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Founder  Joined: 18 Apr 2015
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If |2z| – 1 ≥ 2, which of the following graphs could be a nu [#permalink]
Expert's post 00:00

Question Stats: 75% (01:16) correct 24% (00:59) wrong based on 49 sessions

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

Attachment: #GREpracticequestion which of the following graphs could be a number line representing all the possible values of z.jpg [ 35.38 KiB | Viewed 1337 times ]
[Reveal] Spoiler: OA

_________________ Director Joined: 03 Sep 2017
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Kudos [?]: 356  , given: 66

Re: If |2z| – 1 ≥ 2, which of the following graphs could be a nu [#permalink]
1
KUDOS
The equation is solved as

for x >= 0, $$2z-1\geq 2$$ i.e. $$z\geq \frac{3}{2}$$

for x <= 0, $$-2z-1\geq 2$$ i.e. $$z\leq -\frac{3}{2}$$

Thus, the value of x are $$x\leq -\frac{3}{2} and x \geq \frac{3}{2}$$.

The line number is A because 3/2 is greater than 1 and -3/2 is less than -1 Intern Joined: 12 Oct 2017
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Kudos [?]: 12  , given: 3

Re: If |2z| – 1 ≥ 2, which of the following graphs could be a nu [#permalink]
1
KUDOS
Solutions :-

There Can be multiple ways to think over this problem but whenever we are finding answer from the given choices try to eliminate the wrong choice instead of finding the right one ( this is very useful in most of the cases )

In above problem, from the given choices we need to check the value for z=1 or -1 as for both |z| = 1

for z=1
equation is "1>=2"
which is not correct
so |z| should be greater than 1
or z should be greater than 1 or less than -1

if we check the choices, Choice "B,C,D,E" is eliminated as all are wrong so we are left with only Choice "A" GRE Instructor Joined: 10 Apr 2015
Posts: 1968
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Kudos [?]: 1797  , given: 9

Re: If |2z| – 1 ≥ 2, which of the following graphs could be a nu [#permalink]
1
KUDOS
Expert's post
Carcass wrote:

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

Brent Hanneson – Creator of greenlighttestprep.com  Re: If |2z| – 1 ≥ 2, which of the following graphs could be a nu   [#permalink] 12 Oct 2017, 15:17
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