hey guys, i was doing a mixtures problem and got confused looking at the solutionsPlease read and follow: how-to-improve-the-forum-search-function-for-others-99451.html
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
can anyone please help
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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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