A chemist has two acid solutions in stock: one is a 50% solution, and the other is a 80% solution. How much of each solution should be mixed to obtain 100 milliliters of a 68% solution?
50---68---80% Difference 18 12 Invert order to give ratio of 50% and 80% Ratio 12 18 = 2:3
So by simple proportion: volume of 50% needed = 100*2/(2+3)=40 mL volume of 80% needed = 100*3/(2+3)=60 mL
Algebra way Let x=volume of 50% needed. Then 0.50*x+0.80*(100-x) = 0.68(100) Expand and solve for x 0.5x+80-0.8x = 68 0.8x-0.5x = 80-68 0.3x = 12 x=12/0.3 = 40 mL (volume of 50%) 100-x=100-40=60 mL (volume of 80%)
