Math, asked by vatsal123456, 10 months ago


Two years ago, father was three times as old as his son and two years hence, twice
his age will be equal to five times that of his son. Find their present ages.​

Answers

Answered by vanijamathur
1

Answer:

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

Age of the son 2 years ago = (x - 2) years.

Age of the father 2 years ago = 3(x - 2) years.

Age of the son after 2 years = (x + 2) years

Age of the father after 2 years = 5(x + 2) / 2

⇒ 3(x - 2) + 4 = 5(x + 2) / 2

⇒ 6x -4 = 5x + 10

⇒ x = 14

Therefore, the present age of the son is 14 years.

Age of the father 2 years ago = 3(14 - 2) years = 36 years.

Present age of the father = 36 + 2 = 38 years.

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