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1. 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
- B. 54
- C. 54
- D. 54
Answer: Option D
Explanation:
Let the present age of father be x and that of his Son Be y
Then 18 years ago
Father's age =x−18
Son's age =y−18
But 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=36
Putting y=36 in (2)
x=2y
=2×36
=72
Sum of their present ages
=x+y
72+36
=108
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