SOLUTION: Elementary & Intermediate Algebra: Angela needs 20 quarts of 50% antifreeze solutioin in her radiator. She plans to obtain this by mixing some pure antifreeze with an appropriate a
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-> SOLUTION: Elementary & Intermediate Algebra: Angela needs 20 quarts of 50% antifreeze solutioin in her radiator. She plans to obtain this by mixing some pure antifreeze with an appropriate a
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Question 73318: Elementary & Intermediate Algebra: Angela needs 20 quarts of 50% antifreeze solutioin in her radiator. She plans to obtain this by mixing some pure antifreeze with an appropriate amount of a 40% antifreeze solution. How many quarts of each should she use? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=amount of 40% antifreeze
Then 20-x=amount of pure antifreeze
Now we know that the amount of pure antifreeze in the 40% mixture (0.40x) plus the amount of pure antifreeze added (20-x) equals 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 (note: 1=100% and that's what (20-x) is multiplied by)
0.40x+20-x=10 subtract 20 from both sides
0.40x+20-20-x=-20+10 collect like terms
-0.60x=-10 divide both sides by -0.60
x=16.667 qts----------------------------------amount of 40% antifreeze
20-x=20-16.667=3.333 qts----------------------amount of pure antifreeze
CK
16.667(0.40)+3.333=10
6.668+3.333=10
~10=10
Hope this helps----ptaylor
