Question

The sum of ages of a father and son is 45 years. Five years ago the product of their ages was four times the father's age at that time. The twice the difference of father and son's ?

Answer 1

Answer:

Let father's age be ' x' years & son's age be 'y ' years

Sum of the ages

x+ y = 45

y= 45-x

5 Years ago the product of their ages was 34 years

(x-5)(y-5) = 34

xy -5(x+y) + 25 = 34

x(45-x) -5(45) +25 =34

45x -x^2 - 225 -25 = 34

45x -x^2 -200 = 34

-x^2 +45x - 200–34 = 0

-x^2 +45x -234 = O

x^2-45x +234 = 0

x^2–39x- 6x+234 = 0

x(x-39)-6(x-39) = 0

(x-6)(x-39) = 0

x= 6 or 39

taking x= 39 father's age must be greater than son's age .

Father's age is 39 years & son's age is 6 years.

Verification x+y = 39+6 = 45,

(x-5)(y-5) = 34

(39–5)(6–5)

(34)(1) = 34

Ans : 39 years & 6 years.

Step-by-step explanation: