Sallyzodiac wrote:

Bunuel's solution is pretty awesome, I didn't think one could solve it that simply. I took a different approach and worked with actual numbers:

1) We know that the ratio of water:alcohol in the first solution is \(\frac{3}{1}\) (which give 4 parts) and we know that the ratio of water:alcohol in the second solution is \(\frac{2}{3}\) (which give 5 parts).

2) 20 is the LCM of 4 and 5, so let's assume that both solutions have a volume of 20 units. From the first solution, we have that water is \(\frac{3}{4}\) of the solution (which we said had a volume of 20), which gives 15 units of water and thus 5 units of alcohol. From the second solution, we have that water is \(\frac{2}{5}\) of the solution, which gives 8 units of water and 12 units of alcohol.

3) Since we are mixing both solutions with an equal amount, 20 units, our new solution has a total volume of 40 units. From 2), combining the water and alcohol in both solutions, we get that our new solution has 17 units of alcohol (5 units from the first solution and 12 units from the second solution) and 23 units of water (15 units from the first solution and 8 units from the second solution). Thus, the amount of alcohol in the mixed solution is \(\frac{17}{40}\) = 42.5%.

PS: Time-wise I managed to do this slightly north of 2 minutes, but Bunuel's approach is of course much faster if one can identify the validity of that approach right off the bat.

/S

Took the same approach. What if the question asked for the solutions to be mixed in different proportions instead of the same? How would you use this method with unequal proportions?