Question

a man age is 2 times the square of his son. after 8 years, mans age will be 4 more than 2 times the age of his son. find their present ages

Answer 1

Answer:

18 years, 3 years

Step-by-step explanation:

Let the present age of his son be 'x' years.

Then, the age of man = 2x² years.

After 8 years:

⇒ Age of son = (x + 8) years.

⇒ Age of man = (2x² + 8) years.

According to the given condition,

⇒ (2x² + 8) = 2(x + 8) + 4

⇒ 2x² + 8 = 2x + 16 + 4

⇒ 2x² + 8 = 2x + 20

⇒ 2x² - 2x - 12 = 0

⇒ x² - x - 6 = 0

⇒ x² + 2x - 3x - 6 = 0

⇒ x(x + 2) - 3(x + 2) = 0

⇒ (x - 3)(x + 2) = 0

⇒ x = 3, -2{age cannot be negative}

⇒ x = 3

When x = 3:

Age of father = 2x² = 18 years.

Therefore:

⇒ Present age of Man = 18 years.

⇒ Present age of Son = 3 years

Hope it helps!

Answer 2

• Let present age of his son be (M) years

Then, the age of man = 2(M²) years

After 8 years:

=> Age of son = (M + 8) years.

⇒ Age of man = (2M² + 8) years.

A.T.A.

=> (2M² + 8) = 2(M + 8) + 4

=> 2M² + 8 = 2M + 16 + 4

=> 2M² + 8 = 2M + 20

=> 2M² - 2M - 12 = 0

=> M² - M - 6 = 0

=> M² + 2M - 3M - 6 = 0

=> M(M + 2) - 3(M + 2) = 0

=> (M - 3)(M + 2) = 0

=> M = 3, -2 (beacuseage cannot be negative)

=> M = 3

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Present age of Man = 18 years. [M = 2M² = 2(3)²]

Present age of Son = 3 years (M = 3)

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