Mixing solutions of specified strength While reading one aptitude book, I came across this question:

$35\%$ alcohol mixed with $60\%$ alcohol to get a $50\%$ alcohol. In what ratio were they mixed?

I spent 20 minutes on this question, but I couldn't even figure out what it means! Could anyone explain this to me?
 A: Call the ratio you want to find r=(volume of 35% alcohol):(volume of 60% alcohol).  Suppose that there are r liters of the 35% alcohol (so there are 0.35 L of alcohol and 0.65 L of water), 1 liter of the 60% alcohol (so the ratio is r), and 1+r liters of the 50% alcohol.  From there, you can construct an equation about the alcohol (or about the water) and solve for r.
A: Mixture1 : 60%
Mixture2 : 35%
Let us say you take mixture using a pot of 100 units.
also you take Mixture1 'a' times and Mixture2 'b' times using the pot.
$\rightarrow$
$\frac{(a  60 + b  35)}{(a 100 + b  100)} = \frac{50}{100}$
solving this you get 
$\frac{a}{b} = \frac{3}{2}$
-> 300 units of Mixture1 and 200 units of Mixture2 will give you 500 units of mixture with alcohol concentration of 50% because 3 x 60 + 2 x 35 = 250.  
A: As Mention mixtures Should be $50$%
Assume we take mixture of A is $x$ ($35$% alcohol)and the second mixture in $y$ ($60$% alcohol) volume. So as given
$$\frac{\text{alcohol volume}}{\text{total volume}}=\frac{50}{100}$$
we get $$\frac{35x+60y}{100x+100y}=\frac{50}{100}$$
By simplifying it
$$35x+60y=50(x+y)$$
$$15x=10y$$
we get
$$x/y=2/3$$
A: Given a liters of 35% alcohol solution, and b liters of 60% alcohol solution, what would be the alcohol ratio after we'll mix this two solutions?
The a liters are actually a*35/100 alcohol, and a*(1-(35/100)) water. Likewise the b liters contains b*60/100 alcohol and b*(1-(60/100)) water.
Mixing them together would give us a*35/100+b*60/100 alcohol and a*(1-(35/100))+b*(1-(60/100)) water. The question is for which a,b we have the ratio
((a*35+b*60)/100)/((a*65+b*40)/100) = 50% = 1/2

or
(a*35+b*60)/(a*65+b*40) = 1/2

I believe you can take it from here...
