Question

(ii) 2 years ago, my age was 4 1/2 times the age of my son then. 6 years ago, my age
2
was twice the square of the age of my son then. Find the present age of my son.​

Answer 1

Let 2 years ago, son's age be x.

So my age(2 years ago) =4.5 x

So, 6 years ago,

My age=4.5x-4

Son's age=x-4

B/C,

4.5x-4=2((x-4)^2)

or, 4x^2–41x+72=0

Solving the quadratic, we get-

x=8 or x=2.5(not possible)

So my present age of=4.5x+2=38

My son's present age=x+2=10.(Ans)