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M10-23

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M10-23  [#permalink]

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New post 16 Sep 2014, 00:42
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A
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E

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42% (02:03) correct 58% (02:09) wrong based on 106 sessions

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\(X\) grams of water were added to 80 grams of a strong solution of acid. If as a result, the concentration of acid in the solution became \(\frac{1}{Y}\) times of the initial concentration, what was the concentration of acid in the original solution?


(1) \(X = 80\)

(2) \(Y = 2\)

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Re M10-23  [#permalink]

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New post 16 Sep 2014, 00:42
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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Re: M10-23  [#permalink]

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New post 28 Nov 2014, 07:26
3
let k grams of solute be present.
then,
equating final concentrations:
(1/y)*(k/80)=k/(x+80)
so , k cancel out,
we need to calculate k/80
hence ans is E.
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Re: M10-23  [#permalink]

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New post 26 Apr 2015, 11:07
Bunuel wrote:
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


Bunuel,

Can you please explain this step by step?
Thank you
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Re: M10-23  [#permalink]

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New post 27 Apr 2015, 01:45
propcandy wrote:
Bunuel wrote:
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


Bunuel,

Can you please explain this step by step?
Thank you


Check here: x-grams-of-water-were-added-to-the-80-grams-of-a-strong-73028.html

Hope it helps.
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Re: M10-23  [#permalink]

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New post 19 Feb 2016, 14:19
1
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.
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Re: M10-23  [#permalink]

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New post 19 Apr 2016, 13:27
1
Bunuel wrote:
\(X\) grams of water were added to 80 grams of a strong solution of acid. If as a result, the concentration of acid in the solution became \(\frac{1}{Y}\) times of the initial concentration, what was the concentration of acid in the original solution?


(1) \(X = 80\)

(2) \(Y = 2\)


This procedure is according to here: mixture-problems-made-easy-49897.html
>> !!!

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Re: M10-23  [#permalink]

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New post 18 May 2016, 07:26
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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
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Re: M10-23  [#permalink]

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New post 18 May 2016, 19:30
Lets take both A and B as given and look at it this way:

In the first solution (80 Grams of strong acid mixture) you have certain amount of acid. The concentration will be grams of acid / total weight. Lets say there was A grams of acid in the first mixture, then the concentration will be A/80.

Now you mix 80 grams of water (double the total weight) to the original mixture. The new concentration still contains same amount of Acid (A grams) so the new concentration = A/160.

You can see that the new concentration is 1/2 of the old concentration no matter how much acid you had in the original mixture so we can't know the original concentration. In fact, looking from outside of the box, when you add equal amount of water to any mixture you are making its concentration become 1/2 of original concentration hence B is basically giving us redundant information.
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Re: M10-23  [#permalink]

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New post 29 Jun 2016, 09:01
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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Re: M10-23  [#permalink]

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New post 22 May 2018, 20:25
Bunuel you should try explaining the solutions more thoroughly. We are paying for this service and a lot of questions are answered in a very poor way.
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Re: M10-23  [#permalink]

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New post 22 May 2018, 21:30
mariobc90 wrote:
Bunuel you should try explaining the solutions more thoroughly. We are paying for this service and a lot of questions are answered in a very poor way.


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

Hope it helps.
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Re M10-23  [#permalink]

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New post 10 Jun 2018, 22:16
I think this is a high-quality question and I agree with explanation. Great question. Really tests your ability to solve for the correct variable. The question is asking for the concentration of Acid (i.e. what is Acid/80). Each statement gives you enough info to solve for X/Y, but have these 2 variables by themselves do not give you A/80 (try to find A/80; you can't with the given info).
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Re: M10-23  [#permalink]

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New post 08 Jan 2019, 08:16
1
Bunuel wrote:
\(X\) grams of water were added to 80 grams of a strong solution of acid. If as a result, the concentration of acid in the solution became \(\frac{1}{Y}\) times of the initial concentration, what was the concentration of acid in the original solution?


(1) \(X = 80\)

(2) \(Y = 2\)


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
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Re M10-23  [#permalink]

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New post 15 Mar 2019, 01:03
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Cant We deduce the original concentration to be 100% from statements 1 and 2 ?
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Re M10-23   [#permalink] 15 Mar 2019, 01:03
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