Question 52998: Pleas help me with this problem.
Roy has 8 gallons of 40% antifreeze solution. How much 100% antifreeze mus be added to make a solution that is at least 60% antifreeze.
Thanks =)
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! Let x=amt of pure antifreeze that must be added
Solution (1)
Amt of pure antifreeze in the 8 gals (8)(.40) plus amt of pure antifreeze
added (x)(1) equals the amt of pure antifreeze in the final mixture (x+8)(.60).
Thus, our equation to solve is:
(8)(.4)+(x)(1)=(x+8)(.60) simplifying, we get:
3.2+x=.6x+4.8 or
.4x=1.6
x=4 gal
Solution (2)
If we assume the mixture is antifreeze and water, notice that the amt of water does not change throughout the process.
Amt of water in the 8 gal (8)(.60) plus the amt of water added (0)equals the amt of water in the final mixture (x+8)(.40).
Thus, our equation to solve is:
(8)(.60)=(x+8)(.40) simplifying, we get:
4.8=.4x+3.2 or
.4x=1.6
x=4 gal
Hope this helps---ptaylor
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