Basic algebraic solution of father's and son's ages
Algebra
Level
2
This year a father is the cube of his son's age that was 1/13 of his father's age last year. Please find out how long it would be for the father to be twice his son's age.
21 years time
45 years time
24 years time
20 years time
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1 solution
Firstly, let the son's age last year be 'x', then the father's age this year will be 13x + 1; and the father's age this year = (x + 1)^3 = x^3 + 3x^2 + 1 = 13x +1. Therefore the equation is x^3 + 3x^2 - 10x + 1 = 1; so the 1s on each side cancel leaving the quadratic equation: (x + 5)(x - 2) = 0. The child cannot be - 5, so the father must be 24 years older than him, as per the question information. Multiply 24 x 2 = 48, when he will be half his father's age. So by subtracting the present age of the father, it would take 21 years for him to be twice his son's age.