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Six years hence, a father's age will be twice the age of his son. Eightyears ago, the father was four timesas old as his son. Find their present ages. |
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Answer» Let the present age of father be F years and the age of his SON be S years. Six years hence, a father's age will be twice the age of his son. Six years hence, age of father = (F + 6) years Age of son = (S + 6) years According to question, Six years hence, the age of father = 2(Six years hence, the age of his son) ⇒ F + 6 = 2(S + 6) ⇒ F + 6 = 2S + 12 ⇒ F = 2S + 12 - 6 ⇒ F = 2S + 6 Eight years ago, the father was four times as OLD as his son. Eight years ago, the age of father = (F - 8) years And age of his son = (S - 8) years According to question, ⇒ F - 8 = 4(S - 8) ⇒ F - 8 = 4S - 32 ⇒ F = 4S - 24 On COMPARING we get, ⇒ 4S - 24 = 2S + 6 ⇒ 4S - 2S = 6 + 24 ⇒ 2S = 30 ⇒ S= 15 Substitute value of S in F ⇒ F = 2(15) + 6 ⇒ F = 30 + 6 ⇒ F = 36 Therefore, the present age of the father is 36 years and son is 15 years. |
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