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# The Sum of Ages of a Man and His Son is 45 Years. Five Years Ago, the Product of Their Ages Was Four Times the Man'S Age at the Time. Find Their Present Ages. - Mathematics

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#### Question

The sum of ages of a man and his son is 45 years. Five years ago, the product of their ages was four times the man's age at the time. Find their present ages.

#### Solution

Let the present age of the man be x years

Then present age of his son is  (45 - x) years

Five years ago, man’s age = (x - 5) years

And his son’s age (45 - x - 5) = (40 - x) years

Then according to question,

(x - 5)(40 - x) = 4(x - 5)

40x - x2 + 5x - 200 = 4x - 20

-x2 + 45x - 200 = 4x - 20

-x2 + 45x - 200 - 4x + 20 = 0

-x2 + 41x - 180 = 0

x2 - 41x + 180 = 0

x2 - 36x - 5x + 180 = 0

x(x - 36) -5(x - 36) = 0

(x - 36)(x - 5) = 0

So, either

x - 36 = 0

x = 36

Or

x - 5 = 0

x = 5

But, the father’s age never be 5 years

Therefore, when x = 36 then

45 - x = 45 - 36 = 9

Hence, man’s present age is36 years and his son’s age is 9 years.

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