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ggarr
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some of the %50 intensity red paint is replaced with %25 21 Jun 2007, 18:55
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some of the %50 intensity red paint is replaced with %25 solution of red paint such that the new paint intensity is 30%. What fraction of the original paint was replaced?
1/30
1/5
2/3
3/4
4/5
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I know this was solved earlier but I don't understand the solution. I might be oversimplifying this but I set the prob up like this:

x-P/2+P/4=30. this is obviously wrong. will someone explain?



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some of the %50 intensity red paint is replaced with %25 21 Jun 2007, 19:16
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https://www.gmatclub.com/phpbb/viewtopic ... ity#329033
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some of the %50 intensity red paint is replaced with %25 21 Jun 2007, 20:05
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forgmat
https://www.gmatclub.com/phpbb/viewtopic.php?p=329033&highlight=paint+intensity#329033
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Thanks. but I mentioned that I didn't understand the solutions here. Can anyone else explain this?
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some of the %50 intensity red paint is replaced with %25 21 Jun 2007, 21:25
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Too tired to read the other explanation. Here is what i did

Let there be 100 parts of solution.
50 parts is pure red.

Let X be the amount removed.

Thus, 50 - X/2 + X/4 = 30
X=80

From 100 parts, 80 parts were removed, giving 4/5


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some of the %50 intensity red paint is replaced with %25 22 Jun 2007, 06:15
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GK_gmat

Thus, 50 - X/2 + X/4 = 30 ==> what does x/2 stand for?
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Since X part is removed from the original solution, 1/2 of which is red, we subtract this from 50.
Then X part of new solution is added, 1/4 of which is red. We add this.

Total quantity of new solution is 100, as before, but red has changed in its concentration from 50 to 30.

Hope that helps
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some of the %50 intensity red paint is replaced with %25 22 Jun 2007, 10:01
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I am not a the best gmat guy, but the equation can be found from thinking about the logic of the question.

There is a certain amount of fluid that consists of 50 percent or 1/2 of paint. If we remove Y amount of this fluid and replace it with Y amount of fluid containing 1/4 concentrate paint what will Y be if in the end we want the same amount of fluid with a 30 percent concentrate.

We will make the current amount of fluid (and the desired amount of fluid) X
Since we need to know how much concentrate there will be once we remove y the first part of the equation looks like this

1/2(X-Y)

Then we need to add Y back into the equation to get the amount of fluid back up to X

1/2(X-Y) + Y

The resulting fluid will be X, but the concentrate we want will be 30 percent therefore

1/2(x-Y) + Y= 3/10X

Then we take the resulting number and devide it by X to find out the new ratio.

Easiest way to solve is to pick a number for X.

I made X 20

10-y/2 +y= 6
-y/4=-4
y=16
16/20
4/5



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