SOLUTION: Adjusting Antifreeze. Angela needs 20 quarts of 50% antifreeze solution in her radiator. She plans to obtain this by mixing some pure antifreeze with and appropriate amount of a 4

Algebra ->  Inequalities -> SOLUTION: Adjusting Antifreeze. Angela needs 20 quarts of 50% antifreeze solution in her radiator. She plans to obtain this by mixing some pure antifreeze with and appropriate amount of a 4      Log On


   

Question 83526: Adjusting Antifreeze. Angela needs 20 quarts of 50% antifreeze solution in her radiator. She plans to obtain this by mixing some pure antifreeze with and appropriate amount of a 40% solution. How many quarts of each should she use?

Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
Let x=amount of 40% antifreeze needed
Then 20-x=amount of pure antifreeze needed
Now we know that the amount of pure antifreeze in the 40% solution (0.40x) plus the amount of pure antifreeze (20-x) must equal the amount of pure antifreeze in the final mixture 0.50(20). So our equation to solve is:
0.40x+(20-x)=0.50(20) get rid of parens
0.40x+20-x=10 subtract 20 from both sides
0.40x+20-20-x=10-20 collect like terms
-0.60x=-10 divide both sides by -0.60
x=16.666 gal --------------------------amount of 40% antifreeze
20-x=20-16.666=3.333 gal-------------------amount of pure antifreeze
CK
3.333+0.40(16.666)=0.50(20)
3.333+6.664=10
10=10

Hope this helps---ptaylor