Question

one year ago a man was 8 times as old as his son how his age is equal to the square of his age find their present age​

Answer 1

Correct Question :- One year ago a man was 8 times as old as his son. Now his age is equal to the square of his age find their present age.

$\huge\mathfrak{Solution}$

Let the age of son be x

So, one year ago, son's age = x - 1

Given, then man's age = 8(x - 1)

= 8x - 8

This is the man's age one year ago.

Now given that his present age is equal to the square of son's age.

So, man's present age would be 8x - 8 + 1 = 8x - 7

(Since, 8x - 8 is the age of a year ago)

So, according to the question,

x² = 8x - 7

→ x² - 8x + 7 = 0

→ x² - 7x - x + 7 = 0

→ x(x - 7) - 1(x - 7) = 0

→ (x - 1)(x - 7) = 0

So, either, (x - 1) = 0 or (x - 7) = 0

→ x = 1 or x = 7

Now, if we take x as 1, then one year ago his age would be 0, so one year ago father's age would be 8 × 0 = 0, which doesn't satisfy the conditions. Hence, we will take x = 7

Hence, present age of son = 7 years.

So, present age of man = 7² = 49 years

$\huge\mathfrak{Answer}$

$\textsf{Man's age = 49 years}$

$\textsf{Son's age = 7 years}$

Answer 2

Let the age of the son (1 year ago)= x-1 years

Then according to the question , the age of man = 8(x-1)= 8x-8 years

So their current ages are like :

Age of son = x years

Age of man = 8x-8+1 = 8x-7 years

According to the question ,

x² = 8x-7

x²-8x+7 = 0

x²-x-7x+7 = 0

x(x-1)-7(x-1) = 0

(x-1)(x-7) = 0

So the values of x are :

x = 1 and x = 7

For satisfying our need, we have taken x = 7

The present age of son = x = 7 years

Age of father (man)= 8x-7 = 49 years