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M60-17

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M60-17  [#permalink]

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New post Updated on: 30 Sep 2018, 04:58
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

69% (00:25) correct 31% (00:35) wrong based on 26 sessions

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Is \(x^2-x>0?\)

1) \(x>0\)

2) \(x^3+x>0\)

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Originally posted by MathRevolution on 11 Jun 2018, 06:35.
Last edited by Bunuel on 30 Sep 2018, 04:58, edited 7 times in total.
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Re M60-17  [#permalink]

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New post Updated on: 30 Sep 2018, 04:58
1
Official Solution:


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.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question. 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

Modifying the question:

\(x^2-x>0\)

\(x(x-1) > 0\)

\(x < 0\) or \(x > 1\) by the "LLGG" rule.

Condition 1): \(x > 0\)

Since the solution set of the question does not include the solution set of condition 1), condition 1) is not sufficient.

Condition 2):

\(x^3+x>0\)

\(x(x^2+1)>0\)

\(x>0\), since \(x^2+1 > 0\) is always true.

Condition 2) is equivalent to the condition 1), so it is not sufficient.

Conditions 1) & 2)

We have x >0 only from both conditions together.

Since the solution set of the question does not include the solution set of conditions 1) and 2) together, they are not sufficient.

Therefore, the answer is E.

Answer: E

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.


Answer: E
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Originally posted by MathRevolution on 11 Jun 2018, 06:35.
Last edited by Bunuel on 30 Sep 2018, 04:58, edited 3 times in total.
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Re: M60-17  [#permalink]

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New post 26 Jul 2018, 18:35
I don't agree with the explanation. I believe this question is mistaking one of the solutions x < 0, for the first condition x >0. 1/2 would satisfy condition one but result in NO, but 2 would result in yes.
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Re: M60-17  [#permalink]

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New post 26 Jul 2018, 19:19
4
Is \(x^2-x>0?\)
\(x^2-x>0..........X(x-1)>0\)..
Two cases..
a) if \(x>0\), then \(x-1>0....X>1\)
b) if \(x<0....x-1<0....x<1\)
Hence question is asking
Is x<0 and x>1 or is it false that \(0\leq{X}\leq{1}\)
1) \(x>0\)
0<X<1....no
X>1....yes
Insufficient

2) \(x^3+x>0\)
\(x(x^2+1)>0\).
\(x^2+1>0\), so \(x>0\)..
Again \(0<x<1\)....no
\(x>1\)....yes
Insufficient

Combined..
Same cases remain
Insufficient

E

OA and solution is wrong
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Re: M60-17  [#permalink]

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New post 26 Jul 2018, 20:12
MathRevolution wrote:
Is \(x^2-x>0?\)

1) \(x>0\)

2) \(x^3+x>0\)



To determine whether \(x^2 - x > 0\)

=> \(x * (x - 1) > 0\)

Statement 1

x > 0

x can be 9 => x(x-1) = 9*8 = 72 > 0

x can be 0.5 => x(x-1) = 0.5 * -0.5 = -0.25 < 0

Statement 1 is not sufficient

Statement 2

\(x^3 + x > 0\)

x can be 2 => \(x^3 + x > 0\) = 8 + 2 = 10 > 0

x can be -2 => \(x^3 + x > 0\) = -8 - 2 = -10 < 0

Statement 2 is not sufficient

Combining statements 1 and 2

x can be 2 => \(x^3 + x > 0\) = 8 + 2 = 10 > 0

x can be 0.5 => x(x-1) = 0.5 * -0.5 = -0.25 < 0

Statements 1 and 2 together are not sufficient

Hence option E

MathRevolution Can you please look into this ?
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Re: M60-17  [#permalink]

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New post 26 Jul 2018, 21:38
The Q is basically asking to conclude that: x is NOT a proper fracton, positive or negative?

None of the two statements can confirm or dismiss this conclusively. Hence, Ans is E.

Posted from my mobile device
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Re: M60-17  [#permalink]

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New post 27 Jul 2018, 11:25
I don't agree with the explanation.
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Re: M60-17  [#permalink]

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New post 30 Jul 2018, 05:35
I think this is a high-quality question and I don't agree with the explanation. In case of first statement, if x lies between 0 and 1, lets say x=1/2, then x2-x<0, while if x>1, then x2-x>0, hence A is not sufficient.
The same goes for B if x lies in the range of 0 to 1
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Re: M60-17  [#permalink]

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New post 30 Jul 2018, 11:30
The explaination is incorrect. The correct answer should be Both the conditions are insufficient for x>0 gives us no information whether x>1 or not.
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Re: M60-17  [#permalink]

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New post 05 Aug 2018, 13:19
THE ANS SHOULD BE - E
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Re: M60-17  [#permalink]

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New post 21 Aug 2018, 17:22
I think this is a high-quality question and I don't agree with the explanation. Hello! I believe the correct answer to this solution is E, not D.

A clear counterexample here is x = 1/2. This satisfies statement 1, but it does not satisfy x^2 - x > 0.

Alternatively, the question could be clarified such that x must be an integer.
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Re: M60-17  [#permalink]

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New post 22 Aug 2018, 23:27
I think this the explanation isn't clear enough, please elaborate. if 0<x<1
then x^2-x <0, but x>0 holds and x^3+x >0 holds
if x>1
then x^2-x >0, but x>0 holds and x^3+x >0 holds
i am unable to understand the explanation, where am i going wrong
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Re: M60-17  [#permalink]

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New post 26 Aug 2018, 04:49
I don't agree with the explanation. The answer posted here does not seem correct. If x = 0.5, the expression is not greater than 0 whereas if x = 10, the expression is greater than 0 so the answer should be E, shouldn't it?
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Re: M60-17  [#permalink]

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New post 27 Aug 2018, 10:10
I think this is a high-quality question and I don't agree with the explanation. pls check. answer should be E because both statements given translate to x>0 but the question only has clear solutions if x>1 or x<0. Hence even both statements together arent sufficient
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Re: M60-17  [#permalink]

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New post 28 Aug 2018, 11:00
I think this is a poor-quality question and I don't agree with the explanation. Hello, it seems there is an issue. Would like it if you elaborated on it:
We need either x larger than 1 or x smaller than 0

In condition 1, x is larger than zero, but it could be between 1 and 0 so it should be insufficient
similarly for the second condition.
Please let me know if I am wrong
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Re: M60-17  [#permalink]

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New post 28 Aug 2018, 13:10
I think this is a high-quality question and I don't agree with the explanation. the solution is wrong.
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Re: M60-17  [#permalink]

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New post 31 Aug 2018, 03:47
The solution is not correct for this problem. The correct answer would be E, proving this is as simple as substituting x with 2 which satisfies all equations and the condition in the posted question. Then, substitute x with 1 in both equations, which satisfies both conditions since the result is greater than 0 but substituting x for 1 in the original equation will yield 0 as a result and hence does not satify that condition.
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Re: M60-17  [#permalink]

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New post 01 Sep 2018, 01:00
I think this is a high-quality question and I don't agree with the explanation. it's clearly e
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Re: M60-17  [#permalink]

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New post 02 Sep 2018, 08:13
I think this is a poor-quality question and I don't agree with the explanation. as many have mentioned this solution is wrong and E should be the right choice just try substituting 2 and 1/2
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Re: M60-17  [#permalink]

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New post 05 Sep 2018, 20:16
1
I think this is a high-quality question and I don't agree with the explanation. I do not agree with solution that x>0 is sufficient to judge x2-x>0.

1. for any value of x >0 and x<1, solution x2-x<0
2. for any value of x>1, solution x2-x>0
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Re: M60-17 &nbs [#permalink] 05 Sep 2018, 20:16

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