Question

One year ago, man was 8 times as old as his son. Now, his age is equal to the square of his son’s age. Find their present ages.

Answer 1

Let the present age of the son be x years. 

∴ Present age of the man =`x^2` years 

One year ago,
Age of the son = (x-1) years
Age of the man = (x^2-1) years 

According to the given condition,
Age of the man =`8xx`Age of the son 

∴`x^2-1=8(x-1)` 

⇒`x^2-1=8x-8` 

⇒`x^2-8x+7=0` 

⇒`x^2-7x-x+7=0` 

⇒`x(x-7)-1(x-7)=0` 

⇒`(x-1)(x-7)=0` 

⇒`x-1=0  or  x-7=0` 

⇒`x=1  or x=7` 

∴ `x=7`              (Man’s age cannot be 1 year)
Present age of the son =`7 years `

Present age of the man=`7^2 years=49 years`