SOLUTION: 1.A phramacist needs to obtain a 60% alcohol solution. How many ounces of a 30% alcohol solution must be mixed with 50 ounces of an 80% alcohol solution to obtain a 60% alcohol sol

Question 650872: 1.A phramacist needs to obtain a 60% alcohol solution. How many ounces of a 30% alcohol solution must be mixed with 50 ounces of an 80% alcohol solution to obtain a 60% alcohol solution?

2.A tank can be filled in 2 hours. The same tank can be emptied in 6 hours. If the faucet is turned on and the drain is left open, when will the tank overflow?
3.Tom Kline mixed 60lbs of sunflower seeds at $3.00 per lb with 40lbs of daisy seeds at $2.00 per pound to produce a wild flower mix. What is the unit price of the mix?
Answer by ptaylor(2198) (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 that exists before the mixture takes place (0.30x+0.80*50) has to equal the amount of pure alcohol that exists after the mixture takes place (0.60(50+x)).
Soooo:
0.30x+0.80*50=0.60(50+x) simplify
0.30x+40=30+0.60x or
0.60x-0.30x=40-30
0.30x=10
x=33.333 ounces---amount of 30% alcohol solution needed
CK
0.30*(33 1/3)+40=30+0.60*(33 1/3)
0.30(100/3)+40=30+0.60(100/3)
10+40=30+20
50=50
Hope this helps---ptaylor

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