CLASS-10QUADRATIC EQUATIONS - PROBLEMS ON AGES

Problem On Ages

Example.1) Five years ago, a father’s age was the square of his daughter’s age. Ten years hence, his age will be twice that of his daughter’s age. Find –

(i) The age of the daughter five years ago,

(ii) The present age of the father

Ans.) Let the daughter’s age 5 years ago be x years. Then father’s age 5 years ago = years

Daughter’s present age = (x + 5) years, and father’s present age is (x²+ 5) years

Daughter’s age 10 years hence = {(x + 5) + 10} = (x + 15) years

Father’s age 10 years hence = {(x² + 5) + 10} = (x² + 15) years

As per the given condition, ten years hence father’s age will be two times of daughter’s age, then the equation would be –

x²+ 15 = 2(x + 15)

=> x²+ 15 = 2x + 30

=> x²- 2x + 15 – 30 = 0

=> x²- 2x – 15 = 0

=> x²- (5 – 3)x – 15 = 0

=> x²- 5x + 3x – 15 = 0

=> x (x – 5) + 3 (x – 5) = 0

=> (x – 5) (x + 3) = 0

=> (x – 5) = 0 or (x + 3) = 0

=> x = 5, or x = - 3

=> x = 5             [where x ≠ - 3, since age cannot be negative ]

Daughter’s age 5 years ago = 5 years ……………………..(i) (Ans.)

Father’s present age (x²+ 5) = (5²+ 5) = 25 + 5 = 30 years………………..(ii) (Ans.)

Example.2) The sum of the ages of a father and his son is 45 years. Five years ago, the product of their age was 34. Find the ages of the son and the father.

Ans.) Let the present age of the father is x years

Then, son’s present age = (45 – x) years

Father’s age 5 years ago = (x – 5) years

Son’s age 5 years ago = (45 – x – 5) years = (40 – x) years

As per the given condition-

(x – 5) (40 – x) = 34

=> 40x - 200 - x²+ 5x = 34

=> 45x - x² = 234

=> x²- 45x + 234 = 0

=> x²- (39 + 6)x + 234 = 0

=> x²- 39x – 6x + 234 = 0

=> x (x – 39) – 6 (x – 39) = 0

=> (x – 39) (x – 6) = 0

=> (x – 39) = 0 or (x – 6) = 0

=> x = 39 or x = 6

Present age of father cannot be 6. So, we will reject the value x = 6

So, x = 39

Hence the present age of father is 39 years and the present age of the son is (45 – x) = (45 – 39) = 6 years.      (Ans.)