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M60-17

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fatchimp
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Re: M60-17 [#permalink] New post   13 Sep 2018, 22:53
This is high-quality question with wrong answer and poor solution... I believe it should be E too
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librahoanganh
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Re: M60-17 [#permalink] New post   15 Sep 2018, 01:01
I think this is a poor-quality question and I don't agree with the explanation. x>0: if x=5 then x^2-x>0, if x=1/2 then x^2-x<0 NOT SUFFICIENT
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Re: M60-17 [#permalink] New post   15 Sep 2018, 11:56
I think this is a poor-quality question and I don't agree with the explanation. When X>0 then we can't say X^2 -X >0
Eg: X=1/2 then The above equation results in -1/2
and X=2 then the above equation results in 2.

Hence in decisive
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Re: M60-17 [#permalink] New post   16 Sep 2018, 09:37
The solution should be E.

For statement 1, if x=1, then x^2-x = 0 and NOT greater than 0. If x=2, x^2-x > 0. Insufficient.

Can refute statement 2 on the same lines.

Please check the OE and OA.
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vb1991
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Re: M60-17 [#permalink] New post   Updated on: 18 Sep 2018, 06:21
This is a poor quality question and the OA is wrong.

The correct answer is E, as many people have stated.

Originally posted by vb1991 on 17 Sep 2018, 06:18.
Last edited by vb1991 on 18 Sep 2018, 06:21, edited 1 time in total.
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vb1991
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Re: M60-17 [#permalink] New post   Updated on: 18 Sep 2018, 06:21
I think this is a poor-quality question and I don't agree with the explanation. The correct answer is E. Many people have asked for a clarification here, yet there has been no response from the team at Math Revolution. I was considering buying your product - however, given the poor quality of explanation and lack of response on the forum I have changed my mind.

Originally posted by vb1991 on 17 Sep 2018, 06:20.
Last edited by vb1991 on 18 Sep 2018, 06:21, edited 1 time in total.
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maarroger
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Re: M60-17 [#permalink] New post   18 Sep 2018, 06:00
I think this is a poor-quality question and I don't agree with the explanation. the solution is wrong?
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annievu
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Re: M60-17 [#permalink] New post   21 Sep 2018, 09:29
I think this is a poor-quality question and I don't agree with the explanation. I go with E
as x^2-x>0 only if x<0 or x>1
1) x>0 we are not sure if 0<x<1 or x>1 -->insufficient
2) x^3+x>0, we have x>0, the same case as (1) ---> insufficient
Hence E
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Re: M60-17 [#permalink] New post   22 Sep 2018, 05:42
I think this is a high-quality question and I don't agree with the explanation. Can somebody help with the explanation to this question.??
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Re: M60-17 [#permalink] New post   25 Sep 2018, 12:05
I think this the explanation isn't clear enough, please elaborate. what about 0<x<1
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Re: M60-17 [#permalink] New post   29 Sep 2018, 19:13
I don't agree with the explanation. The answer has been corrected in the forum but hasn't been updated. COMMON! don't waste time!
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Re M60-17 [#permalink] New post   18 Jan 2019, 12:15
I think this is a high-quality question and I don't agree with the explanation.
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Re M60-17 [#permalink] New post   16 Apr 2019, 07:52
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. Kindly elaborate on the "LLGG" rule, plz.
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Re: M60-17 [#permalink] New post   02 Jul 2019, 03:04
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x^2−x>0
---> x(x - 1) > 0???
Illustrating on the number line, we have: +ve 0 -ve 1 +ve

ST1: x > 0
According to the number line, the inequality will be positive when x > 1, and negative when 0 < x < 1.
--->NS

ST2: x^3 + x > 0
-->x( x^2 +1) > 0
( x^2 +1) > 0 is obvious since even exponents always have positive values, we need to check x >0
Since it's the same as the 1st statement---> NS
ST1 + ST2: still unable to solve the question---->The answer is E.
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Re: M60-17 [#permalink] New post   15 Jan 2020, 09:03
I'm not familiar with the "LLGG" rule.

How do we simplify the question to x<0 or x>1?



I had simplified it to x(x-1)>0.

Had it been x(x-1)=0, I know that the roots are x=0,1. But with inequalities, I'm a bit confused.
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Re: M60-17 [#permalink]
15 Jan 2020, 09:03
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