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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?

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

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

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.

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

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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?

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

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