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The salinity of the Atlantic Ocean averages 37 parts per [#permalink]
 10 Oct 2009, 19:17
Question Stats:
66% (03:53) correct
33% (02:42) wrong based on 1 sessions
I'm having a h*ll of a time w/this one (POSSIBLE SPOILER ALERT:Scroll down to see my set-up. Perhaps, someone can tell me where I screwed up?):
The salinity of the Atlantic Ocean averages 37 parts per thousand. If 64 oz of water is collected and placed in the sun, how many oz of pure water would need to evaporate to raise the salinity to 45 parts per thousand?
(x)(45/1000) = (64)(37/1000) - (64-x)(37/1000)
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Re: Salinity (Help!) [#permalink]
10 Oct 2009, 19:42
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Re: Salinity (Help!) [#permalink]
10 Oct 2009, 20:19
Bunuel wrote: Well don't know how you did it, but here is my way:
Salinity=3.7%
Water=64
Salt in it=64*3.7%=2.368
We need 2.368 to be 4.5% --> x*4.5%=2.368 --> x=52.62. So we should evaporate 64-52.62=11.37 This is the exact way that i followed.. But is there a standard formula that i can use for such questions?
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Re: Salinity (Help!) [#permalink]
10 Oct 2009, 20:44
rvthryet wrote: Bunuel wrote: Well don't know how you did it, but here is my way:
Salinity=3.7%
Water=64
Salt in it=64*3.7%=2.368
We need 2.368 to be 4.5% --> x*4.5%=2.368 --> x=52.62. So we should evaporate 64-52.62=11.37 This is the exact way that i followed.. But is there a standard formula that i can use for such questions? You can use formula but the formula changes as the question changes. You can have formula for every problem in gmat but thats not possible to remember. So its better to understand the problem and use the logic to solve it. Trying to remember and use formula in gmat probably costs you too much because there are hundreds of formulas to be remembered. Try to remember as few formulas (that are must) as possible.
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Re: Salinity (Help!) [#permalink]
10 Oct 2009, 21:04
thanks. your solution is much simpler.
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Re: Salinity (Help!) [#permalink]
12 Oct 2009, 13:31
I use a mixure formula: (% difference between the weaker solution and the desired solution) x (amount of weaker solution) = (% difference between the stronger solution and the desired solution) x (amount of stronger solution). In this case: .045x = .008(64) x=11.37 I'm not recommending or endorsing formula use, but since you asked for it, it works well in this instance. Posted from my mobile device 
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Re: Salinity (Help!) [#permalink]
12 Oct 2009, 21:48
I found the explanation of bunuel much easier than memorizing any standard formula for this type of questions. Thank you
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Re: Salinity (Help!) [#permalink]
11 Nov 2009, 02:59
Economist method:( http://gmatclub.com:8080/forum/viewtopic.php?p=618107)
Attachments
Mixture.jpg [ 17.74 KiB | Viewed 1166 times ]
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Re: Salinity (Help!) [#permalink]
12 Nov 2009, 03:39
Calculation is really hard in this example... Nothing can't be simplified...
I don't think it is representative of a GMAT exercise
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Re: Salinity (Help!) [#permalink]
12 Nov 2009, 04:11
ctrlaltdel wrote: Getting very familiar with this table is the key to success regarding mixture problems. Mixture problems used to be very difficult for me, until I started using these tables to organize the facts. I do agree with pierrealexandre77 that this is not representative of what you would see in a GMAT problem...
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Re: Salinity (Help!) [#permalink]
20 Nov 2009, 11:27
HI ctrlaltdel If you can please explain from where you got the values in the table like 0.963 and the other. I also use the table but sometimes it didnt serve the purpose mentioned in the question so i need to use alternate methods. If you can please explain then it'd be great. Thanks
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Re: Salinity (Help!) [#permalink]
13 Jan 2012, 07:13
think of the problem as though we have a mixture of 4.5% salinity and we need to add some water (0%) to make a final solution having 3.7% salinity. this would be reverse of what is stated in the question, but is easier to solve. final ratio would be (weighted average) = 4.5-3.7 : 3.7-0 = 8:37 water needed = \frac{8}{45}*64 = 11.377 oz
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Re: Salinity (Help!)
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13 Jan 2012, 07:13
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