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 1161939: 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?
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Available, 4% and 10% acid
Want 6% and use 12L of the 10%


Account for amount of pure acid.
x of the 4%.
4x%2B10%2A12=6%28x%2B12%29
Simplify and solve for x.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


A typical setup for solving the problem using formal algebra....

10% of the 12L you start with, plus 4% of the x L you add, yields 6% of the (12+x) L:

.10%2812%29%2B.04%28x%29+=+.06%2812%2Bx%29

Solve using basic algebra....

If a formal algebraic solution is not required, here is a quick and easy way to solve mixture problems like this....

You are starting with 10% acid and adding 4% acid until the solution is 6% acid.

Model that by imagining walking on a number line from 10 towards 4, stopping when you get to 6. What fraction of the distance have you gone?

From 10 to 4 is a distance of 6; from 10 to 6 is a distance of 4. The fraction of the distance you have gone is 4/6 = 2/3.

That means 2/3 of the final mixture is what you are adding.

So the 12L you started with is 1/3 of the final mixture; and that means the amount of 4% acid you added (2/3 of the final mixture) was 24L.

ANSWER: 24L