SOLUTION: A paint that is contains 21% green dye is mixed with another that is 15% green dye. If you need 60
gallons of 19% green dye, how much of each should you use? Hint: the problem onl
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gallons of 19% green dye, how much of each should you use? Hint: the problem onl
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Question 1150072: A paint that is contains 21% green dye is mixed with another that is 15% green dye. If you need 60
gallons of 19% green dye, how much of each should you use? Hint: the problem only tells you one variable Found 2 solutions by Alan3354, ikleyn:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! A paint that is contains 21% green dye is mixed with another that is 15% green dye. If you need 60
gallons of 19% green dye, how much of each should you use? Hint: the problem only tells you one variable
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t = 21%
f = 15%
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t + f = 60
21t + 15f = 19*60
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Hint: we don't need hints.
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If you meant to ask to have it done with only 1 variable, "Hint: the problem only tells you one variable" is not a request to do that.
The problem doesn't tell us any variables.
With one variable . . .
Let x be the amount of the 21% mixture, in gallons.
Then the amount of the 15% mixture is (60-x) gallons.
Then the balance equation is
0.21x + 0.15*(60-x) = 0.19*60.
From the equation, express "x" and calculate
x = = 40.
ANSWER. 40 gallons of the 21% mixture and (600-40) = 20 gallons of the 15% mixture.
CHECK. 0.21*40 + 0.15*20 = 11.4 = 0.19*60. ! Correct !
You will find there ALL TYPICAL mixture problems with different methods of solutions,
explained at different levels of detalization, from very detailed to very short.
Read them and become an expert in solution mixture word problems.
