Question 1094741
<br>First, here is an easy way to solve mixture problems like this, if you can understand it....<br>
(1) Look to see how far (or close) the percentage of the mixture is to the percentages of the two ingredients:
{{{90-80 = 10}}}
{{{80-30 = 50}}}<br>
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.<br>
If you want to use the slow traditional algebraic solution method...<br>
let x = liters of 90% antifreeze
80 = liters of 30% antifreeze<br>
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
{{{.30(80)+.90(x) = .80(80+x)}}}
{{{24+.9x = 64+.8x}}}
{{{.1x = 40}}}
{{{x = 400}}}<br>
You need to use 400 gallons of the 90% antifreeze.