Question

Five years ago a man was 7 times as old as his son. Five years later he will be 3 times as old as his son. Find their present ages.​

Answer 1

let 5 years ago age of son=x and age of father=7x

present age of son=x+5

present age of father=(7x+5)

five years later age of son= x+10

and age of father=(7x+10)

(7x+10)=3(x+10)

7x+10=3x+30

7x-3x=30-10

4x=20

x=20/4=5

present age of son=x+5=10

present age of father=7x+5=40

Answer 2

Answer:

Let the age of man be x and the age of his son be y .

ATQ,

Given that five years ago the age of the father is x-5 and the age of the son is y-5 years.

So the equation formed is x-5 = 7(y-5)

⇒ x-5 = 7y-35

⇒ x-7y = -30 --- (i)

After five years the age of the father is x+5 and the age of the son is y+5 years.

⇒x+5=3(y+5)

⇒x+5=3y+15

⇒x-3y=10 (ii)

On applying the substitution method:

⇒ x-7y= -30

⇒ x-3y=  10

  --------------

⇒ -4y=-40

⇒ y= 10 years

Putting the value of y in equation 2 :

We get the age of father as 40 years

Hence the age of the father is 40 years and the age of the son is 10 years. (Ans)

Hope it helps you