SOLUTION: To make a 6% acid solution, a chemist mixes some 4% acid solution with 12 L of 10% acid solution. How much of the 4% solution must be added to the 10% solution to make the 6% acid
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Question 1136765: To make a 6% acid solution, a chemist mixes some 4% acid solution with 12 L of 10% acid solution. How much of the 4% solution must be added to the 10% solution to make the 6% acid solution?
I'm guessing that I need to make x=4% but I'm not sure what I need to use for the other variable. Found 2 solutions by rothauserc, greenestamps:Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! Let x be the amount of the 4% solution needed to mix with the 10% solution to get a 6% solution
:
There are three relations to consider, the third relation is the mixture
:
1) 12 * 0.10 = 1.20 L of acid
:
2) x * 0.04 = 0.04x L of acid
:
3) (x+12) * 0.06 = 0.06(x+12)
:
1.20 +0.04x = 0.06(x+12)
:
1.20 +0.04x = 0.06x +0.72
:
0.02x = 0.48
:
x = 24
:
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The chemist needs 24 L of the 4% solution to make a 6% solution
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:
Here is an alternative to the traditional algebraic solution method shown by the other tutor. If you understand how to use it, it will get you to the answer to any of this kind of mixture problem much faster and with far less effort.
(1) 6% is "twice as close" to 4% as it is to 10%. (6-4=2; 10-6=4)
(2) Therefore, the amount of 4% solution in the mixture must be twice as much as the amount of 10% solution.
(3) Since there are 12L of the 10% solution, you need 24L of the 4% solution.
