Question

1 year ago the father was 8 times as old as his son .Now his age is the square of his sons age find their present ages

Answer 1

HEY Buddy.....!! here is ur answer

Let, the present age of father = x year

and the present age of son = y year

Then, Given that : 1 year ago the father was 8 times as old as his son.

1 year ago their ages :

the age of father = (x–1) years

and the age of son = (y–1) years

According to the question : (x–1) = 8(y–1)......(1)

and also given that : the present age of father is square of his son age.

Then, according to the question : x = y²...(2)

From equation (2) the value of x put in the equation (1),

y²–1 = 8y–8

=> y²–8y+7 = 0

=> y²–7y–y+7 = 0

=> y(y–7)–1(y–7) = 0

=> (y–1)(y–7) = 0

=> y = 1 and 7

According to the question the age of son is 1 year doesn't exist. So the present age of
son = 7 years

On putting the value of y in equation (2)

=> x = 7²

=> x = 49 years

Thus, the present age of father and son = 49 years and 7 years respectively.

I hope it will be helpful for you....!!

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