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?

Indeed, you have two equations. One is that there are $100$ people, i.e.
$$m+w+c=100$$
and the other is that there are $100$ sweets handed out to them with the particular distribution, i.e.
$$10m + 5w + \frac{c}{2} = 100$$
this other equation can be written to avoid fractions, so
$$20m + 10w + c = 200$$
These are the only two equations you have, however there is another piece of information, and that is that $m$, $w$ and $c$ must all be nonnegative integers, i.e. $m,w,c\in\mathbb \{0,1,\dots,100\}$.

Now, using the first equation, we can get rid of one variable, let's say $c$. So
$$c=100-m-w$$ which means that the second equation becomes
$$20m + 10w + (100-m-w)=200$$
which simlifies to 
$$19m + 9w = 100$$
This already tells you that $m\leq 5$, since if $m>5$, $w$ must be negative, which is impossible. So you can now simply test which of the values of $\{0,1,\dots, 5\}$ gives an integer value of $w$.
Or, you can go a little further. Note that since $9w=100-19m$, you need to find a value of $m$ such that $100-19m$ will be divisible by $9$. Now since $100$ leaves a remainder of $1$ when dividing by $9$, while $19$ leaves a remainder of $1$, you have
$$1\cdot 9 + 1 - m(2\cdot 9 + 1) = 9\cdot(1 -2m) + 1 -m$$ which shows that $100-19m$ will leave a remainder of $1-m$ when dividing by $9$, and therefore, $m=1$.
Note that the above calculation is more elegantly done in modular arithmetic, since
$$0=100-19m \equiv 1-19m \equiv 1-m\,(\mathrm{mod}\,9)$$
