X grams of water were added to the 80 grams of a strong : GMAT Data Sufficiency (DS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 19 Jan 2017, 20:31

### 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

# X grams of water were added to the 80 grams of a strong

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Manager
Joined: 08 Aug 2008
Posts: 234
Followers: 1

Kudos [?]: 36 [2] , given: 0

X grams of water were added to the 80 grams of a strong [#permalink]

### Show Tags

18 Nov 2008, 10:16
2
KUDOS
13
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

38% (02:36) correct 62% (01:33) wrong based on 279 sessions

### HideShow timer Statistics

$$X$$ 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
Manager
Joined: 23 Jul 2008
Posts: 203
Followers: 1

Kudos [?]: 118 [0], given: 0

Re: X grams of water were added to the 80 grams of a strong [#permalink]

### Show Tags

18 Nov 2008, 11:54
let us assume that originally 80 gms of solution had a gms of acid

Hence from statement 1

we get a/80 - a/160 = y/100 -------equation can t be solved hence insufficient

from 2 we get

a/80 - a/(80+x) = 50/100 = 1/2 --------------again can t be solved insuff

combining can be solved

Hence C
Current Student
Joined: 28 Dec 2004
Posts: 3384
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 15

Kudos [?]: 282 [0], given: 2

Re: X grams of water were added to the 80 grams of a strong [#permalink]

### Show Tags

19 Nov 2008, 09:15
I would go with E..we dont know the original concentration..
Manager
Joined: 23 Jul 2008
Posts: 203
Followers: 1

Kudos [?]: 118 [0], given: 0

Re: X grams of water were added to the 80 grams of a strong [#permalink]

### Show Tags

19 Nov 2008, 21:34
taking stmts 1 and 2 in consideration and the given statements the solution is

a 100% solution of srong acid (all 80 gms is acid) reduces in concetration to 50 % ( 80 gm acid + 80 gms water)

Hence C
Intern
Joined: 20 Sep 2008
Posts: 15
Followers: 0

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

Re: X grams of water were added to the 80 grams of a strong [#permalink]

### Show Tags

20 Nov 2008, 09:48
Even i would go with E.
We do not know the original concentration of the acid.Strong does not mean concentration is 100%.
Intern
Status: Into the last good fight
Joined: 22 Feb 2012
Posts: 19
GMAT 1: 550 Q42 V24
GMAT 2: 550 Q44 V23
GMAT 3: 540 Q41 V24
Followers: 0

Kudos [?]: 2 [0], given: 264

Re: X grams of water were added to the 80 grams of a strong [#permalink]

### Show Tags

30 Oct 2014, 05:31
Bunuel

Can you please give a detailed explanation to this question ?
Math Expert
Joined: 02 Sep 2009
Posts: 36567
Followers: 7081

Kudos [?]: 93219 [0], given: 10553

Re: X grams of water were added to the 80 grams of a strong [#permalink]

### Show Tags

30 Oct 2014, 07:31
Expert's post
1
This post was
BOOKMARKED
prasun84 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) $$Y = 2$$

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.

_________________
Intern
Status: Into the last good fight
Joined: 22 Feb 2012
Posts: 19
GMAT 1: 550 Q42 V24
GMAT 2: 550 Q44 V23
GMAT 3: 540 Q41 V24
Followers: 0

Kudos [?]: 2 [0], given: 264

Re: X grams of water were added to the 80 grams of a strong [#permalink]

### Show Tags

31 Oct 2014, 01:25
Bunuel

But, I must admit my weakness on "Mixture" type of questions. I didn't find the MGMAT mathbook sufficient to face different types of Mixture problems.

Can you please explain any CORE FORMULA regarding this type of problem, or how to set the equations in such cases ?

Intern
Status: Into the last good fight
Joined: 22 Feb 2012
Posts: 19
GMAT 1: 550 Q42 V24
GMAT 2: 550 Q44 V23
GMAT 3: 540 Q41 V24
Followers: 0

Kudos [?]: 2 [0], given: 264

Re: X grams of water were added to the 80 grams of a strong [#permalink]

### Show Tags

31 Oct 2014, 02:27
Another question:

I set the equation using st1 and st2 as...

80C = C/2 (80+80) [ old weight.old concentration = new weight.new concentration ]
80C= 80C

Is there any flaw/mistake above ?
Math Expert
Joined: 02 Sep 2009
Posts: 36567
Followers: 7081

Kudos [?]: 93219 [2] , given: 10553

Re: X grams of water were added to the 80 grams of a strong [#permalink]

### Show Tags

31 Oct 2014, 05:27
2
KUDOS
Expert's post
4
This post was
BOOKMARKED
Tanvr wrote:
Bunuel

But, I must admit my weakness on "Mixture" type of questions. I didn't find the MGMAT mathbook sufficient to face different types of Mixture problems.

Can you please explain any CORE FORMULA regarding this type of problem, or how to set the equations in such cases ?

All DS Mixture Problems to practice: search.php?search_id=tag&tag_id=43
All PS Mixture Problems to practice: search.php?search_id=tag&tag_id=114

Hope this helps.
_________________
Manager
Joined: 22 Jan 2014
Posts: 138
WE: Project Management (Computer Hardware)
Followers: 0

Kudos [?]: 54 [0], given: 135

Re: X grams of water were added to the 80 grams of a strong [#permalink]

### Show Tags

01 Nov 2014, 07:05
Bunuel wrote:
prasun84 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) $$Y = 2$$

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.

Hi Bunuel

Didn't quite get what you did up there.

This is what I did.

let in the original solution there be U unknown liquid and 80 gms of acid.

combining (1) & (2) --> {80/(80+u)}*(1/y) = {80/(80+80+u)}
=> u = 0
That's how i got C.

_________________

Illegitimi non carborundum.

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7125
Location: Pune, India
Followers: 2136

Kudos [?]: 13655 [3] , given: 222

Re: X grams of water were added to the 80 grams of a strong [#permalink]

### Show Tags

09 Dec 2014, 22:01
3
KUDOS
Expert's post
4
This post was
BOOKMARKED
prasun84 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) $$Y = 2$$

Responding to a pm:
Quote:
I made this equation: A1/(80+x) = 1/Y *(A2/80).
Is this equation right? Where A1 is the new concentration of acid, and A2 is the original one. I realize my initial wrror , which was to put both A1 and A2 as A (lack of concentration I suppose).

I don't know how you get this equation.

Amount of acid remains same after you add water so

Amount of acid = C1*V1 (initial) = C2*V2 (final)
C1*80 = C1/Y * (80+x)
80 = (80+x)/Y

The equation is independent of C1.

Look at the statements now:
(1) $$X = 80$$
If X = 80, Y = 2
But you don't get the value of C1.

(2) $$Y = 2$$
If Y = 2, you get x = 80
But you don't get the value of C1.

For more details, check here: http://www.veritasprep.com/blog/2012/01 ... -mixtures/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Manager
Joined: 08 Oct 2013
Posts: 52
Followers: 1

Kudos [?]: 14 [1] , given: 16

Re: X grams of water were added to the 80 grams of a strong [#permalink]

### Show Tags

26 Jun 2015, 22:56
1
KUDOS
Can someone please tell me if my logic is correct?

According to the statements:
if a is the weight of acid and w is the weight of water then total weight is a + w = 80

Initial concentration of acid is a/(a+w) = a/80 >> this is what we need to find.

If x gms of water is added to the mixture then,

a/(80+x) = (1/y) a/80; which gives
1/80+x = 1/y80

Substituting value of x from statement 1 we get 1/160 = 1/160 >> Not Sufficient
Substituting vale of y from statement 2 again gives the same fraction >> Not sufficient

Combining 1 & 2 does not give us any new info; hence E

This approach is correct?
VP
Joined: 08 Jul 2010
Posts: 1443
Location: India
GMAT: INSIGHT
WE: Education (Education)
Followers: 68

Kudos [?]: 1409 [0], given: 42

Re: X grams of water were added to the 80 grams of a strong [#permalink]

### Show Tags

26 Jun 2015, 23:09
AjChakravarthy wrote:
Can someone please tell me if my logic is correct?

According to the statements:
if a is the weight of acid and w is the weight of water then total weight is a + w = 80

Initial concentration of acid is a/(a+w) = a/80 >> this is what we need to find.

If x gms of water is added to the mixture then,

a/(80+x) = (1/y) a/80; which gives
1/80+x = 1/y80

Substituting value of x from statement 1 we get 1/160 = 1/160 >> Not Sufficient
Substituting vale of y from statement 2 again gives the same fraction >> Not sufficient

Combining 1 & 2 does not give us any new info; hence E

This approach is correct?

This is absolutely fine approach.

Only the reasoning of highlighted part is not very clear.

We must be clear that, We only need to know the value of either a or W to answer the question

Statement 1: Gives us the value of y but no information about values of a or W hence Not Sufficient
Statement 2: Gives us the value of y but no information about values of a or W hence Not Sufficient

Combining the two gives us the value of y only so NOT Sufficient
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com
Call us : +91-9999687183 / 9891333772
http://www.GMATinsight.com/testimonials.html

Feel free to give a Kudos if it is a useful post .

Manager
Joined: 08 Oct 2013
Posts: 52
Followers: 1

Kudos [?]: 14 [0], given: 16

Re: X grams of water were added to the 80 grams of a strong [#permalink]

### Show Tags

27 Jun 2015, 00:15
GMATinsight wrote:
AjChakravarthy wrote:
Can someone please tell me if my logic is correct?

According to the statements:
if a is the weight of acid and w is the weight of water then total weight is a + w = 80

Initial concentration of acid is a/(a+w) = a/80 >> this is what we need to find.

If x gms of water is added to the mixture then,

a/(80+x) = (1/y) a/80; which gives
1/80+x = 1/y80

Substituting value of x from statement 1 we get 1/160 = 1/160 >> Not Sufficient
Substituting vale of y from statement 2 again gives the same fraction >> Not sufficient

Combining 1 & 2 does not give us any new info; hence E

This approach is correct?

This is absolutely fine approach.

Only the reasoning of highlighted part is not very clear.

We must be clear that, We only need to know the value of either a or W to answer the question

Statement 1: Gives us the value of y but no information about values of a or W hence Not Sufficient
Statement 2: Gives us the value of y but no information about values of a or W hence Not Sufficient

Combining the two gives us the value of y only so NOT Sufficient

Thanks very much for the clarification

Regarding the highlighted part : just saying that wen we substitute the values and simplify, in each case we get 1/160 = 1/160....
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13459
Followers: 575

Kudos [?]: 163 [0], given: 0

Re: X grams of water were added to the 80 grams of a strong [#permalink]

### Show Tags

30 Jul 2016, 06:02
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: X grams of water were added to the 80 grams of a strong   [#permalink] 30 Jul 2016, 06:02
Similar topics Replies Last post
Similar
Topics:
1 A pizza recipe calls for cheese and flour in the ratio of 2 grams to 1 1 31 May 2016, 01:44
5 The table above gives the number of calories and grams of 7 01 Apr 2012, 04:26
11 How much water (in grams) should be added to a 35%-solution 11 27 Dec 2009, 14:59
4 How much water (in grams) should be added to a 35%-solution of acid to 8 17 Aug 2009, 07:09
2 The table above gives the number of calories and grams of pr 5 15 Mar 2007, 19:55
Display posts from previous: Sort by

# X grams of water were added to the 80 grams of a strong

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 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®.