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Question

# Eighteen years ago, a father was three times as old as his son. Now the father is only twice as old as his son. Then the sum of the present ages of the son and the father is

A
54
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B
72
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C
105
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D
108
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Solution

## The correct option is D 108Let the present age of father be x and that of his Son Be yThen 18 years ago Father's age =x−18 Son's age =y−18But given that x−18=3(y−18) ⇒x−18=3y−54 ⇒x−3y+36=0 (1)Also given that x=2y (2)From (1) & (2) x−3y=36=0⇒2y−3y+36=0⇒y=36Putting y=36 in (2)x=2y=2×36=72Sum of their present ages =x+y72+36=108

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