Question 1058345: Pure alcohol is added to 50 gallons of coolant mixture that starts out 40% alcohol write an expression for the percentage of alcohol in the mixture (y) in terms of the gallons of alcohol added (X) be sure to clearly explain how your expression represents the given problem show the first quadrant values only assuming that the mixture in the container that holds a maximum of 200 gallons
Answer by jorel555(1290)   (Show Source):
You can put this solution on YOUR website! If x is the amount of pure alcohol added, then your equation looks like:
x+.4(50)=y(x+50)
y=(x+20)/(x+50) where x≤150.
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x represents the amount of pure alcohol to be added. .4(50) represents the original mixture; and y represents the percentage of alcohol in the resulting mixture when x is added to the original 50 gallons of mixture. ☺☺☺☺
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x represents pure alcohol. The original mixture contains 50 x .4, or 20 gallons of alcohol (The other 30 gallons are composed of something else). So x gallons of pure alcohol mixed with 50 gallons of 40% alcohol gives you 50+x gallons of mixture made of y% alcohol. The amount of alcohol in the final mixture varies as the amount of pure alcohol added. Since the limit of total mixture is 200 gallons, you can add no more than 150 gallons of pure alcohol to the 50 gallons of 40% mixture you already have.
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