Question

How much of a 29% solution is required to mix with 44mL of a 51% solution to make a 37% solution?

Answer 1

Let me take a stab at this...these are quite confusing and difficult! I made a table like this:
           mL sol.                %                  amt%
29%      x                      .29      =        .29x
51%      44                    .51      =      22.44
--------------------------------------------------------
37%    44+x                  .37      =  .37(44+x)
You have to add the first two rows together to give you what you want in the end. In other words, how much (x) of a 29% solution (.29) + 44mL of 51% (.51) solution will give you a solution of 37% (.37). The word "of" in those phrases mean to multiply:  .29x + 22.44 = .37(44+x). Distribute the .37 into the parenthesis to get .29x + 22.44 = 16.28 + .37x. Solve for x to get x=77 mL