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# Mixtures problems!

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Intern
Joined: 04 Sep 2010
Posts: 10
Followers: 0

Kudos [?]: 3 [1] , given: 6

Mixtures problems! [#permalink]  02 Oct 2010, 05:46
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50% (01:00) correct 50% (07:35) wrong based on 17 sessions
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 -
[Reveal] Spoiler:
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

OA -
[Reveal] Spoiler:
E

Any suggestions as to how these are solved would be helpful. Thank you!
Math Expert
Joined: 02 Sep 2009
Posts: 30376
Followers: 5088

Kudos [?]: 57271 [1] , given: 8811

Re: Mixtures problems! [#permalink]  02 Oct 2010, 06:01
1
KUDOS
Expert's post
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 -
[Reveal] Spoiler:
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

OA -
[Reveal] Spoiler:
E

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: $$0.9*4=0.75(4+w)$$ --> $$w=0.8$$

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: $$25+w=0.25(125+w)$$ --> $$w=\frac{25}{3}\approx{8.33}$$.

Hope it helps.
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6062
Location: Pune, India
Followers: 1596

Kudos [?]: 8936 [1] , given: 195

Re: Mixtures problems! [#permalink]  26 Feb 2011, 20:10
1
KUDOS
Expert's post
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]
tough-ds-105651.html#p828579[/url]
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6062 Location: Pune, India Followers: 1596 Kudos [?]: 8936 [1] , given: 195 Re: Mixtures problems! [#permalink] 26 Feb 2011, 20:51 1 This post received KUDOS Expert's post 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 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Intern
Joined: 04 Sep 2010
Posts: 10
Followers: 0

Kudos [?]: 3 [0], given: 6

Re: Mixtures problems! [#permalink]  02 Oct 2010, 06:29
Thank you for the reply Bunuel. I was unable to come up with the correct equations earlier. This makes it a lot more simpler.
Manager
Joined: 22 Aug 2008
Posts: 186
Followers: 5

Kudos [?]: 75 [0], given: 11

Re: Mixtures problems! [#permalink]  03 Oct 2010, 03:08
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
Director
Status: Matriculating
Affiliations: Chicago Booth Class of 2015
Joined: 03 Feb 2011
Posts: 920
Followers: 13

Kudos [?]: 257 [0], given: 123

Re: Mixtures problems! [#permalink]  26 Feb 2011, 19:26
3.6 / (4.0 + x) = 0.75
x = 0.8
Director
Status: Matriculating
Affiliations: Chicago Booth Class of 2015
Joined: 03 Feb 2011
Posts: 920
Followers: 13

Kudos [?]: 257 [0], given: 123

Re: Mixtures problems! [#permalink]  26 Feb 2011, 20:33
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]
tough-ds-105651.html#p828579[/url]
Director
Status: Matriculating
Affiliations: Chicago Booth Class of 2015
Joined: 03 Feb 2011
Posts: 920
Followers: 13

Kudos [?]: 257 [0], given: 123

Re: Mixtures problems! [#permalink]  26 Feb 2011, 20:55
Ohh wow ! Balancing... Cool I get it ! Kudos
Re: Mixtures problems!   [#permalink] 26 Feb 2011, 20:55
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