}=\frac{P\times R\times T}{100}$

Let's calculate the interest received from Scheme A ($I_A$):
$I_A=\frac{X\times \; 10\; \times \; 4}{100}=\frac{40X}{100}=\; 0.4$X

Now, let's calculate the interest received from Scheme B ($I_B$):
$I_B=\frac{\left(X+\; 400\right)\times \; 12\; \times \; 2}{100}=\frac{24\left(X+\; 400\right)}{100}=\; 0.24$(X + 400)
Simplifying this:
$I_B=\; 0.24$X + 96

According to the question, the sum of the interest received from both schemes is Rs 640:
$I_A+I_B=\; 640$

Substituting the values of $I_A$ and $I_B$:
$0.4X+0.24X+\; 96\; =\; 640$

Combine the terms containing X:
$0.64X+\; 96\; =\; 640$

Subtract 96 from both sides of the equation:
$0.64X=\; 640\; -\; 96$
$0.64X=\; 544$

Now, solve for X:
$X=\frac{544}{0.64}$
$X=\frac{54400}{64}$
$X=\; 850$

Thus, the value of X is Rs 850.