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Mixture X contains 60% water and 40% chloride. If 120 gallon

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grindcore
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Mixture X contains 60% water and 40% chloride. If 120 gallon [#permalink] New post   02 Dec 2010, 15:12
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Mixture X contains 60% water and 40% chloride. If 120 gallons of Mixture Y is added to 80 gallons of mixture X in order to produce a solution that contains 75% water and 25% chloride, how many gallons of water must mixture Y contain?

My answer
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there are 72 gal of water and 48 gal of chloride in mixture X
mix x: 72+48=120
add 80 gal of mixture y
mix y: (80-x)+x = 80
total: 150 + 50 = 200
total water: 72+80-x=150
total chloride: 48+x=50
x=2 in both cases
therefore: mixture y should have 78 gallons of water

the solutions say mixture y should have 102 gallons of water

i'm confused
can anyone please help
thanks!
grind
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Bunuel
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Re: mixtures problem [#permalink] New post   02 Dec 2010, 15:43
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grindcore wrote:
hey guys, i was doing a mixtures problem and got confused looking at the solutions

Question
Mixture X contains 60% water and 40% chloride. If 120 gallons of Mixture Y is added to 80 gallons of mixture X in order to produce a solution that contains 75% water and 25% chloride, how many gallons of water must mixture Y contain?

My answer
there are 72 gal of water and 48 gal of chloride in mixture X
mix x: 72+48=120
add 80 gal of mixture y
mix y: (80-x)+x = 80
total: 150 + 50 = 200
total water: 72+80-x=150
total chloride: 48+x=50
x=2 in both cases
therefore: mixture y should have 78 gallons of water

the solutions say mixture y should have 102 gallons of water

i'm confused
can anyone please help
thanks!
grind


Please read and follow: how-to-improve-the-forum-search-function-for-others-99451.html

So please:
Provide answer choices for PS questions.

Also:
Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/
Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/

No posting of PS/DS questions is allowed in the main Math forum.

As for the question:
Mixture X contains 60% water and 40% chloride. If 120 gallons of Mixture Y is added to 80 gallons of mixture X in order to produce a solution that contains 75% water and 25% chloride, how many gallons of water must mixture Y contain?

Final mixture is Y+X=120+80=200 gallons. We want it to contain 75% water or 0.75*200=150 gallons of water. Now, 80 gallons of mixture X will provide 0.6*80=48 gallons of water, so the rest or 150-48=102 gallons of water should provide mixture Y.

You confused mixture X with mixture Y.

Hope it's clear.
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Re: Mixture X contains 60% water and 40% chloride. If 120 gallon [#permalink] New post   26 Oct 2013, 22:04
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Refer the attached Image
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grindcore
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Re: mixtures problem [#permalink] New post   02 Dec 2010, 15:53
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im so sorry i feel stupid now.
lesson learned.
READ CAREFULLY
thanks
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gracie
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Re: Mixture X contains 60% water and 40% chloride. If 120 gallon [#permalink] New post   22 Aug 2016, 07:48
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let w=% of water in mixture Y
.6*80+w*120=.75*200
w=.85
.85*120=102 gallons water
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Re: Mixture X contains 60% water and 40% chloride. If 120 gallon [#permalink] New post   02 Jun 2014, 08:36
Another approach:
Mixture X provide 32 gallons (40%*80) of chloride of the 25%*200 gallons of chloride of the final mixture (=50 gallons of chloride).
Then mixture Y must provide the rest of 18 gallons to complete that 50 gallons of chloride.
If total of mixture Y (water + chloride) is 120 gallons, then gallons of water in mixture Y = 120 (total mixture) -18 (chloride gallons) = 102 water gallons.
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maggie27
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Re: Mixture X contains 60% water and 40% chloride. If 120 gallon [#permalink] New post   29 Jun 2014, 09:05
Narenn wrote:
Refer the attached Image


Hi Narenn,

How did you get the 2 there in Qty X? Kindly explain.
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Re: Mixture X contains 60% water and 40% chloride. If 120 gallon [#permalink] New post   10 Jul 2014, 02:20
maggie27 wrote:
Narenn wrote:
Refer the attached Image


Hi Narenn,

How did you get the 2 there in Qty X? Kindly explain.



He's using allegation method to solve the problem

Mixture X = 80 gallon & Mixture Y = 120

There ratio = 80:120 = 2:3

From here, the values 2 & 3 are put in the allegation expression constructed
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Re: Mixture X contains 60% water and 40% chloride. If 120 gallon [#permalink] New post   21 Aug 2016, 22:29
Expert Reply
grindcore wrote:
Mixture X contains 60% water and 40% chloride. If 120 gallons of Mixture Y is added to 80 gallons of mixture X in order to produce a solution that contains 75% water and 25% chloride, how many gallons of water must mixture Y contain?

My answer
Show Spoiler
there are 72 gal of water and 48 gal of chloride in mixture X
mix x: 72+48=120
add 80 gal of mixture y
mix y: (80-x)+x = 80
total: 150 + 50 = 200
total water: 72+80-x=150
total chloride: 48+x=50
x=2 in both cases
therefore: mixture y should have 78 gallons of water

the solutions say mixture y should have 102 gallons of water

i'm confused
can anyone please help
thanks!
grind


80 gallons of Mix X containing 40% chloride + 120 gallons of Mix Y ----> 25% chloride mixture

wA/wB = (cB - cAvg)/(cAvg - cA)

80/120 = (Y - 25)/(25 - 40)
2 * (-15) = 3 * (Y - 25)
Y = 15

So the concentration of chloride in Mix Y is 15%.
In 120 gallons of mix Y, you will have (85/100)*120 = 102 gallons

Check:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/04 ... -mixtures/
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Re: Mixture X contains 60% water and 40% chloride. If 120 gallon [#permalink] New post   03 Nov 2021, 06:46
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Re: Mixture X contains 60% water and 40% chloride. If 120 gallon [#permalink]
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