SOLUTION: John runs a garage. He needs to mix antifreeze in such a way as to produce 20 quarts of 50% solution. He has on hand ample supplies of 25% and 75% solutions. How much of 25% and 7

Question 133669: John runs a garage. He needs to mix antifreeze in such a way as to produce 20 quarts of 50% solution. He has on hand ample supplies of 25% and 75% solutions. How much of 25% and 75% solutions must he mix together to produce the needed 20 quarts of 50% solution?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=amount of 75% solution needed
Then 20-x=amount of 25% solution needed
Now we know that the amount of pure antifreeze in the 25% solution (0.25(20-x)) plus the amount of pure antifreeze in the 75% solution (0.75x) has to equal the amount of pure antifreeze in the final mixture (0.50*20). So our equation to solve is:
0.25(20-x)+0.75x=0.50*20 simplify
5-0.25x+0.75x=10 subtract 5 from both sides
5-5-0.25x+0.75x=10-5 collect like terms
0.50x=5 divide both sides by 0.50
x=10 qts------------------------------amount of 75% solution needed
20-x=20-10=10 qts------------------------------amount of 25% solution needed
CK
0.25*10+0.75*10=0.50*20
2.5+7.5=10
10=10

Hope this helps----ptaylor
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