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M09-19

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Bunuel
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M09-19 [#permalink] New post  16 Sep 2014, 00:40
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Is \(|x - 5| > 4\)?


(1) \(x^2-4 < 0\)

(2) \(x^2-1 < 0\)
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Bunuel
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Re M09-19 [#permalink] New post  16 Sep 2014, 00:40
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Is \(|x - 5| > 4\)?

Is \(|x-5| > 4\)? Is \(x - 5 < -4\) or \(x - 5 > 4\)?

Is \(x < 1\) or \(x > 9\)?

(1) \(x^2-4 < 0\)"

\(x^2 < 4\);

\(-2 < x < 2\).

We can have an YES answer as well as a NO answer (consider \(x=0\) and \(x=1.5\)). Not sufficient.

(2) \(x^2 - 1 < 0\):

\(x^2 < 1\);

\( -1 < x < 1\).

Therefore, the answer to the question is NO. Sufficient.


Answer: B
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Re: M09-19 [#permalink] New post  31 Aug 2022, 09:36
Bunuel wrote:
Is \(|x - 5| \gt 4\)?


(1) \(x^2-4 \lt 0\)

(2) \(x^2-1 \lt 0\)

-----------------------------

Please guide me where I'm making a mistake.

For an inequality such as this:
|x - 5| > 4?

We have two cases:
Case 1: If x>=0, then
x-5 > 4
x> 9

Case 2: If x<0, then
x-5 < -4
x < 1
But x should be less than 0, right? So this case should be discarded then.
If this happens, then D is my correct answer.

I ain't sure where I'm going wrong in my approach!

Bunuel KarishmaB ScottTargetTestPrep BrentGMATPrepNow
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ScottTargetTestPrep
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Re: M09-19 [#permalink] New post  06 Sep 2022, 12:04
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summerbummer wrote:
Bunuel wrote:
Is \(|x - 5| \gt 4\)?


(1) \(x^2-4 \lt 0\)

(2) \(x^2-1 \lt 0\)

-----------------------------

Please guide me where I'm making a mistake.

For an inequality such as this:
|x - 5| > 4?

We have two cases:
Case 1: If x>=0, then
x-5 > 4
x> 9

Case 2: If x<0, then
x-5 < -4
x < 1
But x should be less than 0, right? So this case should be discarded then.
If this happens, then D is my correct answer.

I ain't sure where I'm going wrong in my approach!

Bunuel KarishmaB ScottTargetTestPrep BrentGMATPrepNow


Recall that |z| = z if z ≥ 0, and |z| = -z if z < 0. If you substitute z = x - 5, then you'll obtain |x - 5| = x - 5 if x - 5 ≥ 0, and |x - 5| = -(x - 5) if x - 5 < 0. When you substitute z = x - 5 for z within the absolute value bars, you need to remember to substitute z = x - 5 in z ≥ 0 and z < 0 as well. Thus, your two cases should be:

Case 1: If x ≥ 5, then

x - 5 > 4
x > 9

Case 2: If x < 5, then

-(x - 5) > 4
-x + 5 > 4
-x > -1
x < 1
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Bunuel
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Re: M09-19 [#permalink] New post  11 Mar 2023, 14:29
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M09-19 [#permalink]
11 Mar 2023, 14:29
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