A simple problem on ratio. If a solution A which has milk to water in the ratio $7:4$ is mixed with solution B which has the ratio of water to milk as $9:2$ such that the ratio becomes $1:1$,then in what ratio were they mixed?
My work:
in solution A: milk=$7x$,water=$4x$ 
in solution B: milk=$9x$,water= $2x$
Now I can't think how to use third part(mixing of both solution.)
 A: Don't use $x$ for both solution $A$ and solution $B$ - use a different variable, because they refer to different things.
If we take $x$ units of solution $A$, we get $(7/11)x$ units of milk and $(4/11)x$ units of water.
If we take $y$ units of solution $B$, we get $(9/11)y$ units of milk and $(2/11)y$ units of water.
Suppose that mixing $x$ units of $A$ and $y$ units of $B$ results in a combination of milk and water that is in a ratio of $1:1$. Use this information to relate $x$ and $y$.
A: Things to make things simpler:
Watch for sadistic problem setters who explicitly reverse the order of ratios. 
Pick a concrete amount of one of the starting solutions;  say 100 ml of the first solution.
Track only one component, say milk.  The other will follow automatically.
Choose the volume of the second solution as the only unknown, say V.
So the amount of milk you have in the final mixture, M, is$$M=\frac{7}{11} \times 100 + \frac{2}{11} \times V$$
and the amount of final solution is simply $V+100$, with half of it milk. So the equation for the final solution is:$$\frac{7}{11} \times 100 + \frac{2}{11} \times V=0.5 \times (V+100)$$Re-arrange and solve.  The desired answer is $\frac{V}{100}$
