Question 8150: A man is 4 times as old as his son. Eighteen years from now, he will be only twice as old as his son. Find their present ages. Answer by kritikapd(4) (Show Source):
You can put this solution on YOUR website! Let present age of son = X years
So, Present age of father = 4X years
Age of son after 18 years = X+18 years
Age of father after 18 years = 4X+18 years ……… (i)
Father will be twice as old his son after 18 years
i.e. 2 (X+18) ……….(ii)
Equating age of father i.e. (i) and (ii)
4X+18 = 2(X+18)
4X+18 = 2X + 36
4X – 2X = 36 – 18
2X = 18
X = 9
So the present age of son and father is 9 years and 4 x 9= 36 years respectively.
Verify- Fathers age will be twice the age of son after 18 years.
Age of son after 18 years = 9+ 18 = 27 years
Age of son after 18 years = 36 + 18 = 54 years which is twice of 27 years.
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