SOLUTION: Mix a solution that is 30% alcohol with a solution that is 80% alcohol to make 800mL of a solution that is 60% alcohol. How much of each solution should you use?
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Question 164388: Mix a solution that is 30% alcohol with a solution that is 80% alcohol to make 800mL of a solution that is 60% alcohol. How much of each solution should you use? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=amount of 30% solution needed
Then 800-x=amount of 80% solution needed
Now we know that the amount of pure alcohol in the 30% solution (0.30x) plus the amount of pure alcohol in the 80% solution (0.80(800-x)) has to equal the amount of pure alcohol in the final mixture (0.60*800). So, our equation to solve is:
0.30x+0.80(800-x)=0.60*800 get rid of parens (distributive law)
0.30x+640-0.80x=480 subtract 640 from each side
0.30x+640-640-0.80x=480-640 collect like terms
-0.50x=-160 divide each side by -0.50
x=320ml------------------------------------amount of 30% alcohol needed
800-x=800-320=480ml----------------------------amount of 80% alcohol needed
CK
0.30*320+0.80*480=0.60*800
96+384=480
480=480
This problem could just as well have been worked using two unknowns
Let x=amount of 30% needed
And let y=amount of 80% needed
x+y=800--------------------------------------eq1
0.30x+0.80y=0.60*800----------------------------eq2
From eq1, we see that y=800-x; substitute this into eq2 and then we will have the same equation as before
