F; Son = STen years ago(F - 10) = 3.5 (S - 10)F = 3.5S - 35 + 103.5S - F = 25 - (i)Ten years henceF + 10 = 2.25 (S + 10)F - 2.25S = 22.5 - 10F - 2.25S = 12.5 i.e.2F - 4.5S = 25 (II)That implies that3.5 S - F = 2F - 4.5S8S = 3FMultiplying (i) by 310.5S - 3F = 75 10.5S - 8S = 752.5 S = 75S = 30; Hence 3F = 240, F = 80Hence, SUM of ages of father and son will be 80 + 30 = 110.

"> F; Son = STen years ago(F - 10) = 3.5 (S - 10)F = 3.5S - 35 + 103.5S - F = 25 - (i)Ten years henceF + 10 = 2.25 (S + 10)F - 2.25S = 22.5 - 10F - 2.25S = 12.5 i.e.2F - 4.5S = 25 (II)That implies that3.5 S - F = 2F - 4.5S8S = 3FMultiplying (i) by 310.5S - 3F = 75 10.5S - 8S = 752.5 S = 75S = 30; Hence 3F = 240, F = 80Hence, SUM of ages of father and son will be 80 + 30 = 110.

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10 years ago, a father’s age was \(3\frac{1}{2}\) times that of his son, and 10 years from now, the father’s age will be \(2\frac{1}{4}\) times that of the son. What will be the sum of the ages of the father and the son at present?

Logical and Verbal Reasoning Ranking Puzzle in Logical and Verbal Reasoning . 3 years ago

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Current age of Father = F; Son = STen years ago(F - 10) = 3.5 (S - 10)F = 3.5S - 35 + 103.5S - F = 25 - (i)Ten years henceF + 10 = 2.25 (S + 10)F - 2.25S = 22.5 - 10F - 2.25S = 12.5 i.e.2F - 4.5S = 25 (II)That implies that3.5 S - F = 2F - 4.5S8S = 3FMultiplying (i) by 310.5S - 3F = 75 10.5S - 8S = 752.5 S = 75S = 30; Hence 3F = 240, F = 80Hence, SUM of ages of father and son will be 80 + 30 = 110.

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