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Is X > 1 ?

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Is X > 1 ?  [#permalink]

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New post 24 Aug 2019, 07:22
1
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

75% (00:56) correct 25% (00:52) wrong based on 56 sessions

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Is X > 1 ?

1. 1/x > 1
2. x > 0
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Re: Is X > 1 ?  [#permalink]

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New post 24 Aug 2019, 07:40
GloryBoy92 wrote:
Is X > 1 ?

1. 1/x > 1
2. x > 0


(1) \(\frac{1}{x}\)>1
Multiply x to both the sides
\(\frac{x}{x}\)>1x
x<1 .................x can never be greater than 1.

SUFFICIENT!

(2) x>0
x can be \(\frac{1}{2}\), \(\frac{2}{3}\) or any number greater than 1. x>0 does not imply that the number must be greater than 1.

INSUFFICIENT!

IMO answer is option A

Kindly recheck the question and OA. I am sure either of them is incorrect!
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Re: Is X > 1 ?  [#permalink]

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New post 24 Aug 2019, 09:09
Yes, using Statement 1 alone, if we know 1/x > 1, we can be completely certain that x is positive (since that inequality is never true if x is negative), so we can safely multiply by x on both sides without reversing the inequality to find that 1 > x. So Statement 1 is certainly sufficient alone; it must be true that 0 < x < 1. We don't need Statement 2, and the answer is A. The OA provided is C, which is not correct.
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Re: Is X > 1 ?  [#permalink]

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New post 24 Aug 2019, 12:17
GloryBoy92 wrote:
Is X > 1 ?

1. 1/x > 1
2. x > 0


Asked: Is X > 1 ?

1. 1/x > 1
x<1
x is NOT >1
SUFFICIENT

2. x > 0
If x=.1 -> x<1
But if x=2>1
NOT SUFFICIENT

IMO A

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Re: Is X > 1 ?  [#permalink]

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New post 24 Aug 2019, 13:22
GloryBoy92 wrote:
Is X > 1 ?

1. 1/x > 1
2. x > 0


#1
1>x
sufficient to say that x>1 not possible
#2
x>0 x can be fraction or >1 insufficient
IMO A
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Is X > 1 ?  [#permalink]

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New post 25 Aug 2019, 09:50
GloryBoy92 wrote:
Is X > 1 ?

1. 1/x > 1
2. x > 0


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have 1 variable (\(x\)) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
We should simplify a condition if necessary.

\(\frac{1}{x} > 1\)
\(⇔ x > x^2\) by multiplying by x^2 since x^2 is positive
\(⇔ x^2 - x < 0\)
\(⇔ x(x-1) < 0\)
\(⇔ 0 < x < 1\)

Since 'no' is also a unique answer by CMT (Common Mistake Type) 1, condition 1) is sufficient, when used together.

Condition 2)

In inequality questions, the law “Question is King” tells us that if the solution set of the question includes the solution set of the condition, then the condition is sufficient.

However, since the solution set \(x > 1\) does not include the solution set \(x > 0\) of condition 2, condition 2) is not sufficient.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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Re: Is X > 1 ?  [#permalink]

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New post 25 Aug 2019, 11:26
MathRevolution wrote:
GloryBoy92 wrote:
Is X > 1 ?

1. 1/x > 1
2. x > 0


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have 1 variable (\(x\)) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
We should simplify a condition if necessary.

\(\frac{1}{x} > 1\)
\(⇔ x > x^2\) by multiplying by x^2 since x^2 is positive
\(⇔ x^2 - x < 0\)
\(⇔ x(x-1) < 0\)
\(⇔ 0 < x < 1\)

Since 'no' is also a unique answer by CMT (Common Mistake Type) 1, condition 1) is sufficient, when used together.

Condition 2)

In inequality questions, the law “Question is King” tells us that if the solution set of the question includes the solution set of the condition, then the condition is sufficient.

However, since the solution set \(x > 1\) does not include the solution set \(x > 0\) of condition 2, condition 2) is not sufficient.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.


Thanks for the answer, I'm curious as to why the lecturer in your video gave a different answer
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Re: Is X > 1 ?  [#permalink]

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New post 26 Aug 2019, 01:20
IanStewart wrote:
Yes, using Statement 1 alone, if we know 1/x > 1, we can be completely certain that x is positive (since that inequality is never true if x is negative), so we can safely multiply by x on both sides without reversing the inequality to find that 1 > x. So Statement 1 is certainly sufficient alone; it must be true that 0 < x < 1. We don't need Statement 2, and the answer is A. The OA provided is C, which is not correct.


Edited the OA. C is the answer to the similar question also from Math Revolution: https://gmatclub.com/forum/is-x-1-1-1-x ... 84240.html
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Re: Is X > 1 ?   [#permalink] 26 Aug 2019, 01:20
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