Question

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 differs by $40$ years, find A's age.

Answer 1

Step 1: Form a system of equation based on the given statement of problem

Let the present age of A is $x$,

The present age of B is $y$,

The present age of F is $z$

The present age of S is $t$

According to the question,

A is elder to B by $2$ years$\Rightarrow x=y+2................\left(i\right)$

F is twice as old as A$\Rightarrow z=2x......................\left(ii\right)$

B is twice as old as S$\Rightarrow y=2t.........................\left(iii\right)$

The age of the father and sister differ by $40$ years$\Rightarrow z-t=40....................\left(iv\right)$

Step 2: Solve above system of equations

From equation (i) and (iii) we get $\Rightarrow x=2t+2.......................\left(v\right)$

From equation (iv) we have $t=z-40$ put this value in equation (v) we get,

$$\begin{gathered} x=2\left(z-40\right)+2 \\ x=2z-80+2 \\ x=2z-78.........................\left(vi\right) \end{gathered}$$

Put value of $z$ from equation (ii) in equation (vi)

$$\begin{gathered} x=2\left(2x\right)-78 \\ x=4x-78 \\ 4x-x=78 \\ 3x=78 \\ x=\frac{78}{3} \\ x=26 \end{gathered}$$

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