GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 15 Oct 2019, 06:47 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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  Is |x − z − y| > x − z + y?

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

Hide Tags

Intern  Joined: 05 Mar 2013
Posts: 8
Is |x − z − y| > x − z + y?  [#permalink]

Show Tags

13 00:00

Difficulty:   75% (hard)

Question Stats: 60% (02:20) correct 40% (02:27) wrong based on 277 sessions

HideShow timer Statistics

Is |x − z − y| > x − z + y?

(1) 0 < x < z < y
(2) (x – z – y) is negative

Hi,

I have been doing various GMAT problems, and one major confusion was pertaining to absolute sign problems in data sufficiency (Specifically the one mentioned below)

Using the standard approach for absolute problems, I usually interpreted the question as if x-z-y >0, Then y<0 OR if x-z-y<0, Then x<z.
However my point of confusion starts, when i start looking at the two given statements.

As per the above problem, statement 1 states that x<z - Is this only information sufficient or do we also need the condition that x-z-y<0. I am really confused at this part, in some problems, both the statements are used, and in some statements only one statement is sufficient. I am really not sure how should i go about dealing with the DS statements for these type of absolute problems.

Please clarify my confusion or let me know if i am missing out on something. Thanks you.

Originally posted by gmat66 on 12 Jun 2013, 06:01.
Last edited by Bunuel on 15 Feb 2018, 09:33, edited 3 times in total.
Renamed the topic, edited the question and added the OA.
Math Expert V
Joined: 02 Sep 2009
Posts: 58332
Re: Is |x − z − y| > x − z + y?  [#permalink]

Show Tags

6
3
Is |x−z−y| > x−z+y?

(1) 0<x<z<y. This implies that x-z-y must be negative, which means that |x-z-y| = -(x-z-y) = y + z - x (recall that when x<0, then |x| = -x). So, the question becomes: is y + z - x > x - z + y? --> is 2z > 2x? is z > x? We are told that this is in fact so (x < z). Sufficient.

(2) (x–z–y) is negative --> |x-z-y| = -(x-z-y) = y + z - x. The question becomes: is y + z - x > x - z + y? --> is 2z > 2x? is z > x? But here we cannot answer this question. For example, if x=y=0 and z=1, then the answer would be YES but if x=1, y=2 and z=0, then the answer would be NO. Not sufficient.

_________________
General Discussion
Math Expert V
Joined: 02 Sep 2009
Posts: 58332
Re: Is |x − z − y| > x − z + y?  [#permalink]

Show Tags

3
3
Is |x−z−y|>x−z+y?

(1) 0<x<z<y. This implies that $$x-z-y$$ is negative (positive number x minus greater number z minus greater number y must be less than 0). Thus $$|x-z-y|=-(x-z-y)=y+z-x$$. So, the question becomes: is $$y+z-x>x-z+y$$? --> is $$2z>2x$$? --> is $$z>x$$? The statement gives an YES answer to the question. Sufficient.

(2) (x–z–y) is negative. Almost the same here: $$|x-z-y|=-(x-z-y)=y+z-x$$. So, the question becomes: is $$y+z-x>x-z+y$$? --> is $$2z>2x$$? --> is $$z>x$$? We don't know that. Not sufficient.

Theory on absolute value is HERE.
PS absolute value questions are HERE.
DS absolute value questions are HERE.
700+ absolute value questions HERE.
_________________
Senior Manager  Joined: 13 May 2013
Posts: 405
Re: Is |x − z − y| > x − z + y?  [#permalink]

Show Tags

1
Is |x−z−y|>x−z+y?

(1) 0<x<z<y
|x−z−y|>x−z+y
Because x is less than z is less than y, |x-z-y| is negative, so:
-(x-z-y)>x-z+y
-x+z+y>x-z+y
z+y>2x-z+y
2z>2x
z>x
z>x which is given to us in the statement.
SUFFICIENT

(2) (x–z–y) is negative
|x−z−y|>x−z+y
-(x-z-y)>x-z+y
-x+z+y>x-z+y
2z>2x
z>x
However, we are not given any more information.
INSUFFICIENT

(A)
Manager  S
Joined: 29 May 2016
Posts: 93
Re: Is |x − z − y| > x − z + y?  [#permalink]

Show Tags

1
Bunuel I have one doubt here
|x−z−y|>x−z+y? in normal case open the mode by multiplying + and another by multiplying with negative

x−z−y>x−z+y --(1) and -x+z+y>x−z+y --(2)
solving 1 we y<0
solving second we get x<z

now if we see option A it gives us hint that basically
|x-z-y| = -(x-z-y) = y + z - x. The question becomes: is y + z - x > x - z + y?--- so I thought Its sufficient
Now in case of B we already know that mode is negative ... so why cant it be sufficient

Manager  S
Joined: 28 Feb 2011
Posts: 66
Location: India
Concentration: Strategy, Marketing
Schools: ISB, IIMA , IIMC , IIM
GPA: 3.77
WE: Analyst (Computer Software)
Re: Is |x − z − y| > x − z + y?  [#permalink]

Show Tags

mbaprep2016 wrote:
Bunuel I have one doubt here
|x−z−y|>x−z+y? in normal case open the mode by multiplying + and another by multiplying with negative

x−z−y>x−z+y --(1) and -x+z+y>x−z+y --(2)
solving 1 we y<0
solving second we get x<z

now if we see option A it gives us hint that basically
|x-z-y| = -(x-z-y) = y + z - x. The question becomes: is y + z - x > x - z + y?--- so I thought Its sufficient
Now in case of B we already know that mode is negative ... so why cant it be sufficient

I think the key here is to note that statement one DOES NOT satisfy the condition of y < 0 (the statement says that 0<x<y<z) and thus, we need to only take care of the other case and can ignore the y < 0 case which makes statement 1 sufficient.

In other words, if we solve the |x-z-y|, we can have two cases - one positive and one negative as you specified above.

Now, if we talk about statement 2, it says that (x–z–y) is negative (less than zero). So, this basically means that our expression in the absolute expression (in the question stem is negative) or in other words, |x−z−y| = - (x-z-y). BUT, this does not answer the main question we have in the question stem.

Knowing that the absolute value of an expression is negative does NOT answer the main question at hand which is whether |x−z−y| > x−z+y?

Hope the information helps. All the best to you !!
_________________
Best Regards,
RS
Senior SC Moderator V
Joined: 14 Nov 2016
Posts: 1348
Location: Malaysia
Re: Is |x − z − y| > x − z + y?  [#permalink]

Show Tags

AmoyV wrote:
Is $$|x−z−y| > x−z+y$$?

(1) $$0<x<z<y$$
(2) $$(x–z–y)$$ is negative

Official solution from Veritas Prep.

Explanation: For this type of difficult inequality problem, you want to do as much algebraic manipulation as possible. First you should consider the two possible cases for the absolute value in the question (use the two case approach), and then simplify the question. If $$x - z - y$$ is positive, then the question is really “Is $$x - z - y > x - z + y$$?" which can be simplified by adding/subtracting terms to “Is $$y < 0$$” In other words, if you can prove that x-z-y is positive you then only need to determine the sign of y to answer the question. However if $$x - z - y$$ is negative then you need to answer this question: “Is $$- (x - z - y) > x - z + y$$?” Simplifying that question by adding/subtracting terms you see that the question is really “Is $$x < z$$?” In statement 1, you learn cleverly that indeed $$(x – z – y)$$ is negative because if you subtract progressively bigger positive numbers from an original positive number, the result will always be negative. Given that $$(x – z – y)$$ is negative then you need to answer the question “Is $$x < z?$$” and this is expressly given in the statement. Statement (1) is sufficient. Statement (2) is tempting you with information that you already know from statement (1). By itself, it is not sufficient because you do not know if $$x < z$$. Answer is (A).
_________________
"Be challenged at EVERY MOMENT."

“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”

"Each stage of the journey is crucial to attaining new heights of knowledge."

Rules for posting in verbal forum | Please DO NOT post short answer in your post!

Manager  G
Joined: 20 Jun 2018
Posts: 53
Location: Japan
Concentration: Strategy, Finance
GMAT 1: 660 Q49 V31 WE: Analyst (Energy and Utilities)
Re: Is |x − z − y| > x − z + y?  [#permalink]

Show Tags

Let's pick a number:
0<x<z<y
X=1; Z=2; y=10
Is |x−z−y|>x−z+y?
|1-2-10|>1-2+10
or,11>11
NO
Again,
Let's pick a number:
0<x<z<y
X=1; Z=2; y=3
Is |x−z−y|>x−z+y?
|1-2-3|>1-2+3
or,4>2
Yes
Thus, Statement 1 is insufficient, I think.
Intern  B
Joined: 07 Aug 2017
Posts: 9
Re: Is |x − z − y| > x − z + y?  [#permalink]

Show Tags

akash7gupta11 wrote:
Let's pick a number:
0<x<z<y
X=1; Z=2; y=10
Is |x−z−y|>x−z+y?
|1-2-10|>1-2+10
or,11>11
NO
Again,
Let's pick a number:
0<x<z<y
X=1; Z=2; y=3
Is |x−z−y|>x−z+y?
|1-2-3|>1-2+3
or,4>2
Yes
Thus, Statement 1 is insufficient, I think.
Retired Moderator P
Joined: 22 Aug 2013
Posts: 1430
Location: India
Re: Is |x − z − y| > x − z + y?  [#permalink]

Show Tags

1
akash7gupta11 wrote:
Let's pick a number:
0<x<z<y
X=1; Z=2; y=10
Is |x−z−y|>x−z+y?
|1-2-10|>1-2+10
or,11>11
NO
Again,
Let's pick a number:
0<x<z<y
X=1; Z=2; y=3
Is |x−z−y|>x−z+y?
|1-2-3|>1-2+3
or,4>2
Yes
Thus, Statement 1 is insufficient, I think.

Hello

Please check the highlighted part. 1-2+10 will be 9, not 11.
Manager  S
Joined: 24 Sep 2018
Posts: 137
Re: Is |x − z − y| > x − z + y?  [#permalink]

Show Tags

For this type of difficult inequality problem, you want to do as much algebraic manipulation as possible. First you should consider the two possible cases for the absolute value in the question (use the two case approach), and then simplify the question. If x - z - y is positive, then the question is really “Is x - z - y > x - z + y?" which can be simplified to “Is y < 0?” by adding/subtracting terms. In other words, if you can prove that x-z-y is positive you then only need to determine the sign of y to answer the question. However if x - z - y is negative then you need to answer this question: “Is - (x - z - y) > x - z + y?” Simplifying that question by adding/subtracting terms, you see that the question is really “Is x < z?”

Quote:
In statement 1, you learn cleverly that indeed (x – z – y) is negative because if you subtract progressively bigger positive numbers from an original positive number, the result will always be negative. Given that (x – z – y) is negative then you need to answer the question “Is x < z?” and this is expressly given in the statement. Statement (1) is sufficient.

Quote:
Statement (2) is tempting you with information that you already know from statement (1). By itself, it is not sufficient because you do not know if x < z. Answer is (A).

_________________
Please award kudos, If this post helped you in someway. Director  V
Joined: 12 Feb 2015
Posts: 917
Re: Is |x − z − y| > x − z + y?  [#permalink]

Show Tags

When you simplify |x − z − y| > x − z + y? You get two possibilities i.e. “Is y<0” or “Is x<z?”

St1 is in confirmation with Y is not less than zero so we have a definite "No" hence st 1 is sufficient.

St 2 leads us nowhere hence insufficient

Option A is the correct answer
_________________
________________
Manish "Only I can change my life. No one can do it for me" Re: Is |x − z − y| > x − z + y?   [#permalink] 27 May 2019, 23:11
Display posts from previous: Sort by

Is |x − z − y| > x − z + y?

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

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  