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Math Expert V
Joined: 02 Sep 2009
Posts: 57281
M10-23  [#permalink]

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13 00:00

Difficulty:   95% (hard)

Question Stats: 43% (02:03) correct 57% (02:04) wrong based on 118 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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Math Expert V
Joined: 02 Sep 2009
Posts: 57281
Re M10-23  [#permalink]

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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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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.
Intern  Joined: 29 Jun 2014
Posts: 8
Re: M10-23  [#permalink]

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 57281
Re: M10-23  [#permalink]

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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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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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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
>> !!!

You do not have the required permissions to view the files attached to this post.

Manager  Joined: 25 Nov 2009
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Location: India
Re: M10-23  [#permalink]

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1
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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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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GMAT 1: 570 Q40 V28 GMAT 2: 740 Q49 V41 Re: M10-23  [#permalink]

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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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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 57281
Re: M10-23  [#permalink]

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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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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).
Intern  B
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GMAT 1: 650 Q46 V34 Re: M10-23  [#permalink]

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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
Intern  B
Joined: 09 Apr 2018
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Re M10-23  [#permalink]

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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]

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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) $$Y = 2$$

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.

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Re M10-23  [#permalink]

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I think this is a high-quality question and I agree with explanation.
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Re: M10-23  [#permalink]

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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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+1 Kudos if you like my post pls! Re: M10-23   [#permalink] 02 Aug 2019, 22:01
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