Last visit was: 24 Dec 2024, 11:02 It is currently 24 Dec 2024, 11:02

Close

GRE Prep 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
Your Progress

every week, we’ll send you an estimated GRE 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.

Close

Request Expert Reply

Confirm Cancel
Verbal Expert
Joined: 18 Apr 2015
Posts: 30484
Own Kudos [?]: 36842 [23]
Given Kudos: 26105
Send PM
Most Helpful Community Reply
avatar
Retired Moderator
Joined: 20 Apr 2016
Posts: 1307
Own Kudos [?]: 2281 [9]
Given Kudos: 251
WE:Engineering (Energy and Utilities)
Send PM
General Discussion
avatar
Intern
Intern
Joined: 14 Jun 2018
Posts: 36
Own Kudos [?]: 13 [0]
Given Kudos: 0
Send PM
Verbal Expert
Joined: 18 Apr 2015
Posts: 30484
Own Kudos [?]: 36842 [0]
Given Kudos: 26105
Send PM
Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y > [#permalink]
1
Expert Reply
Questions involving pure number properties and absolute value are really tough.

Start always from what you do know and the stem says to you.

Now, \(y > 0\) and \(xz > 0\), which means that y is positive and x and y can be both positive or negative.

At the same time\(y > x\) or\(y < x\). At this point, pick numbers that satisfy all the conditions you consider.

A) \(x=1,y=5,z=9\)

B)\(y=1,x=2,z=5\)

testing them in all answer choices and you can quickly see the only C is always false.

Hope this helps

Originally posted by Carcass on 08 Jul 2018, 10:17.
Last edited by Carcass on 14 Jul 2018, 00:56, edited 1 time in total.
typo
avatar
Intern
Intern
Joined: 14 Jun 2018
Posts: 36
Own Kudos [?]: 13 [0]
Given Kudos: 0
Send PM
Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y > [#permalink]
Carcass wrote:
Questions involving pure number properties and absolute value are really tough.

Start always from what you do know and the stem says to you.

Now, \(y > 0\) and \(xy > 0\), which means that y is positive and x and y can be both positive or negative.

At the same time\(y > x\) or\(y < x\). At this point, pick numbers that satisfy all the conditions you consider.

A) \(x=1,y=5,z=9\)

B)\(y=1,x=2,z=5\)

testing them in all answer choices and you can quickly see the only C is always false.

Hope this helps


How y > 0 and xy > 0, means that both could be positive or negative? For y>0, it means that x must be positive in order for xy to be greater that 0?
Verbal Expert
Joined: 18 Apr 2015
Posts: 30484
Own Kudos [?]: 36842 [0]
Given Kudos: 26105
Send PM
Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y > [#permalink]
Expert Reply
Sorry I made a mistake. \(xz > 0\) as the stems says.

Regards
avatar
Intern
Intern
Joined: 03 Sep 2018
Posts: 10
Own Kudos [?]: 22 [0]
Given Kudos: 0
Send PM
Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y > [#permalink]
What is the significance of this - |x| < |y + 2| < |z| -in the question ..?
avatar
Manager
Manager
Joined: 19 Mar 2018
Posts: 64
Own Kudos [?]: 38 [0]
Given Kudos: 0
Send PM
Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y > [#permalink]
1
well need to increase speed on this!
if I take y =2 ,x = 1, z = 3 --> the condition is not even upheld
This means y has to be less than 1 but positive
so y =1, x = 2, z = 3 --> so D holds true and so does E. A and B are possible but not a 'must'. Hence a,b,d,e.
Kudos if you like this explanation
Intern
Intern
Joined: 08 Aug 2022
Posts: 49
Own Kudos [?]: 38 [1]
Given Kudos: 98
Send PM
Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y > [#permalink]
1
Assume y=10.
We also know x and z are either both negative or both positive, since xz>0.
And then we have the big inequality given by the problem.

a. y=10, x=11, z=13 --> yes, possible
b. y=10, x=5, z=13 --> yes, possible
c. If x and z are both negative, the large inequality in the problems tells us that that z needs to be further from zero than x, in which case z will be more negative, or less than x. --> no, this option is not possible
d. y=10, x = -11, z = -13 --> yes, possible
e. y=10 --> y+1.5 = 11.5 and y+2 = 12, so we could have x=11.75, z=13 --> yes, possible
User avatar
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 5092
Own Kudos [?]: 76 [0]
Given Kudos: 0
Send PM
Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y > [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep 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.
Prep Club for GRE Bot
Re: Suppose that |x| < |y + 2| < |z|. Suppose further that y > [#permalink]
Moderators:
GRE Instructor
88 posts
GRE Forum Moderator
37 posts
Moderator
1115 posts
GRE Instructor
234 posts

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