Question
14 years ago, the age of a father was three times the
age of his son. Now, the father is twice as old as his son. What is the sum of the present ages of the father and the son?Solution
ATQ, Let the son's present age = x, and the father's present age = y. 14 years ago: Father = 3 Γ Son = yβ14 = 3(xβ14) βΉ y = 3xβ28 Now: Father = 2 Γ Son = y = 2x On Solving we get: From y=2x and y=3xβ28: 2x=3xβ28βΉx=28 y=2x=56. Sum of ages = x + y = 28+56 = 84 years
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