mixture problem

[gmatusa2010]

Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. what part of the original solution was replaced?

1/5
1/4
1/2
3/4
4/5


Answer:

(1-X)*0.5 + 0.3*x = 0.4
0.5 - 0.5X +0.3X = 0.4
-0.2X = -0.1
X=1/2

This question doesn't make any sense to me or am I missing something? The answer here seems to be dictated by the choices. The overwhelming majority of the people on gmatclub used the method in pasted above. Why does it have to be 1-X? To get 40% all you need is the ratio between the 2 solutions to be 1 to 1 (ie 50% and 50%). The beginning is meaningless.

Ex: Original coulda been 3/4 50% solution and 1/4 30% solution. You take 1/4 away from the 50% and redistribute to the 30% you will get 1/2 and 1/2 which is 1 to 1. So the the answer should be 1/4 divided by 3/4 which is 1/3 of the original solution (of the 50% solution) or 1/4 of the total solution. In fact, I can proly go choice by choice and make up the original where all of the answer choices can be right. No where does it dictate that the original solution was completely "50% solution".


Is this just a bad question?

[Rahul@gurome]

Maybe this will be clearer.

Solution:
Let the original volume of 50% acid solution be V.
Let x volume be removed from this .
So V-x volume of original solution is left.
This contains 50% acid.
Or volume of acid is 50%*(V-x).
Another x volume of 30% acid solution is being replaced in the original solution.
Volume of acid in this replaced part is 30%*x.
Or total volume of acid now is 50%*(V-x) + 30%*x = 50%*V-20%*x.
Or percentage of acid is [(50%*V - 20%*x)/V]*100 = 40%.
Or 5 - 2x/V = 4.
Or x/V = ½.
So ½ of original solution was replaced.