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

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

Bunuel · Accepted Answer

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.

Narenn · Answer

Refer the attached Image

grindcore · Answer

im so sorry i feel stupid now.
lesson learned.
READ CAREFULLY
thanks

Maxirosario2012 · Answer

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.

maggie27 · Answer

Hi Narenn,

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

PareshGmat · Answer

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

KarishmaB · Answer

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/

gracie · Answer

let w=% of water in mixture Y
.6*80+w*120=.75*200
w=.85
.85*120=102 gallons water

bumpbot · Answer

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

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GMAT Problem Solving (PS) Questions

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

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