Bunuel wrote:

How many ounces of pure acid must be added to 20 ounces of a solution that is 5% acid to strengthen it to a solution that is 24% acid?

(A) 5/2

(B) 5

(C) 6

(D) 15/2

(E) 10

To find the volume of one solution in a resultant mixture, weighted average works well. I use this formula:

\((Concen_{A})(Vol_{A}) + (Concen_{B})(Vol_{B} = (Concen_{A+B})(Vol_{A+B})\) Per the formula, in setting up the equation:

A = the solution that is 5% acid

B = the solution that is pure acid (100% = 1)

Let

\(x\)= # of ounces of pure acid (B) to be added

Desired concentration of resultant mix: 24%=.24

A + B (volume) = 20 + x

\(.05(20) + 1(x) = .24 (20 + x)\)

\(1 + x = .24(20) +.24(x)\)

\(1 + x = 4.8 + .24x\)

\(.76x = 3.8\)

\(x =\frac{3.8}{.76}=\frac{380}{76}=5\)We need 5 ounces of pure acid.

Answer B