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Director
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Don't have options...Please help [#permalink] New post 29 Oct 2009, 07:21
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Forty liters of a 60% salt solution are reduced to a 45% solution. How much must be drained off and replaced with distilled water to that the resulting solutions will contain only 45% salt?
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Re: Don't have options...Please help [#permalink] New post 29 Oct 2009, 07:52
In a mixture problem, it helps to separate out the "water" from the other element.

so with 40 liters, 60% means there are 24 liters of "salt" and 16 liters of water.

If you want to have 45% mixture when you're done, you'll have 18 liters of salt (45% of 40 = 18). So, you need to get rid of 6 liters of salt. If for every liter of mixture, you have 0.6 liters of salt, then you need to get rid of 10 liters of mixture and replace all 10 liters with water.

reply2spg wrote:
Forty liters of a 60% salt solution are reduced to a 45% solution. How much must be drained off and replaced with distilled water to that the resulting solutions will contain only 45% salt?

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Director
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Joined: 12 Oct 2008
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Re: Don't have options...Please help [#permalink] New post 29 Oct 2009, 08:02
Thanks, but how did you get this one?
jallenmorris wrote:
If for every liter of mixture, you have 0.6 liters of salt, then you need to get rid of 10 liters of mixture and replace all 10 liters with water.
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Re: Don't have options...Please help [#permalink] New post 29 Oct 2009, 08:15
Because the mixture to start with (before draining any off and adding water) is 60% so 60% of 1 is 0.6 The way mixtures work is that the 60% represents what portion of the whole is made up of salt. So if 40 liters means that we have 24 liters of salt, and we got the number 24 by multiplying 40 x 0.6, (60% in decimal form is 0.6) then when we find out how much salt is in 1 liter, we multiply 1 x 0.6 = 0.6 salt in each.

Now, we know we need less salt to go from 60% mixture to 45% mixture, but we have to figure out how much salt to get rid of, so that when we add back in water, the salt that remains is 45% of the whole solution.

There are a couple of different ways to do this, figure out how much salt we have at the start (24 liters) then figure out how much we will have once we have 45% of 40 liters, whic is 18 and then the difference is 6. So if 1 liter has 0.6 liters of salt (60% of 1) then 60% of 10 = 6 liters of salt,t he difference we need to get rid of. So drain off 10 liters of mixture, and you get rid of 6 liters of salt. But now the mixture has 30 liters with 18 liters of salt. Add in 10 liters of distilled water to go back to 40 liters and we still have 18 liters of salt (we did not add any).

Hope this helps.

reply2spg wrote:
Thanks, but how did you get this one?
jallenmorris wrote:
If for every liter of mixture, you have 0.6 liters of salt, then you need to get rid of 10 liters of mixture and replace all 10 liters with water.

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Re: Don't have options...Please help   [#permalink] 29 Oct 2009, 08:15
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