SOLUTION: What quantity of a 75% acid solution must be mixed with a 40% solution to produce 300mL of a 50% solution?

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Question 851934: What quantity of a 75% acid solution must be mixed with a 40% solution to produce 300mL of a 50% solution?
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
If the percents are weight per weight, then the densities or specific gravities of the two solutions to be mixed are needed. Why do students have to answer questions like this one?

If the percents are as volume per volume, then:
u = how much of 40%
v = how much of 75%

%2840u%2B75v%29%2F300=50 and u%2Bv=300
Solve this system for u and v.

40u%2B75v=15000
8u%2B15v=3000
And then using the volume sum equation, a substitution,
8u%2B15%28300-u%29=3000
8u%2B4500-15u=3000
8u-15u=3000-4500
7u=1500
u=1500%2F7
u=214 ml.
Again using the volume sum equation, v=300-u=300-214=86