Six year before, the age of mother was equal to the square of her son's age. Three year hence, her age will be thrice the age of her son then. Find the present ages of the mother and son.

Advertisement Remove all ads

#### Solution

Suppose, the age of the son six year before was x

∴ mother’s age six year before was x^{2}

∴ present age of the son is (x + 6) and

present age of the mother is (x^{2} + 6)

Three years hence, son’s age will be (x + 9) and mother’s age will be (x^{2} + 9)

by given condition,

x^{2} + 9 = 3(x + 9)

∴ x^{2} - 3x + 9 - 27 = 0

∴ x^{2} - 3x - 18 = 0

∴ (x - 6) (x + 3) = 0

∴ x = 6 or x = -3

But age cannot be negative ∴ x ≠ -3

∴ son’s present age = x + 6 = 6 + 6 = 12 years.

mother’s present age = x2 + 6 = 36 + 6 = 42 years.

Concept: General Term of an Arithmetic Progression

Is there an error in this question or solution?

Advertisement Remove all ads