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Is x>0? 1) x-3/|x-3|<1 2) 3x<y<2x

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Is x>0? 1) x-3/|x-3|<1 2) 3x<y<2x  [#permalink]

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New post Updated on: 27 Mar 2019, 13:37
1
2
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

50% (02:01) correct 50% (02:09) wrong based on 28 sessions

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Is \(x>0?\)
1) \(\frac{x-3}{|x-3|}<1\)
2) \(3x<2x\)

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Originally posted by Asad on 14 Mar 2019, 08:13.
Last edited by Asad on 27 Mar 2019, 13:37, edited 1 time in total.
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Re: Is x>0? 1) x-3/|x-3|<1 2) 3x<y<2x  [#permalink]

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New post 14 Mar 2019, 09:53
1
In Statement 1, if x-3 were positive, then the absolute value wouldn't change x-3, and (x-3)/|x-3| would equal (x-3)/(x-3) which is 1. Statement 1 tells us that fraction is not equal to 1, so the absolute value must change x-3, and x-3 must be negative, so x-3 < 0, and x < 3. But x can be 2 or -5 so this is not sufficient.

I'm not sure what the "y" is doing in Statement 2, but ignoring it, we have 3x < 2x, which, subtracting 2x on both sides, tells us x < 0, so this is sufficient to give a 'no' answer to the question, and the answer is B.
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Re: Is x>0? 1) x-3/|x-3|<1 2) 3x<y<2x  [#permalink]

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New post 27 Mar 2019, 13:39
IanStewart wrote:
In Statement 1, if x-3 were positive, then the absolute value wouldn't change x-3, and (x-3)/|x-3| would equal (x-3)/(x-3) which is 1. Statement 1 tells us that fraction is not equal to 1, so the absolute value must change x-3, and x-3 must be negative, so x-3 < 0, and x < 3. But x can be 2 or -5 so this is not sufficient.

I'm not sure what the "y" is doing in Statement 2, but ignoring it, we have 3x < 2x, which, subtracting 2x on both sides, tells us x < 0, so this is sufficient to give a 'no' answer to the question, and the answer is B.

removed y from statement 2.
Thanks (with +1) for your explanation...
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Re: Is x>0? 1) x-3/|x-3|<1 2) 3x<y<2x   [#permalink] 27 Mar 2019, 13:39
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