# The sum of ages of a man and his son is 45 years.

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,

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

$40 x-x^{2}+5 x-200=4 x-20$

$-x^{2}+45 x-200=4 x-20$

$-x^{2}+45 x-200-4 x+20=0$

$-x^{2}+41 x-180=0$

$x^{2}-41 x+180=0$

$x^{2}-36 x-5 x+180=0$

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

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

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 is $=36$ years and his son's age is $=9$ years