SOLUTION: A garage owner wants to fill a 55-gallon drum with a 20% winter mixture of antifreeze for his customers. How many gallons of 100% antifreeze should he mix with some 10% antifreeze

Question 260651: A garage owner wants to fill a 55-gallon drum with a 20% winter mixture of antifreeze for his customers. How many gallons of 100% antifreeze should he mix with some 10% antifreeze mixtures in order to fill the drum?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!

Let x=amount of 100% antifreeze needed
Then 55-x=amount of 10% antifreeze mixtures needed
Now we know that the amount of pure antifreeze used (x) plus the amount of pure antifreeze in the 10% antifreeze mixtures (0.10(55-x)) has to equal the amount of pure antifreeze in the final mixture (0.20*55). So our equation to solve is:
x+0.10(55-x)=0.20*55 simplify
x+5.5-0.10x=11 subtract 5.5 from each side
x+5.5-5.5-0.10x=11-5.5 collect like terms
0.9x=5.5 divide each side by 0.9
x=6.1 gal-------------------------amount of 100% antifreeze needed
CK
6.1+0.10(48.9)=11
6.1+4.9=11
11=11
Hope this helps---ptaylor
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