100 persons 100 sweets problem 
There are $100$ persons, including men,women and children.
  Then there are $100$ sweets.
  
  
*
  
*Each man will get $10$ sweets
  
*Each woman will get $5$ sweets
  
*Each child will get $.5$ (i.e half) of the sweet 
  

At the end of sharing every 100 person should get sweets, and there should be no sweets left.
How many men, women and children are present?
I have tried to make two equations by the way


*

*$M + W + C = 100$

*$10M + 5W + .5C = 100$
were $M$ => no. of men
$W$ => no. of women
and $C$ => no. of children
But in order to solve this equation of three variables, I think I need one more equation 
But is there anything else that I miss here?

The part you are ignoring is that this is a diophantine equation -- that is, all of the variables must be non-negative integers.  That will eliminate a large number of solutions to the two equations you listed.
Let's do it intuitionistically to start.  We will imagine that there were $100$ children.  That obviously doesn't work, because we have only given out $50$ candies.  Now, every child we replace with a woman adds $4.5$ (i.e. $5-0.5$) candies, and every child we replace by a man adds $9.5$ (i.e. $10-0.5$) candies.  So if $M$ is the number of men and $W$ is the number of women, to add in the missing fifty candies we must have 
$$4.5W+9.5M=50\\9W+19M=100$$ 
(or you could have doubled your second equation and subtracted it from your first one to get here if you like algebra more than not-algebra. ^_^)
This again must be solved with non-negative integers.  There must be between zero and five men since $19\cdot5=95$, and so doing the math we need $9W\in\{5,24,43,62,81\}$.  Obviously, only $81$ is a multiple of $9$, so the unique solution is $M=1,W=9$. Thinking about the children again, this leads to one man ($10$ candies), nine women ($45$ candies), and ninety children ($45$ candies), which properly adds to $100$.
