SOLUTION: HOW MANY LITERS OF A 30% ALCHOHOL SOLUTION MUST BE MIXED WITH 40LITERS OF 90% SOLUTION TO GET A 50%SOLUTION?

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Question 195302: HOW MANY LITERS OF A 30% ALCHOHOL SOLUTION MUST BE MIXED WITH 40LITERS OF 90% SOLUTION TO GET A 50%SOLUTION?
Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
Let x=amount of 30% solution needed
Now we know that the amount of pure alcohol in the 30% mixture (0.30x) plus the amount of pure alcohol in the 40 liters (0.90*40) has to equal the amount of pure alcohol in the final mixture (0.50(40+x), so:
0.30x+0.90*40=0.50(40+x) get rid of parens
0.30x+36=20+0.50x subtract 36 and also 0.50x from each side
0.30x-0.50x+36-36=20-36
-0.20x=-16 divide each side by -0.20
x=80 liters-------------------amount of 30% solution needed
CK
0.30*80+0.90*40=0.50(120)
24+36=60
60=60
Hope this helps---ptaylor