Official Solution:


Statements (1) and (2) combined are insufficient. Denote the original concentration as \(C\). Construct an equation using S1 and S2:
\(80C + 80*0 = \frac{160C}{2}\)
\(80C = 80C\)

\(C\) cancels out, so we cannot determine the answer.


Answer: E
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Bunuel,

Can you please explain this step by step?
Thank you

I could be wrong but I solved it like this:

Original Solution = (Acid+whatever) = 80
After Adding X = (Acid/y + whatever + X) = 80 + X

With both statements. ((Acid/2) + Whatever + 80) = 160

We had at least three unknowns and were only given 2 equations.

Answer E.

Hey, if Algebra looks a bit confusing here (because of so many variables X, Y, Original Concentration) let's start from Statements.

Statement I: X = 80 gms. So if you add 80 gms water to 80 gms of acid solution, of whatever original concentration, then definitely the resultant concentration will be halved (Since in 1:1 ratio mixing resultant concentration is just average of the concentrations). Thus we get Y = 2. But this will never lead to the original concentration of acid. It can be anything from 1% to 100%. Not sufficient. A and D ruled out!

Statement II: Putting Y = 2, it says that the concentration is becoming 1/2 (as 1/Y = 1/2) of the original concentration. This is only possible if you mix Acid with Water in 1: 1 ratio. So, X = 80 gms. But once again it tells nothing about original concentration. Not sufficient. B ruled out!

As we have seen X helps us to find out Y and Y helps us to find X. Thus considering both these statements here is useless. So, C is also ruled out!

Answer: E

By figuring out that if you add 80grams of pure water to an 80gr solution the solution would become half as concentrated as before you can conclude that y=2 but you don't know anything about the initial concentration (Half of what? Could be 50% or 80% given the "highly concentrated" statement on the prompt) so A and D gone. In statement 2 you get y=2 so still you end up with "half of what?", so B out. Since statement 2 does not add any further information from what we have concluded from statement 1 C is out.

E.
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Check more solutions here: http://gmatclub.com/forum/x-grams-of-wa ... 73028.html

Hope it helps.
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Simple Logical Explanation:

1) X = 80 G. This means that 80G of water was added to 80G of strong acid solution.
In this scenario, the proportion of acid in the final solution will reduce by half i.e. (If initially the proportion was 80% - it will become 40%).
However, Initial proportion cannot be determined.
Not Suff.

2) Y = 2. This means that the proportion of acid in the final solution reduces by half.
For this to happen exactly the same amount of water must be added i.e. 80G
However, Initial proportion cannot be determined.
Not Suff.

1 & 2 Together:
Not Sufficient. Just proves that the initial concentration was halved. Cannot determine initial concentration.

OA: E

If such a question appears on the GMAT, the best approach, instead of panicking and trying to write equations, is to take a deep breath and just go through this slowly and thoroughly.

It's a VERY LOGICAL question.

If the same amount of water is added to some amount of acid to dilute it (in this case, 80 grams of water is added to 80 grams of a strong solution of acid), NO MATTER what the concentration of acid in the original solution, it will ALWAYS be halved.

Hence, the concentration of acid in the original solution can have any (infinite) number of values.

HIT KUDOS IF YOU LIKE THE SOLUTION

We are told that X grams of water are added to 80 grams of strong acid (we don't know the concentrate) to form 1/yY times the concentration of acid. So we can construct a table
% acid Grams grams of acid
Water 0% X 0
Acid C/100 80 4C/5
1/Y (x+80)

(4C/5)/(X+80) = 1/Y* 4C/5
(4C/5)/(X+80) = 4C/5Y

(1) x= 80
(4C/5)/160 = 4C/5Y
(4C*160)/5 = 4C/5Y
can determine Y but can't use to determine the original concentrate.
Insufficient
(
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