Question 674243
.
Bob spent 1/6 of his life as a child, 1/12 as an adolescent, and 1/7 as a bachelor, 
five years after he was married, he had a son who died 4 years before his father 
at half his father's final age. How long did bob live?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~



Wording in your post is not perfect, but don't despair. There is a classic
well known story problem about Diophantus, famous mathematician of ancient Greece.


<pre>
    Diophantus's youth lasts 1/6 of his life. He grew a beard after 1/12 more of his life. 
    After 1/7 more of his life, Diophantus married. Five years later, he had a son. 
    The son lived exactly half as long as his father, and Diophantus died just four years 
    after his son's death. What was his age when he died?
</pre>


This story problem is precisely what you want to follow literally.  So, I will solve it for you.
Again, it is classic of about 2200 years old.


<pre>
Let x be the Diophantus' age when he died.


As we read the problem, we write this equation

    x = {{{x/6}}} + {{{x/12}}} + {{{x/7}}} + 5 + {{{x/2}}} + 4.


To solve, multiply all the terms by the Greatest Common Denominator, which is 12*7 = 84.


You will get

    84x = 14x + 7x + 12x + 5*84 + 42x + 4*84.


Group like terms in right side

    84x = (14 + 7 + 12 + 42)x + (5*84 + 4*84),


combine like terms

    84x = 75x + 756,


simplify and find x

    84x - 75x = 756,

       9x     = 756,

        x     = 756/9 = 84.


Thus, Diophantus died at the age of 84 years.
</pre>

Solved.


About Diophantus, read this remarkable Wikipedia article


<A HREF=https://en.wikipedia.org/wiki/Diophantus>https://en.wikipedia.org/wiki/Diophantus</A>


https://en.wikipedia.org/wiki/Diophantus



Enjoy &nbsp;(&nbsp;!&nbsp;)