Bunuel wrote:

What amount (in milliliters) of a 1% sulfuric acid solution must be added to 60 milliliters of a 6% sulfuric acid solution to yield a 5% sulfuric acid solution?

(A) 12

(B) 15

(C) 18

(D) 20

(E) 45

Let x = amount of 1% sulfuric acid to be added

A = solution with 1% sulfuric acid

B = solution with 6% sulfuric acid

(% A)(Vol A) + (% B)(Vol B) = (% of A+B)(Vol of A+B)\(.01(x) + .06(60) = .05(x + 60)\)

Multiply all by 100 to make math quicker

\(1x + 6(60) = 5(x + 60)\)

\(1x + 360 = 5x + 300\)

\(4x = 60\)

\(x=\frac{60}{4}=15\)

Answer B

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