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A solution is 90% glycerin. If there are 4 gallons of the solution, ho [#permalink]
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02 Oct 2010, 06:46
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Hi, Im posting two basic mixture problems below. The questions seem to be quite similar, but i could not get the hang of the main approach used to solve these particular kind of mixture problems. Also, i'm assuming that for these problems, the usual mixtures ratio method of solving does not hold. Do Correct me if i am wrong. 1) A solution is 90% glycerin. If there are 4 gallons of the solution, how much water, in gallons must be added to make a 75% glycerin solution? a) 1.8 b) 1.4 c) 1.2 d) 1.0 e) 0.8 OA  2) A mixture of 125 gallons of wine and water contains 20% water. How much water must be added to the mixture in order to increase the percentage of water to 25% of the new mixture? a) 10 glas b) 8.5 gals c) 8 gals d) 6.66 gals e) 8.33 gals OA 



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Re: A solution is 90% glycerin. If there are 4 gallons of the solution, ho [#permalink]
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02 Oct 2010, 07:01
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nehab2011 wrote: Hi, Im posting two basic mixture problems below. The questions seem to be quite similar, but i could not get the hang of the main approach used to solve these particular kind of mixture problems. Also, i'm assuming that for these problems, the usual mixtures ratio method of solving does not hold. Do Correct me if i am wrong. 1) A solution is 90% glycerin. If there are 4 gallons of the solution, how much water, in gallons must be addded to make a 75% glycerin solution? a) 1.8 b) 1.4 c) 1.2 d) 1.0 e) 0.8 OA  2) A mixture of 125 gallons of wine and water contains 20% water. How much water must be added to the mixture in order to increase the percentage of water to 25% of the new mixture? a) 10 glas b) 8.5 gals c) 8 gals d) 6.66 gals e) 8.33 gals OA  Any suggestions as to how these are solved would be helpful. Thank you! 1. A solution is 90% glycerin. If there are 4 gallons of the solution, how much water, in gallons must be addded to make a 75% glycerin solution?a) 1.8 b) 1.4 c) 1.2 d) 1.0 e) 0.8 In 4 gallons of the solution there are \(0.9*4=3.8\) gallons of glycerin. We want to add \(w\) gallons of water to 4 gallons of solution so that these 3.6 gallons of glycerin to be 75% of new solution: \(0.9*4=0.75(4+w)\) > \(w=0.8\) Answer: E. 2. A mixture of 125 gallons of wine and water contains 20% water. How much water must be added to the mixture in order to increase the percentage of water to 25% of the new mixture?a) 10 glas b) 8.5 gals c) 8 gals d) 6.66 gals e) 8.33 gals In 125 gallons of the solution there are \(0.2*125=25\) gallons of water. We want to add \(w\) gallons of water to 125 gallons of solution so that \(25+w\) gallons of water to be 25% of new solution: \(25+w=0.25(125+w)\) > \(w=\frac{25}{3}\approx{8.33}\). Answer: E. Hope it helps.
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Re: A solution is 90% glycerin. If there are 4 gallons of the solution, ho [#permalink]
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02 Oct 2010, 07:29
Thank you for the reply Bunuel. I was unable to come up with the correct equations earlier. This makes it a lot more simpler.



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Re: A solution is 90% glycerin. If there are 4 gallons of the solution, ho [#permalink]
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03 Oct 2010, 04:08
the amount of glycerin in the solution 4*.9=3.6 let x gallon water is added so (x+4)*.75 = 3.6 x=.8
the second one: water in the soln = 125*.2=25 gallon let x gallon water needs to be added so (25+x)=.25*(125+x) = 8.33



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Re: A solution is 90% glycerin. If there are 4 gallons of the solution, ho [#permalink]
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26 Feb 2011, 20:26
Let x gallons added. 3.6 / (4.0 + x) = 0.75 x = 0.8



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Re: A solution is 90% glycerin. If there are 4 gallons of the solution, ho [#permalink]
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26 Feb 2011, 21:10
nehab2011 wrote: Hi, Im posting two basic mixture problems below. The questions seem to be quite similar, but i could not get the hang of the main approach used to solve these particular kind of mixture problems. Also, i'm assuming that for these problems, the usual mixtures ratio method of solving does not hold. Do Correct me if i am wrong.
There are many ways to solve such problems, though, the 'usual mixtures ratio method of solving' definitely works here too. Why wouldn't it? There are two solutions and they have to be mixed to get a third solution. It doesn't matter if one of the solutions is pure water. 4 gallons of 90% glycerin has to be mixed with 0% glycerin solution (pure water) to give 75% glycerin solution. So glycerin solution:water should be in the ratio 5:1. Since glycerin solution is 4 gallons, water will be 4/5 = 0.8 gallons For more on this, check: [url] toughds105651.html#p828579[/url]
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Re: A solution is 90% glycerin. If there are 4 gallons of the solution, ho [#permalink]
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26 Feb 2011, 21:33
karishma  Pardon me. please rephrase this in red. Thanks VeritasPrepKarishma wrote: nehab2011 wrote: Hi, Im posting two basic mixture problems below. The questions seem to be quite similar, but i could not get the hang of the main approach used to solve these particular kind of mixture problems. Also, i'm assuming that for these problems, the usual mixtures ratio method of solving does not hold. Do Correct me if i am wrong.
There are many ways to solve such problems, though, the 'usual mixtures ratio method of solving' definitely works here too. Why wouldn't it? There are two solutions and they have to be mixed to get a third solution. It doesn't matter if one of the solutions is pure water. 4 gallons of 90% glycerin has to be mixed with 0% glycerin solution (pure water) to give 75% glycerin solution. So glycerin solution:water should be in the ratio 5:1.Since glycerin solution is 4 gallons, water will be 4/5 = 0.8 gallons For more on this, check: [url] toughds105651.html#p828579[/url]



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Re: A solution is 90% glycerin. If there are 4 gallons of the solution, ho [#permalink]
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26 Feb 2011, 21:51
gmat1220 wrote: karishma  Pardon me. please rephrase this in red. Thanks
Check out the link provided and then consider this: \(C_1 = 90%, C_2 = 0%\) and \(C_{avg} = 75%\) Ratio of solution 1:solution 2 = (75  0) : (90  75) = 5:1
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Re: A solution is 90% glycerin. If there are 4 gallons of the solution, ho [#permalink]
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26 Feb 2011, 21:55
Ohh wow ! Balancing... Cool I get it ! Kudos



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