SOLUTION: This has been troubling me for some time now: How many gallons of an 8% acid solution and a 22% acid solution need to be mixed to make 14 gallons of a 13% acid solution?

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Question 221109: This has been troubling me for some time now: How many gallons of an 8% acid solution and a 22% acid solution need to be mixed to make 14 gallons of a 13% acid solution?
Found 3 solutions by rfer, stanbon, Alan3354:
Answer by rfer(16322)   (Show Source):
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.08x+.22(14-x)=.13*14
.08x+3.08-.22x=1.82
-.14x=-1.26
x=9 gal of 8%
14-x=5 gal of 22%

Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website!
How many gallons of an 8% acid solution and a 22% acid solution need to be mixed to make 14 gallons of a 13% acid solution?
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Equation:
acid + acid = acid
0.08x + 0.22(14-x) = 0.13*14
Multiply thru by 100 to get:
8x + 22*14 - 22x = 13*14
-14x = -9*14
x = 9 gallons (amt. of 8% solution in the mixture)
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14-x = 5 gallons (amt. 22% solution in the mixture)
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Cheers,
Stan H.

Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website!
This has been troubling me for some time now: How many gallons of an 8% acid solution and a 22% acid solution need to be mixed to make 14 gallons of a 13% acid solution?
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14 gal of 13% = 1.82 gal of acid
E + T = 14 (8% + 22%)
E*0.08 + T*0.22 = 1.82 (acid in solution)
8E + 22T = 182
E + T = 14 -->T = 14-E
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8E + 22(14-E) = 182
8E + 308 - 22E = 182
-14E = -126
E = 9 gallons
T = 5 gallons
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If they're mixed in equal amounts, the % will be the average, = 15%
Using more 8% will make the result less than 15%, as it does.