Question

One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Their present ages are ?​

Answer 1

Answer:

Let son's age be x

One year ago

Son's age = x-1

Therefore man's age = 8(x-1) = 8x-8

Now

Man's age = x^2

Difference between both ages of man = 1 year

So

X^2 -(8x-8) = 1

X^2 -8x + 8 - 1 = 0

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, 7

But son cannot be one year old

Therfore x = 7

Son's age = 7 years

Father's age = 49 years

Answer 2

Given :-

• One year ago,a man was 8 times as old as his son.

• Now his age is equal to the square of his son's age.

To find :-

• Their present ages = ?

Solution :-

Let the present age of son be x years

and the present age of man be x² years

$\dag\bf{\underline{One\:year\:ago}}:-$

• Son's age = (x - 1) years
• Man's age = (x² - 1) years

According to the question,

x² - 1 = 8(x - 1)

⤇ x² - 1 = 8x - 8

⤇ x² = 8x - 8 + 1

⤇ x² = 8x - 7

⤇ x² - 8x + 7 = 0

⤇ x² - 7x - x + 7 = 0

⤇ x(x - 7) - 1(x - 7) = 0

⤇ (x - 7) (x - 1) = 0

Now,

x - 7 = 0

⤇ x = 0 + 7

⤇ x = 7

Then,

⤇ x - 1 = 0

⤇ x = 0 + 1

⤇ x = 1 [ Not possible because here man's age cannot be 1 years]

So, here only x will be 7,

Present age of man = x²

Present age of man = (7)²

Present age of man = 49

Hence,the present age of son will be 7 years and present age of man will be 49 years.