# 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. If the age of the father and sister differ by 40 years, find the age of A.

We have to find the age of A

### Solution

Assuming that, the present age of A = x

the present age of B = y

the present age of F = z

the present age of S = t

It’s understood from the question that,

A is elder to b by 2 years. ⇒ x = y + 2

F is twice as old as A. ⇒ z = 2x

B is twice as old as S. ⇒ y = 2t

Also given that the ages of F and S is differing by 40 years. ⇒ z – t = 40.

So, the four equations are:

x = y + 2 … (i)

z = 2x … (ii)

y = 2t … (iii)

z – t = 40 …(iv)

It’s clearly seen from the equations obtained that x, y, z and t are unknowns.

And we have to find the value of x.

So, by using equation (iii) in (i),

(i) Becomes x = 2t + 2

From (iv), we have t = z – 40

Hence, we get

x = 2(z – 40) + 2

= 2z – 80 + 2

= 2z – 78

Using the equation (ii), we have

x = 2×2x – 78

⇒ x = 4x − 78

⇒ 4x – x = 78

⇒ 3x = 78

⇒ x = 78/3

⇒ x = 26

Hence, the age of A is 26 years.

The age of A is 26 years