The sum of the 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.

Given:

The sum of the 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.

To do:

We have to find their present ages.

Solution:

Let the age of the man be $x$ years.

This implies, the age of the son$=45-x$ years.

Age of the man 5 years ago$=x-5$ years

Age of the son 5 years ago $=45-x-5=40-x$ years

According to the question,

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

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

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

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

Solving for $x$ by factorization method, we get,

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

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

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

$x-5=0$ or $x-36=0$

$x=5$ or $x=36$

But father's age cannot be $5$ years. Therefore, the value of $x$ is $36$.

$45-x=45-36=9$

The present age of the man is $36$ years and the present age of the son is $9$ years.

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Updated on: 10-Oct-2022

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