Question

Five years ago, A's age was four times the age of B. Five years hence, A's age will be twice the age of B. Find their present age

Answer 1

Step 1. Assume the age five year ago

Let five year ago B's age was $x$

Then age of A's is $4x$

Step 2. Find the present age

The present age of B is $x+5$

The present age of A is $4x+5$

Step 3. Find the age five year from present time

After $5$ years, B's age will be $x+10$

A's age will be $4x+10$

It is given that A's age will be twice the age of B.

$$\begin{gathered} \therefore 4x+10=2\left(x+10\right) \\ \Rightarrow 4x+10=2x+20 \\ \Rightarrow 2x=10 \\ \Rightarrow x=5 \end{gathered}$$

Step 4. Calculate the present age

The present age of B is $x+5$

$\therefore$Age of B is

$x+5=5+5=10\left[\because x=5\right]$

So Age of B is $10\mathrm{years}$

The present age of A is $4x+5$

$\therefore$Age of A is

$4x+5=20+5=25\left[\because x=5\right]$

So Age of A is $25\mathrm{years}$.

Hence, the age of A is $25\mathrm{years}$ and age of B is $10\mathrm{years}$.