Last visit was: 14 Nov 2024, 09:15 It is currently 14 Nov 2024, 09:15

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: 29958
Own Kudos [?]: 36219 [3]
Given Kudos: 25903
Send PM
avatar
Director
Director
Joined: 03 Sep 2017
Posts: 518
Own Kudos [?]: 702 [0]
Given Kudos: 0
Send PM
avatar
Intern
Intern
Joined: 12 Oct 2017
Posts: 23
Own Kudos [?]: 19 [0]
Given Kudos: 0
Send PM
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12188 [0]
Given Kudos: 136
Send PM
Re: If |2z| – 1 ≥ 2, which of the following graphs could be a nu [#permalink]
1
Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project



If |2z| – 1 ≥ 2, which of the following graphs could be a number line representing all the possible values of z?

Image


Another approach is to TEST SOME values.

When I scan the answer choices, I see that some have 0 as part of the solution, and others don't.
So, let's see what happens when z = 0

Plug z = 0 into original equation to get: |2(0)| – 1 ≥ 2
Evaluate to get: |0| – 1 ≥ 2
We get: 0 – 1 ≥ 2
And then: -1 ≥ 2
NOT TRUE.
So, z = 0 is NOT a solution.
Therefore, we'll ELIMINATE D and E, since they include z=0 as a solution.

Among the three remaining answer choices, two have have 1 as part of the solution, and one doesn't.
So, let's see what happens when z = 1
Plug z = 1 into original equation to get: |2(1)| – 1 ≥ 2
Evaluate to get: |2| – 1 ≥ 2
We get: 2 – 1 ≥ 2
And then: 1 ≥ 2
NOT TRUE.
So, z = 1 is NOT a solution.
Therefore, we'll ELIMINATE B and C, since they include z=1 as a solution.

By the process of elimination, we're left with A, the correct answer.

Cheers,
Brent
avatar
Intern
Intern
Joined: 02 Jul 2020
Posts: 26
Own Kudos [?]: 22 [1]
Given Kudos: 11
Send PM
Re: If |2z| 1 2, which of the following graphs could be a nu [#permalink]
1
Well, let´s see:
[2z] = 2z ou -2z, right?

Therefore,

2z -1 >= 2
z>= 3/2

OU

-2z -1 >= 2
-2z >= 3 multiplied by(-1)
2z <= -3
z <= -3/2

Option A
Piece of cake!
User avatar
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 5014
Own Kudos [?]: 74 [0]
Given Kudos: 0
Send PM
Re: If |2z| 1 2, which of the following graphs could be a nu [#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: If |2z| 1 2, which of the following graphs could be a nu [#permalink]
Moderators:
GRE Instructor
78 posts
GRE Forum Moderator
37 posts
Moderator
1111 posts
GRE Instructor
234 posts

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