I'll assume the percentage is that **by volume**.

We're asked to find the amount of alcohol that needs to be added to #400# #"mL"# of a #15%# solution to yield a #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"#