SOLUTION: Weighted Averages. A car radiator has a capacity of 14 quarts and is filled with a 20% antifreeze solution. How much must be drained off and replaced with pure antifreeze to o

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Question 164965This question is from textbook
: Weighted Averages.

A car radiator has a capacity of 14 quarts and is filled with a 20% antifreeze solution. How much must be drained off and replaced with pure antifreeze to obtain a 40% antifreeze solution? This question is from textbook

Found 2 solutions by nerdybill, ptaylor:
Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website!
.
Let x = amount drained and replaced w/pure antifreeze
.
"antifreeze left" + "added antifreeze" = "antifreeze in new mixture"
.20(14-x) + x = .40(14)
2.8 - .20x + x = 5.6
.80x = 2.8
x = 3.5 quarts

Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website!
Let x=amount that needs to be drained off and replaced with pure antifreeze
Now we know that the amount of pure antifreeze left after x amount is drained off (0.20(14-x)) plus the amount of pure antifreeze added(x) has to equal the amount of pure antifreeze in the final mixture (0.40(14)). So, our equation to solve is:
0.20(14-x) + x=0.40*14 get rid of parens (distributive)
2.8-0.20x+x=5.6 subtract 2.8 from each side
2.8 - 2.8-0.20x+x=5.6-2.8 collect like terms
0.80x=2.8 divide each side by 0.80
x=3.5 qts-----------------------------------ans
CK
0.20*10.5+3.5=5.6
2.1+3.5=5.6
5.6=5.6
Hope this helps---ptaylor