Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 01 May 2016, 16:43

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# m10q23

Author Message
Manager
Joined: 16 Feb 2010
Posts: 224
Followers: 2

Kudos [?]: 197 [0], given: 16

### Show Tags

21 Nov 2010, 20:15
 $$X$$ 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. $$Y = 2$$ (C) 2008 GMAT Club - m10#23 * 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 sufficientStatements (1) and (2) combined are insufficient. Denote the original concentration as $$C$$ . Construct an equation using S1 and S2:$$\begin{eqnarray*} 80C + 80*0 &=& \frac{160C}{2}\\ 80C &=& 80C\\ \end{eqnarray*}$$$$C$$ cancels out, so we cannot determine the answer.The correct answer is E.

still doesnt get it...anyoen to help with a more thorough explanation?
Kaplan GMAT Instructor
Joined: 21 Jun 2010
Posts: 148
Location: Toronto
Followers: 44

Kudos [?]: 162 [1] , given: 0

### Show Tags

22 Nov 2010, 14:12
1
KUDOS
zisis wrote:
[quote2]$$X$$ 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. $$Y = 2$$

(C) 2008 GMAT Club - m10#23

still doesnt get it...anyoen to help with a more thorough explanation?

While statement (2) might seem new and exciting, it actually provides identical information to (1).

To the explanation!

From (1), we know that we doubled the volume of the solution. There's a handy formula to remember for mixture questions:

(concentration 1)(volume 1) = (concentration 2)(volume 2)

or:

(C1)(V1) = (C2)(V2)

for short.

We know that the original volume is 80 grams (we're measuring in grams, so technically it's weight, not volume, but that doesn't matter) and that there's some acid in there as well. "Strong" has no specific meaning, so we don't know the quantity of acid.

So, based on (1), we know that:

(C1)(80) = (C2)(80 + 80)

(C1)(80) = (C2)(160)

(C1) = (C2)(160/80)
C1 = 2(C2)

so, from the original statement, we know that C1 is twice as strong as C2. Of course, this doesn't give us the actual value of C1, so it's insufficient.

From (2), we know that the new solution is 1/Y times as strong as the original. Since Y = 2, we know that the new solution is 1/2 times as strong. Also nothing about numbers, insufficient.

Now let's look at the statements together:

(1) C1 is twice as much as C2

(2) C2 is half as much as C1

When we stack them together like that, we quickly realize that each statement gives us identical information! Since neither statement was good enough by itself, they certainly won't be sufficient together: choose (E).
Re: m10q23   [#permalink] 22 Nov 2010, 14:12
Similar topics Replies Last post
Similar
Topics:
17 M10:Q23 - DS 21 22 Aug 2008, 15:38
Display posts from previous: Sort by

# m10q23

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.