SOLUTION: A container contained 500mL of a solution that was 52% water. How much water should be removed from the solution so that the remainder would be only 40% water?

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Question 1109169: A container contained 500mL of a solution that was 52% water. How much water should be removed from the solution so that the remainder would be only 40% water?
Found 4 solutions by addingup, TeachMath, josgarithmetic, greenestamps:
Answer by addingup(3677)   (Show Source):
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.52(500) = .40(500+x)
260 = 200+.40x
.40x = 60
x = 150
You have to remove 150ml of water to get a solution with only 40% water

Answer by TeachMath(96)   (Show Source):
You can put this solution on YOUR website!
100 mL should be removed.

Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website!
48% is not water, at the start.
60% would be not water , at the end.

, for removing x mL of water.
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Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!

Tutor addingup worked the wrong problem; he added more of the other ingredient to make the solution 40% water.

Tutor teachmath gave his usual useless response -- a right answer with no work shown. That's a strange way of "teaching" math.

Tutor teachmath: if you are by chance reading this, please stop providing useless answers to people who are nearly always interested in learning HOW to solve a problem -- not just in seeing the answer.

The other tutor provided a valid solution to the problem.

My approach would be slightly different:

The 500mL is 52% water, so it is 48% of the other ingredient.
48% of 500mL is 240mL.
If the final solution is to be only 40% water, it must be 60% of the other ingredient.
If 60% of the final mixture is the original 240mL of the other ingredient, then the volume of the final mixture is 240/.60 = 400mL.
So, since the original mixture was 500mL, 100mL of water needs to be removed.