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# M10:Q23 - DS

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22 Aug 2008, 15:38
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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. $$Y = 2$$

[Reveal] Spoiler: OA
E

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24 Jul 2010, 16:13
7
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Good Question,

Let m% be the concentration
Initially :
acid = m% of 80 = 4m/5
water = 80-4m/5
Concentration = m

Now :
acid = m% of 80 = 4m/5
water = 80-4m/5 +X
Concentration = m/y

$$\frac{m}{y} =\frac{4m}{5} * \frac{1}{(80+x)}$$

=> 80y = 80+X

Using either of the equation we can get the value of y or x, but we can not calculate the value m, as it was canceled earlier.

Thus even if we take both the statement's together , it is not sufficient to answer the question.

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10 Jun 2010, 16:15
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Let initial acid concentration = a
amount of acid = 80a
after adding water, amount of acid = (80 + x)*a/Y
=> 80a = (80+x)*(a/Y)
=> 80 = (80+x)*(1/Y)
right from the stem it is not possible to find a-the initial acid concentration.
Am I doing something wrong, or is the question suspect?
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15 Jun 2012, 08:38
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E. Definite answer.. as 1 and 2 tell the same thing and it is not enough to answer the q.
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13 Jun 2011, 08:36
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Very simple problem if you take a deep breath initially and find out what they are asking for and what we know.

ACID/80 = Old Concentration of Acid

ACID/x + 80 = New Concentration of Acid

ACID/80 * (1/Y) = NEW Concentration

1. Tells us nothing about the original acid levels - NS
2. We still have to know the original acid level. We only know it will be half as concentrated ..... NS

E
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13 Jun 2011, 18:40
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Let initial concentration = z

So initial quantity of acid = z/100 * 80 = 4z/5

After adding x grams of water, Total amount of solution = (80 + x)

So concentraion of acid = 4z/5/(80+x) = 1/y (z)

Clearly, z cancels out even if x and y are known

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18 Jul 2011, 20:03
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Answer has to be E. Both 1 and 2 give you the same piece of information. In order for either of these to be helpful we would need to know the original concentration or potency of the acid solution therefore we are left with E.
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15 Jun 2012, 08:23
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Straightforward E. Setting up the equation involving pre and post addition of X grams of water renders an equation in which initial amount of acid (a) cancels out. Thus 'a' cannot be found.

(1/Y)(a/80)=a/(80+X)

As you can see, a cancels out, thus cannot be relevant to the question and can't be found.

Cheers.

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Last edited by OldFritz on 16 Jun 2012, 09:37, edited 1 time in total.
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23 Aug 2008, 03:27
Ans should be E. We don't have enough information to find out the concentration in the original solution.

1 is insuff because it just gives the value of X so the new solution = 80 + 80 = 160 but we cant find out the concentration of acid.

2 is insuff as well since we dont know what is X.

Both taken together are insuff as we dont know the amount of acid in the new solution.
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10 Jun 2010, 04:48
coz it sys strong acid
nt pure acid
n after 80 gm of water acid conc is 1/2 of original
so we can get conc
thus erlr was 100%
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10 Jun 2010, 07:32
I guess strong acid does not necessarily mean 100% concentration so even C is not possible. It should be E
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16 Jun 2010, 15:00
IMO .
80 grams of strong solution
Amount of acid in solution initially : p/80
X grams water added so now acid becomes : p/(80+X)
Concentration of acid now is 1/y times of original
p/(80+X) =(1/y)* (p/80 )
==> 80Y =80+X
Answer choices give value of x and Y so either should be sufficient .
What am I missing here ?
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28 Sep 2010, 14:28
A "strong solution of acid" generally means that the acid is 100% concentrate, or as strong as the solution can be. This threw me off a little, but I reasoned it out through the context the question (there would be no DS question if original concentration were already given as 100%).

Still got it wrong, though—didn't write this out on paper and thus didn't cancel out the initial concentrations.
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24 Oct 2010, 05:10
Why doesn't the amalgamation method work for this? does anyone know?

It leads to original concentration as 100% and that leads you to C

C (ORIGINAL ACID CONCENTRATION)--------------0 (CONCENTRATION OF ACID in WATER)

-----------------------------------0.50 (from stem 2)----------------------------------------

----------------0.50------------------------------(C-0.50)

So,

concentration of original acid in new solution = .50 / .50 + C - .50 = 1/2C

1/2C * 160 = 80 (from stem 1 and 2)

C = 1 or 100% acid

I get it that the question says initial acid concentration was "strong" (which can or can not be 100 pc). But usually this method never goes wrong. Only way you can get E based on this method if there is a chance that the original solution was less than 50% acid and addition of water in fact brought the concentration of the new solution up to 50% which doesn't seem right to me as water I think we can assume has no acid.

Thanks.
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24 Oct 2010, 05:19
Ahhh. I get it - even with this this method you do get E. 80 = 80 as C cancels out.

stem 2 provides the value of Y = 2 for 1/Y of the "original concentration" (which is C) and not 1/Y of the new solution

I didn't read that part carefully... damn - little tuffie.
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13 Jun 2011, 06:02
Ans E

"what was the concentration of acid in the original solution", said the initial solution did not have 100% acid concentration.
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13 Jun 2011, 06:27
This question is a little convulated but if read properly is very easy. Firstly let's make a table

Old new
Acid. 80. 80
Water 0. X
Total 80+z. 80/80+x
Additional info : 1/y ( 80+z)
Statement 1
Insufficient since we only have one variable

Statement 2
Insufficient

Both together still insufficient hence E

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13 Jun 2011, 09:27
Good one...I got it wrong first, my bad...Could not find anything wrong in my approach then saw Yes, it should be (E) and not (D)

Let a = initial concentration

So
80a=(80+x).(a/y)
=>a[80-(80+x)/y]=0

a will never be 0...Because the question tells us strong acid..So neither of the choices provides sufficient data to give the Ans. ==> (E)
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13 Jun 2011, 17:14
Can someone explain to me where you are pulling the letter choices from? I get E...but how do you know what the other letter answers are?
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13 Jun 2011, 17:58
jw wrote:
Can someone explain to me where you are pulling the letter choices from? I get E...but how do you know what the other letter answers are?

The general pattern in GMAT is

A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statement (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed

This is what we follow here in GmatClub..So tthe answer choices are implied.

Hope this helps!
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Re: M10:Q23 - DS   [#permalink] 13 Jun 2011, 17:58

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# M10:Q23 - DS

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