Question 100503: How many liters of 80% alcohol must be mixed with 40 liters of 70% alcohol to produce a mixture that is 75% alcohol? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=amount of 80% alcohol needed
Now we know that the amount of pure alcohol in the 80% solution (0.80x) plus the amount of pure alcohol in the 70% solution ((0.70)(40)) has to equal the amount of pure alcohol in the final mixture (0.75)(40+x)). So our equation to solve is:
0.80x+0.70*40=0.75(40+x) get rid of parens
0.80x+28=30+0.75x subtract 28 and also 0.75x from both sides
0.80x-0.75x+28-28=30-28+0.75x-0.75x collect like terms
0.05x=2 divide both sides by 0.05
x=40----------------------------number of gal of 80% alcohol needed
