Question 1017708: Kelly, a chemist, was asked to prepare 24 gallons of a solution that is 50% acidic by mixing a solution that is 20% acidic with another solution that is 70% acidic. How much of each solution should she use?
Found 3 solutions by stanbon, Boreal, ikleyn:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Kelly, a chemist, was asked to prepare 24 gallons of a solution that is 50% acidic by mixing a solution that is 20% acidic with another solution that is 70% acidic. How much of each solution should she use?
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Equation::
acid + acid = acid
0.20x + 0.70(24-x) = 0.50*24
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20x + 70*24 - 70x = 50*24
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-50x = -20*24
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x = (2/5)*24
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x = 48/5 = 9.6 gallons (amt. of 20% solution needed)
24-x = 13.4 gallons (amt. of 70% solution needed)
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Cheers,
Stan H.
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Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! x=20%
24-x=70%
.20x+.70(24-x)=0.50*24=12 gallons of pure acid.
.20x+16.8-.70x=12
-0.50x=-4.8
x=9.6 gallons of 20%=1.92 gallons pure acid.
24-x=14.4 gallons of 70%=10.08 gallons pure acid.
Sum is 12 gallons pure acid, checks.
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! Kelly, a chemist, was asked to prepare 24 gallons of a solution that is 50% acidic by mixing a solution that is 20% acidic with another solution that is 70% acidic. How much of each solution should she use?
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It is a typical mixture word problem.
See the lesson Solving typical word problems on mixtures for solutions in this site, Problem 1.
This problem is very similar to yours.
In this lesson you will find the detailed solution.
Consider it as a sample. Read it attentively.
Then solve your problem by substituting your data.
In this way you will learn how to solve problems like this, once and for all .
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