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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 became \(\frac{1}{Y}\) times of the initial concentration. What was the concentration of acid in the original solution? 1. \(X = 80\) 2. \(Y = 2\) (C) 2008 GMAT Club  m24#4Source: GMAT Club Tests  hardest GMAT questions Denote the original concentration \(C\) . Then the new concentration is \(\frac{80C}{80 + X}\) . We can compose an equation for \(C\) : \(\frac{80C}{80 + X} = \frac{C(100  Y)}{100}\) . This equation cannot be solved even if \(X\) and \(Y\) are known. No matter what the original concentration is, \(X\) and \(Y\) are bound with the equality \(\frac{80}{80 + X} = \frac{100  Y}{100}\) . The correct answer is E. I don't understand why the original concentration of acid is bound by the above equation. If the amount of acid in the original concentration is unknown but a given quantity of water is added and there is a known drop in the concentration of the acid in the solution, then the original concentration can be solved for. Where am I going wrong here?



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Re: M24 #4 [#permalink]
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05 Oct 2008, 03:25
lets assume Orignal concentration is C. and New concentration is C1. Given C = C1 + Y,  1st Eq
Amount of Acid in Original solution = 80C Amount of Acid in new solution = (80 + X)C1
Now both amount should be equal 80C = (80 + X)C1  2nd Eq
So you can easily solve both the equation for C, if you have values of X & Y. Answer is C



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Re: M24 #4 [#permalink]
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05 Oct 2008, 10:28
Thank you for the confirmation. I just now noticed that the correct answer is stated as C, but in the explanation the correct answer is stated as E. Thanks!



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Re: M24 #4 [#permalink]
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17 Jan 2009, 16:41
botirvoy wrote: bump. Detailed solution to this question would be appreciated Its little long cuz I wanted to spell out every tiny step. Before: solution weight = 80 grams water's ratio = a. so water = 80a acid ratio = 1a and acid ratio = 80 (1a) After: water weight = x grams water ratio = 1 and water weight = x acid ratio = 0 and acid weight = 0 After water added to the solution: solution weight = (80 + x) grams water ratio = (80a+x)/(x+80) and water weight = (80a+x) acid ratio = 80(1a)/(x+80) and acid weight = 80(1a) (1a)  [80(1a)/(x+80)] = y (1a) [1  80/(x+80)] = y (1a) [(x+8080)/(x+80)] = y x(1a)/(x+80) = y xxa = y (x+80) 1. if x = 80, a and y are not known. 2. if y = 50, a and x are not known. 1&2: x  xa = y (x + 80) 80  80a = 50 (80+80) a = 80/{50*160) a = 1% so it is C. snowy2009 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. \(Y = 50\)
(C) 2008 GMAT Club  m24#4
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Re: M24 #4 [#permalink]
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24 Jan 2009, 05:56
GMAT TIGER wrote: botirvoy wrote: bump. Detailed solution to this question would be appreciated Its little long cuz I wanted to spell out every tiny step. Before: solution weight = 80 grams water's ratio = a. so water = 80a acid ratio = 1a and acid ratio = 80 (1a) After: water weight = x grams water ratio = 1 and water weight = x acid ratio = 0 and acid weight = 0 After water added to the solution: solution weight = (80 + x) grams water ratio = (80a+x)/(x+80) and water weight = (80a+x) acid ratio = 80(1a)/(x+80) and acid weight = 80(1a) (1a)  [80(1a)/(x+80)] = y (1a) [1  80/(x+80)] = y (1a) [(x+8080)/(x+80)] = y x(1a)/(x+80) = y xxa = y (x+80)
1. if x = 80, a and y are not known. 2. if y = 50, a and x are not known.
1&2: x  xa = y (x + 80) 80  80a = 50 (80+80) a = 80/{50*160) a = 1%
so it is C. snowy2009 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. \(Y = 50\)
(C) 2008 GMAT Club  m24#4 Fantastic effort  thank you. Now even I understand the question! +1 One point: Because acid concentration dropped by "Y percent", you need to write "y/100". This will result in a=1. This means, original solution was 100% acid and doubling the solution to 180 grams would naturally half the acid concentration.



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Re: M24 #4 [#permalink]
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11 Mar 2009, 22:30
Just came across this Q and I got C. Answer was E.
I was under the impression that GMAT Club challenges were devoid of mistakes. not the case!



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Re: M24 #4 [#permalink]
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30 Mar 2009, 06:36
Sorry guys. I remember changing this question but now I see the old version live again. The OA is still E. Here's what the question is now: \(X\) 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 became \(\frac{1}{Y}\) times of the initial concentration. What was the concentration of acid in the original solution? 1. \(X = 80\) 2. \(Y = 2\) (C) 2008 GMAT Club  m24#4 * Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient * Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient * BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient * EACH statement ALONE is sufficient * Statements (1) and (2) TOGETHER are NOT sufficient
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Re: M24 #4 [#permalink]
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23 Jun 2009, 16:10
It is E....
You don't know the actual concentration of the original solution  you just know that it was cut in half when adding 80 grams of water.



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Re: M24 #4 [#permalink]
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26 Jun 2009, 22:20
botirvoy wrote: bump. Detailed solution to this question would be appreciated seofah wrote: bump. Detailed solution to this question would be appreciated Any difference between the two posts? Why & how? botirvoy = seofah ?
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Re: M24 #4 [#permalink]
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26 Nov 2009, 05:37
dzyubam wrote: Sorry guys. I remember changing this question but now I see the old version live again. The OA is still E. Here's what the question is now:
\(X\) 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 became \(\frac{1}{Y}\) times of the initial concentration. What was the concentration of acid in the original solution?
1. \(X = 80\) 2. \(Y = 2\)
(C) 2008 GMAT Club  m24#4
* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient * Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient * BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient * EACH statement ALONE is sufficient * Statements (1) and (2) TOGETHER are NOT sufficient I do not see why the answer would be change to E. Let Z be the concentration of original acid. (80+X) * Z/Y = 80Z (Acid balance) Given X and Y, you can find Z=1 which means 1% was the initial acid concentration. The answer should be C IMO.



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Re: M24 #4 [#permalink]
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28 Jan 2010, 06:14
Let A be the Acid component in 80 grams of the solution.
So the original concentration = A/80.
When X grams of water is added to the solution the concentration changes to 
A/(80+W) = Y/100
A = Y (80 +X) /100
So if we have Y and X we can find the Acid component and hence the original concentration.
IMO C is the answer.



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Re: M24 #4 [#permalink]
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28 Jan 2010, 07:08
I know this is a late reply but still... Can you elaborate how you solve the equation you made? This one: \((80+X) * \frac{Z}{Y} = 80Z\) After plugging in 80 for \(X\) and 2 for \(Y\), we are left with this after simplifying: \(80Z = 80Z\) \(Z\) can be any number, so S1+S2 is insufficient. Am I mistaken anywhere? lonewolf wrote: dzyubam wrote: Sorry guys. I remember changing this question but now I see the old version live again. The OA is still E. Here's what the question is now:
\(X\) 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 became \(\frac{1}{Y}\) times of the initial concentration. What was the concentration of acid in the original solution?
1. \(X = 80\) 2. \(Y = 2\)
(C) 2008 GMAT Club  m24#4
* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient * Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient * BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient * EACH statement ALONE is sufficient * Statements (1) and (2) TOGETHER are NOT sufficient I do not see why the answer would be change to E. Let Z be the concentration of original acid. (80+X) * Z/Y = 80Z (Acid balance) Given X and Y, you can find Z=1 which means 1% was the initial acid concentration. The answer should be C IMO.
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Re: M24 #4 [#permalink]
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28 Jan 2010, 07:15
I'm probably missing something here, but I can't understand what the part in red is based upon. Can you please elaborate? Please note that the question has been revised and it now differs from the one posted in the first post of the thread. Here's the updated question: dzyubam wrote: Sorry guys. I remember changing this question but now I see the old version live again. The OA is still E. Here's what the question is now:
\(X\) 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 became \(\frac{1}{Y}\) times of the initial concentration. What was the concentration of acid in the original solution?
1. \(X = 80\) 2. \(Y = 2\)
(C) 2008 GMAT Club  m24#4
* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient * Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient * BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient * EACH statement ALONE is sufficient * Statements (1) and (2) TOGETHER are NOT sufficient rohityes wrote: Let A be the Acid component in 80 grams of the solution.
So the original concentration = A/80.
When X grams of water is added to the solution the concentration changes to 
A/(80+W) = Y/100
A = Y (80 +X) /100
So if we have Y and X we can find the Acid component and hence the original concentration.
IMO C is the answer.
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Re: M24 #4 [#permalink]
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13 Feb 2010, 22:39
dzyubam, I think you are right mate. The answer is E. But i don't see difference between the original and the "corrected" version of the question! Original: "the concentration of acid in the solution dropped by Y percent" where Y is given as 50 this implies that the concentration after adding water halved. "Corrected": "the concentration of acid in the solution became 1/Y times of the initial concentration" where Y is given as 2 this also means that the concentration after adding water halved. Moreover, imo the answer couldn't be C, because both A and B are giving essentially same information. A. 80 grams of water is added to 80 grams of solution. Implying that the volume of solution doubled while the acid remained the same, meaning the concentration of acid halved. B. As explained above, the information in B along with the question implies that the concentration of acid halved. And the question is asking for original concentration. But as shown we can only infer that the original concentration halved but we can't derive halved from what. Hence, both statements are insufficient. E. All this can be explained mathematically, but inference is much quicker Let me know if I missed something somewhere. Cheers.
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Re: M24 #4 [#permalink]
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15 Feb 2010, 06:17
You're right. We decided to change the question that way because some people understood "dropped by 50 percent" incorrectly. It might have been understood as "the concentration of acid in the solution has decreased by 50%" (it could be 80% and it was changed to 80%  50% = 30%). As you see, thinking along those lines might confuse some people. This is why we decided to change the wording the way we did. I hope it cleared the possible confusion. scarish wrote: dzyubam, I think you are right mate. The answer is E. But i don't see difference between the original and the "corrected" version of the question! Original: "the concentration of acid in the solution dropped by Y percent" where Y is given as 50 this implies that the concentration after adding water halved. "Corrected": "the concentration of acid in the solution became 1/Y times of the initial concentration" where Y is given as 2 this also means that the concentration after adding water halved. Moreover, imo the answer couldn't be C, because both A and B are giving essentially same information. A. 80 grams of water is added to 80 grams of solution. Implying that the volume of solution doubled while the acid remained the same, meaning the concentration of acid halved. B. As explained above, the information in B along with the question implies that the concentration of acid halved. And the question is asking for original concentration. But as shown we can only infer that the original concentration halved but we can't derive halved from what. Hence, both statements are insufficient. E. All this can be explained mathematically, but inference is much quicker Let me know if I missed something somewhere. Cheers.
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Re: M24 #4 [#permalink]
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24 Jun 2010, 17:19
Thanks scarish. never noticed that S1 and S2 basically gives the same info.



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Re: M24 #4 [#permalink]
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28 Jun 2011, 09:33
Posted in another forum 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 Answer  E
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Re: M24 #4  A solution to the solution problem [#permalink]
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Updated on: 10 Aug 2012, 16:57
Let me take a go at this :
First of all, lets establish the formula to calculate concentration : This is the mass of the solute divided by the mass of the solution (mass of solute plus mass of solvent), multiplied by 100.. So if you mix 20 grams of salt (NaCl) in 80 grams of water, you have a 20% NACL solution. (20/(80+20)) = 20%
Now that we have go this out of the way, lets deep dive into the problem : Lets assume that the Solution contained "a" grams of acid and "w" grams of water. As per the question stem ==> a+w = 80 gms. The Concentration of acid prior to the addition of water ==> a/(a+w) = a/80 ==> Lets call this (1) In order to answer the question  we need to figure of the value of "a".
Per the question stem, when "X" grams of water is added, the new concentration of the acid becomes (1/Y) OF THE ORIGINAL CONCENTRATION. When x grams of water is added, the new concentration = a/(a+w+x) = a/80+x ==> Lets call this (2)
This new concentration (ie (2) is now (1/Y) times the original concentration, ie (1). This implies the following a/(80+x) = 1/Y (a/80) ==> 80 + x = 80y (The "a"s cancel out)
Inspite of answer options being provided(1) and (2), we have no way of deriving "a", which is the key piece of information that is being asked. Hence (E)
Originally posted by hafizkarim on 02 Jul 2012, 06:20.
Last edited by hafizkarim on 10 Aug 2012, 16:57, edited 2 times in total.



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Re: M24 #4 [#permalink]
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02 Jul 2012, 08:10
if we are given how many grams of water were added and how much it was reduced by... can we not solve the problem?
you're given 2 eqns and you defined the 2 variables, which means you can solve no?



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Re: M24 #4 [#permalink]
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02 Jul 2012, 23:33
Let the amount of acid in 80 gram solution(acid+water) be m grams: So concentration= [m][/80]
water added = x gram So new conc. = [m][/80+x]
Now Acc. to question [m][/80+x] = [1][/y] of [m][/80]
Solving we get: 80y = 80+x
Therefore each statement is sufficient.
Answer should be (D)







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