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Question 360575: Find the amount of a 6% acid solution and the amount of a 14% acid solution that should be combined to prepare 50cc of a 12% solution.
I believe the equation is set up this way:
1st 6% = .06x
2nd 14% = .14y
mix 12% = .12(50)
x + y = 50
.06x + .14y = 6
Then subtract .14y from both sides, then multiply both by 100?
Am I on the right track, because I am way off on the correct answer.
Thanks in advance!
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Find the amount of a 6% acid solution and the amount of a 14% acid solution that should be combined to prepare 50cc of a 12% solution.
I believe the equation is set up this way:
x = amount of 6%
y = amount of 14%
1st 6% = .06x ( = amount of acid in x cc's of 6%)
2nd 14% = .14y ( = amount of acid in y cc's of 14%)
mix 12% = .12(50)
x + y = 50
0.06x + 0.14y = 0.12*50
.06x + .14y = 6 This is not right.
Then subtract .14y from both sides, then multiply both by 100?
Am I on the right track, because I am way off on the correct answer.
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Solve these 2 eqns
x + y = 50
0.06x + 0.14y = 0.12*50
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