SOLUTION: How many gallons of a 90​% antifreeze solution must be mixed with 80 gallons of 30​% antifreeze to get a mixture that is 80​% ​antifreeze?
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Question 1094741: How many gallons of a 90% antifreeze solution must be mixed with 80 gallons of 30% antifreeze to get a mixture that is 80% antifreeze?
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
First, here is an easy way to solve mixture problems like this, if you can understand it....
(1) Look to see how far (or close) the percentage of the mixture is to the percentages of the two ingredients:
The percentage of the mixture, 80%, is "5 times as close" to 90% as it is to 30%. That means there must be 5 times as much of the 90% ingredient as the 30% ingredient. Since there are 80 gallons of the 30% antifreeze, the number of gallons of 90% antifreeze needed is 5*80 = 400.
If you want to use the slow traditional algebraic solution method...
let x = liters of 90% antifreeze
80 = liters of 30% antifreeze
The total mixture is (80+x) liters; the amount of antifreeze is 30% of the 80, plus 90% of the x. You want the antifreeze to be 80% of the total mixture, so
You need to use 400 gallons of the 90% antifreeze.
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