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

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Manager
Joined: 22 May 2007
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X grams of water were added to the 80 grams of a strong [#permalink]

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09 Jul 2008, 21:37
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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

Generally on mixture problems , I do individual balance.

In this case, acid balance or water balance

Acid balance

80(A)=(A-Y)(80+W)

Where

A= Original % of Acid
W= Amt of water

I can solve for A if I use both the equations.

What am I doing wrong? OA is E
Director
Joined: 27 May 2008
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Re: Mixture problem - Am I correct? [#permalink]

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09 Jul 2008, 22:17
if you are adding same amount of water to a solution of acid of any cocentration, there will be a 50% drop in concentration.

earlier concentration = pure Acid / (acid + water) = a/80
new concentration = a/(80+80) = 1/2 (a/80)

so statement 1 and 2 are telling the same thing.

Thus not suff.
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Re: Mixture problem - Am I correct? [#permalink]

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10 Jul 2008, 07:21
Is there a problem with the wording? I'm not so sure there is. The questions says the concentration dropped by 50%. The concentration could have started out at 80%, and then dropped to 40%, which is a drop of 50%.

Let's say it's ethanol. If you have 80% concentration in 80 grams (which is an odd measurement unit for a liquid), 80% would be 64 grams of ethenol and 16 grams of water. If you add 80 grams of water to it, you now have 160 grams total, 64 grams of ethanol. This is 40%, a drop of 50%, rather than a drop to 50%.

I think the answer should be E because you don't need to know the original concentration in order to know that adding 80 grams of water to 80 grams of solution will result in a drop of 50% in the concentration. There also ISN'T a way to determine the original concentration. It could have been any % of concentration and the dropped to 1/2 of that % concentration.

aaron22197 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

_________________

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

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Re: Mixture problem - Am I correct?   [#permalink] 10 Jul 2008, 07:21
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