I'll assume the percentage is that by volume.
We're asked to find the amount of alcohol that needs to be added to 400400 "mL"mL of a 15%15% solution to yield a 32%32% solution.
What we can first do is determine the amount of water present, because this quantity doesn't change (only alcohol is added):
"water" = overbrace((0.85))^(100%-15% = 85% "water")(400color(white)(l)"mL") = 340 "mL"
We want to make a 32% by volume solution of alcohol, so we can realize that 340 "mL H"_2"O" is
100%-overbrace(32%)^"desired alcohol percentage"=68%
of the solution. Using this, the total volume of the new solution is
"total volume" = (340color(white)(l)"mL")/(0.68) = color(red)(500 color(red)("mL"
The original volume is 400 "mL", so the amount of alcohol needed to be added is
"alcohol added" = color(red)(500 color(red)("mL" - 400 "mL" = color(blue)(100 color(blue)("mL"
