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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?

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? _________________

KUDOS me if you feel my contribution has helped you.

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 ?

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. _________________

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.

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

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) _________________

Labor cost for typing this post >= Labor cost for pushing the Kudos Button http://gmatclub.com/forum/kudos-what-are-they-and-why-we-have-them-94812.html

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! _________________

Fight till you succeed like a gladiator..or doom your life which can be invincible

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.

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.

Posted from my mobile device _________________

+1 Kudos me - I'm half Irish, half Prussian.

Last edited by OldFritz on 16 Jun 2012, 09:37, edited 1 time in total.