Question 1112892
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The desired percentage of the mixture, 80, is 2/5 of the way from 70 to 95.  (95-70 = 25; 80-70 = 10; 10/25 = 2/5)<br>
That means 2/5 of the mixture must be the 95% antifreeze.<br>
95% antifreeze: 2/5 of 140 gallons = 56 gallons
70% antifreeze: 3/5 of 140 gallons = 84 gallons<br>
Answer: 56 gallons of 95%, 84 gallons of 70%<br>
or you can use the much more difficult traditional algebraic solution method:<br>
Let x be the number of gallons of 95% antifreeze
Then 140-x = number of gallons of 70% antifreeze<br>
The 140 gallons of the mixture is 80% antifreeze:
{{{.95(x)+.70(140-x) = .80(140)}}}<br>
A relatively easy equation to solve; but far more work than is required by the first method....