Math, asked by Ashutosh121, 1 year 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


Ashutosh121: somebody give answer

Answers

Answered by TishaKalia
2
Sol:
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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Answered by Happiness07
1

Answer:

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