Question

the sum of ages of a man and his son is 100 years;one tenth of the product of their age exceeds the fathers age by 180. how old are they?

Answer 1

See the attachment for detail solution
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Answer 2

Concept

An algebraic equation could also be defined as an announcement within which two expressions are set capable each other.

Given

We are on condition that the sum of ages of a person and his son is $100$ years; one-tenth of the merchandise of their age exceeds the father's age by $180$.

Find

We have to seek out the age of son and his father.

Solution

Let the father's age is $x$ and therefore the son age is $y.$

According to first condition,

$\begin{aligned}x+y&=100\\ x&=100-y\end{aligned}$                             ......(1)

According to second condition,

$\begin{aligned}\frac{1}{10}xy&=x+180\\ xy&=10x+1800\\ x(y-10)&=1800\\ x&=\frac{1800}{y-10}\end{aligned}$                      .....(2)

Substitute value of equation (1) in equation (2), we get

$\begin{aligned}100-y&=\frac{1800}{y-10}\\ 100y-1000-y^2&=1800\\ y^2-110y+2800&=0\\ y&=70,40\end{aligned}$

From equation one we are saying that son's age can not be over father's age.

So, son age is $40$ and father age is $100-40=60$.

Hence, the father age is $60yrs$ and son age is $40yrs.$

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