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

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Mixture X contains 60% water and 40% chloride. If 120 gallon [#permalink]

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02 Dec 2010, 14:12
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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?

[Reveal] 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
thanks!
grind
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Re: mixtures problem [#permalink]

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02 Dec 2010, 14:43
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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?

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
thanks!
grind

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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: mixtures problem [#permalink]

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02 Dec 2010, 14:53
im so sorry i feel stupid now.
lesson learned.
thanks
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Re: Mixture X contains 60% water and 40% chloride. If 120 gallon [#permalink]

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25 Oct 2013, 02:00
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Re: Mixture X contains 60% water and 40% chloride. If 120 gallon [#permalink]

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26 Oct 2013, 21:04
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Re: Mixture X contains 60% water and 40% chloride. If 120 gallon [#permalink]

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02 Jun 2014, 07:36
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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Re: Mixture X contains 60% water and 40% chloride. If 120 gallon [#permalink]

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29 Jun 2014, 08:05
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]

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10 Jul 2014, 01:20
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]

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Re: Mixture X contains 60% water and 40% chloride. If 120 gallon [#permalink]

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21 Aug 2016, 20:27
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Re: Mixture X contains 60% water and 40% chloride. If 120 gallon [#permalink]

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21 Aug 2016, 21:29
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?

[Reveal] 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
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:
http://www.veritasprep.com/blog/2011/04 ... -mixtures/
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Re: Mixture X contains 60% water and 40% chloride. If 120 gallon [#permalink]

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22 Aug 2016, 06:48
let w=% of water in mixture Y
.6*80+w*120=.75*200
w=.85
.85*120=102 gallons water
Re: Mixture X contains 60% water and 40% chloride. If 120 gallon   [#permalink] 22 Aug 2016, 06:48
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