SOLUTION: Beth wants to make 11.6 fl. oz. of a 45% acid solution by mixing together a 22% acid solution and a 80% acid solution. How much of each solution must she use?

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Question 275795: Beth wants to make 11.6 fl. oz. of a 45% acid solution by mixing together a 22% acid solution and a 80% acid solution. How much of each solution must she use?
Found 4 solutions by mananth, ikleyn, josgarithmetic, greenestamps:
Answer by mananth(16949)   (Show Source):
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Beth wants to make 11.6 fl. oz. of a 45% acid solution by mixing together a 22% acid solution and a 80% acid solution. How much of each solution must she use?

Let the 22% solution required be x
The total required = 11.6 oz
22%------------------80%-----------45% strength
x--------------------11.6-x---------11.6 Quantity
0.22x+0.8(11-x) = 0.45*11.6
0.22x+8.8-0.8x=5.22
-0.58x==-3.58
x=-3.58/-0.58
= 6.17 oz of 22% acid

11.6-6.17 = 5.43 oz

Answer by ikleyn(53742)   (Show Source):
You can put this solution on YOUR website!
.
Beth wants to make 11.6 fl. oz. of a 45% acid solution by mixing together a 22% acid solution and a 80% acid solution.
How much of each solution must she use?
~~~~~~~~~~~~~~~~~~~~~~~

        Calculations in the post by @mananth are incorrect, leading to wrong answer.
        I came to bring a correct solution.

Let x be the volume of the 80% solution, in oz. Then the volume of the 22% solution is (11.6-x) oz. The balance equation is 0.8x + 0.22(11.6-x) = 0.45*11.6. (1) It says that the combined mass of pure acid in ingredients (left side) is the same as the mass of pure acid in the mixture (right side). Simplify and find 'x' 0.8x + 0.22*11.6 - 0.22x = 0.45*11.6, 0.8x - 0.22x = 0.45*11.6 - 0.22*11.6 0.58x = 2.668 x = 2.668/0.58 = 4.6. ANSWER. 4.6 oz of the 80% solution is needed and (11.6-4.6) = 7 oz of the 22% solution. CHECK. Let's check the final solution for its concentration = 0.45, which is precisely correct.

Solved.

Answer by josgarithmetic(39790)   (Show Source):
You can put this solution on YOUR website!
Beth wants to make 11.6 fl. oz. of a 45% acid solution by mixing together a 22% acid
solution and a 80% acid solution. How much of each solution must she use?
----------------------------------------------------------------------
Make M fl. oz. of a T% acid solution by mixing together a L% acid
solution and a H% acid solution. How much of each solution to use?
----------------------------------------------------------------------

Use an unknown v fl. oz. of the H% solution and M-v fl. oz. of the L% solution.

Plug in the given values when you are ready and compute.

Answer by greenestamps(13326)   (Show Source):
You can put this solution on YOUR website!

Tutor @josgarithmetic provides a response showing her favorite multiple-variable formula for solving 2-part mixture problems like this. Use that method if you love using formulas without having any understanding of how you are solving the problem.

Tutor @ikleyn provides a response showing a typical formal algebraic method for solving such problems. That is perfect if what you want is a formal algebraic solution method.

An informal (and usually faster) method for solving this kind of problem uses the logical fact that the ratio in which the two ingredients need to be mixed is exactly determined by where the target percentage lies between the two given percentages.

For this particular problem....
(1) Using a number line if it helps, observe/calculate that 45 is 23/58 of the distance from 22 to 80 (22 to 45 is a difference of 23; 22 to 80 is a difference of 58)
(2) That means that 23/58 of the mixture must be the higher percentage ingredient

ANSWER: The mixture should be made using 4.6 fl. oz. of the 80% acid solution and (11.6-4.6) = 7.0 fl. oz. of the 22% acid solution