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    Question

    The present ages of a father and his son are such that

    10 years ago, the father was 3 times as old as his son. Eight years from now, the father will be twice as old as his son. Find their present ages.
    A Father = 64 yrs and son = 28 yrs Correct Answer Incorrect Answer
    B Father = 74 yrs and son = 22 yrs Correct Answer Incorrect Answer
    C Father = 64 yrs and son = 18 yrs Correct Answer Incorrect Answer
    D Father = 54 yrs and son = 28 yrs Correct Answer Incorrect Answer

    Solution

    Let present ages be: Father = F, Son = S 10 years ago: F − 10 = 3(S − 10) F − 10 = 3S − 30 F = 3S − 20 …(1) After 8 years: F + 8 = 2(S + 8) F + 8 = 2S + 16 F = 2S + 8 …(2) Equate (1) and (2): 3S − 20 = 2S + 8 S = 28 From (2): F = 2×28 + 8 = 64 Answer: Father is 64 years old, son is 28 years old.

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