Can someone please help with the attached problem?
1 answer:
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440 gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get a mixture that is 70% antifreeze
<em><u>Solution:</u></em>
Let "x" be the gallons of 80 % antifreeze added
Therefore, "x" gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze
Final mixture is x + 80
Therefore, we can frame a equation as:
"x" gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get (x + 80) gallons of 70 % antifreeze
Thus, we get,
x gallons of 80 % + 80 gallons of 15 % = (x + 80) gallons of 70 %
Thus 440 gallons of 80 % antifreeze solution must be mixed