Question

A and B are friends and their ages differ by $2$ years. A's father D is twice as old as A and B is twice as old as his sister C. The age of D and C differ by $40$ years. Find the ages of A and B.

Answer 1

Let the ages of A and B be x and y years respectively. Then,
$x-y=\pm 2$ [Given]

D's age $=2x$ years and, C's age $=\frac{y}{2}$ years.

Clearly, D is older than C
$2x-\frac{y}{2}=40\Rightarrow 4x-y=80$

Thus, we have the following two systems of linear equations
$x-y=2$ .(i)
and, $4x-y=80$ (ii)

or
$x-y=-2$ (iii)
and, $4x-y=80$ ..(iv)

Substracting equation (i) from equation (ii), we get
$3x=78\Rightarrow x=26$Putting $x=26$ in equation (i), we get $y=24$

or

Substracting equation (iv) from equation (iii), we get
$-3x=-82\Rightarrow x=\frac{82}{3}=27\frac{1}{3}$

Putting $x=\frac{82}{3}$ in equation (iii), we get

$y=\frac{82}{3}+2=\frac{88}{3}=29\frac{1}{3}$

Hence, A's age $=26$ years and B's age $=24$ years
or
A's age $=27\frac{1}{3}$ years and B's age $=29\frac{1}{3}$ years.