SOLUTION: A chemist needs 120 milliliters of a 59% solution but has only 32% and 68% solutions available. Find how many milliliters of each that should be mixed to get the disired solution.

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Question 69848: A chemist needs 120 milliliters of a 59% solution but has only 32% and 68% solutions available. Find how many milliliters of each that should be mixed to get the disired solution.

Found 2 solutions by checkley75, ankor@dixie-net.com:
Answer by checkley75(3666) (Show Source):
You can put this solution on YOUR website!
120*.59=.32X+.68(120-X)
70.8=.32X+81.6-.68X
.68X-.32X=81.6-70.8
.36X=10.8
X=10.8/.36
X=30 MILLILETERS OF 32% SOLUTION &
120-30=90 MILLILETERS OF 68% SOLUTION
PROOF
70.8=.32*30+.68*61.2
70.8=9.6+61.2
70.8=70.8

Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
A chemist needs 120 milliliters of a 59% solution, but has only 32% and 68% solutions available. Find how many milliliters of each, that should be mixed to get the desired solution.
:
Let x = amt of 68% solution
(120-x) = amt of 32% solution
:
The percent equation:
.68x + .32(120-x) = .59(120)
:
Multiply what's inside the brackets:
.68x + 38.4 - .32x = 70.8
:
.68x - .32x = 70.8 - 38.4
:
.36x = 32.4
:
x = 32.4/36
:
x = 90 milliliters of the 68% solution
:
I'll let you figure out how much 32% solutions is needed