Question

A father is 7 times as old as his son. Two years ago the father was 13 times as old as his son. Find the present age

Answer 1

let father age be x and son age be y.
x=7y (1)
x-2=13(y-2). (2)
substituting x value in case 2
7y-2=13y-26
6y=24
y=4.
substituting y value in case 1
x=7y
x=7*4
x=28

Answer 2

GiveN:-

A father is 7 times as old as his son, two years ago the Father was 13 times as old as his son.

To FinD:-

The present age of son.

SolutioN:-

Let the present age of the son be x years.

Let the present age of the father be 7x years.

2 years ago their ages were:-

Son = (x - 2) years.

Father = (7x - 2) years.

It is said that two years ago the Father was 13 times as old as his son.

According to the question,

$\large\implies{\sf{7x-2=13(x-2)}}$

$\large\implies{\sf{7x-2=13x-26}}$

$\large\implies{\sf{-2+26=13x-7x}}$

$\large\implies{\sf{24=6x}}$

$\large\implies{\sf{\dfrac{24}{6}=x}}$

$\large\implies{\sf{\dfrac{\cancel{24}}{\cancel{6}}=x}}$

$\large\implies{\sf{4=x}}$

$\large\therefore\boxed{\bf{x=4.}}$

Their present ages:-

Son's age = x = 4 years.

Father's age = 7x = 7 × 4 = 28 years.

The present age of the son is 4 years.