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X grams of water were added to the 80 grams of a strong

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Director
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X grams of water were added to the 80 grams of a strong [#permalink] New post 03 Jan 2008, 23:02
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X grams of water were added to the 80 grams of a strong solution of acid. As a result, the concentration of acid in the solution dropped by Y percent. What was the concentration of acid in the original solution?

1. X = 80
2. Y = 50

Please explain your work.
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Re: DS: Mixture [#permalink] New post 03 Jan 2008, 23:40
GK_Gmat wrote:
X grams of water were added to the 80 grams of a strong solution of acid. As a result, the concentration of acid in the solution dropped by Y percent. What was the concentration of acid in the original solution?

1. X = 80
2. Y = 50

Please explain your work.


80=a+w

a/(a+w+x)=? a/(80+x) = y/100*a/(80)

a/(80+x)=a/160 --> 160a=80a+ax --> 80a=ax --> x=80

I know you can solve it from here, but I can't finish it. Brain dead right now
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Re: DS: Mixture [#permalink] New post 04 Jan 2008, 05:31
Let's suposse z is the quantity of acid in the original mixture. the initial concentration will be z/80. Know we add x grams of water so the final concentration will be z/[80+x]. Therefore Y=[ z/80 - z/(80+x) ] / [ z/80 ]=1-80/[80+x]

Either if you know x or Y you can calculate Y or X. These values are x=80 and Y=50. Therefore there are only two possible solutions: D and E

Lets pick numbers: z=1 => 0.5*1/80=1/160 or z=40 => 0.5*40/80=40/160
Both values of z are ok so we can not calculate the original concentration of acid. E
Re: DS: Mixture   [#permalink] 04 Jan 2008, 05:31
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X grams of water were added to the 80 grams of a strong

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