Question

The present age of a father is $3$ years more than thrice the age of his son.$3$ years hence ,the father will be $10$ years more than twice the age of his son. After how many years, the fathers age will be double the age of the son?

Answer 1

Step 1: Assuming the present ages of father and son by variables $\mathit{x}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{y}$

Let the present age of father be $x$ years and that of son be $y$ years.

Step-2 Framing the equations to find their present ages

Case 1

$x=3y+3........................\left(1\right)$

Case 2

$$\begin{gathered} (x+3)=10+2(y+3) \\ \Rightarrow x+3=10+2y+6 \\ \Rightarrow x-2y=13..................................\left(2\right) \end{gathered}$$

Substituting (1) in (2)

$$\begin{gathered} \left(2\right)\Rightarrow 3y+3-2y=13 \\ \Rightarrow y=13-3=10 \\ \Rightarrow x=33 \end{gathered}$$

Therefore the present age of father =33 years and that of son =10 years

Step-3: Applying condition for fathers age after ‘n’ years

Let after ‘$n$’ years, the fathers age will be double of the age of the son.

After ‘$n$’ years,

father's age =$x+n=33+n$

son's age=$y+n=10+n$

Given,

$$\begin{gathered} 33+n=2(10+n) \\ \Rightarrow 33+n=20+2n \\ \Rightarrow 33-20=2n-n \\ \Rightarrow n=13 \end{gathered}$$

Therefore after 13 years, father's age will be double of sons age.

Checking

after 13 years,

father's age =$33+13=46$

son's age =$10+13=23$

$46=2\times 23$