A is elder to B by 2 years. A's father F is twice as old as A and B is twice as old as his sister S.

Solution:

Let the present ages of $A, B, F$ and $S$ be $x, y, z$ and $t$ years respectively.

$A$ is elder to $B$ by 2 years. Thus, we have $x=y+2$

$F$ is twice as old as $A$. Thus, we have $z=2 x$

$B$ is twice as old as $S$. Thus, we have $y=2 t$

The ages of $F$ and $S$ is differing by 40 years. Thus, we have $z-t=40$

So, we have four equations

$x=y+2, \ldots \ldots(1)$

$z=2 x, \ldots \ldots(2)$

$y=2 t, \ldots \ldots(3)$

 

$z-t=40 \ldots \ldots(4)$

Here x, y, z and t are unknowns. We have to find the value of x.

By using the third equation, the first equation becomes $x=2 t+2$

 

From the fourth equation, we have $t=z-40$

Hence, we have

$x=2(z-40)+2$

$=2 z-80+2$

 

$=2 z-78$

Using the second equation, we have

$x=2 \times 2 x-78$

$\Rightarrow x=4 x-78$

$\Rightarrow 4 x-x=78$

 

$\Rightarrow 3 x=78$

$\Rightarrow x=\frac{78}{3}$

$\Rightarrow x=26$

Hence, the age of $A$ is 26 years.