If $$\frac{30}{x-y} + \frac{44}{x+y} = 10 \quad \text{and} \quad \frac{40}{x-y} + \frac{55}{x+y} = 13$$, then xy is equal to :  

Correct option is D

Given:

$$\frac{30}{x-y} + \frac{44}{x+y} = 10 \quad \text{and} \quad \frac{40}{x-y} + \frac{55}{x+y} = 13$$​

Let $$\frac{1}{x-y} = a \quad \text{and} \quad \frac{1}{x+y} = b$$​​

So, equations are: 30a + 44b = 10 and 40a + 55b = 13

From first equation, we get

$$a = \frac{10 - 44b}{30}$$​

​$$a = \frac{13 - 55b}{40}$$​​

Equating these two equations, we get

$$\frac{10 - 44b}{30} = \frac{13 - 55b}{40}$$​

​$$400 - 1760b = 390 - 1650b$$​​

$$-1760b + 1650b = 390 - 400$$​

​$$b = \frac{10}{110} = \frac{1}{11}$$​​

$$a = \frac{13 - 55 \times \frac{1}{11}}{40} = \frac{13 - 5}{40} = \frac{8}{40} = \frac{1}{5}$$​

​$$\frac{1}{x-y} = \frac{1}{5}$$​​

x – y = 5

x = y + 5a (1)

$$\frac{1}{x+y} = \frac{1}{11}$$​

x = 11 – y(2)

Solving these two equations, we get

y + 5 = 11 – y

y + y = 11 – 5

2y = 6

$$y = \frac{6}{2} = 3$$​

So, xy = 8 × 3 = 24