Solving 600L of 30% Acid Algebra Problem

[emma3163]

Homework Statement

A lab has a 20% acid solution and a 50% acid solution. How many liters of each are required to obtain 600 liters of a 30% acid solution?

Homework Equations

The Attempt at a Solution

.20x + .50(600 - x) = .30(600)
.20x + 300 - .50x = 180
-0.3x + 300 = 180
- 300 -300
-0.3x = -120
----- -----
-0.3 -0.3
x = 400 liters

I am sure that the answer is wrong. Can anyone tell what I am doing wrong, or how to set the problem up properly?
Thank you

 

[gabbagabbahey]

Your solution for x looks good to me, what does that mean the quantity of the 20% solution is? How about the 50% solution?

 

[HallsofIvy]

The only thing I see wrong with it is that you haven't yet answered the question.

"x= 400 liters" does not answer "How many liters of each are required to obtain 600 liters of a 30% acid solution?" For one thing, there is no "x" in the question and, for another, there are obviously two quantities required, not just one.

I presume that x represents the amount of 20% acid solution used. It is very good practice to SAY that at the beginning of the solution (if nothing else, it will really shock your professor!). You also have (600- x) in your formula and that, I guess, is the amount of 50% acid solution used. Well, 600- 400= 200, so your answer should be "400 liters of the 20% acid solution and 200 liters of the 50% acid solution must be used to make 600 liters of 30% acid solution."

Now, let's see if that is correct. Certainly 400+ 200= 600 liters so that gives the correct amount of solution. 400 liters of 20% acid solution contains .20(400)= 80 liters of pure acid. 200 liters of 50% acid solution contains .5(200)= 100 liters of pure acid so the 600 liters of mixture contains 180 liters of acid. 180/600= 18/60= 3/10= .30. Yes, that is a 30% solution.

 

[emma3163]

thanks soo much!