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Is x^2 greater than x ?

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Is x^2 greater than x ?  [#permalink]

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New post 10 Feb 2014, 00:24
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

Is x^2 greater than x ?

(1) x^2 is greater than 1.
(2) x is greater than -1.

Data Sufficiency
Question: 81
Category: Arithmetic; Algebra Exponents; Inequalities
Page: 158
Difficulty: 650


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Re: Is x^2 greater than x ?  [#permalink]

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New post 10 Feb 2014, 00:25
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SOLUTION

Is x^2 greater than x ?

Is \(x^2 > x\)? --> is \(x(x-1)>0\)? --> is \(x\) in the following ranges: \(x<0\) or \(x>1\)?

(1) x^2 is greater than 1 --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Sufficient.

(2) x is greater than -1 --> \(x>-1\). Not sufficient.

Answer: A.


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Re: Is x^2 greater than x ?  [#permalink]

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New post 10 Feb 2014, 13:44
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(A) alone is sufficient.
Since its given x^2 is greater than 1, so assume x^2 as 1.21/1.44/ 1.69/1.96/2.25 etc. Whether the square root is +/- tive, x^2 will always be greater than x given condition (A).

B alone is insufficient by the same logic as 0.25 is less than 0.5 and greater than -0.5. So no certainty whether X^2 greater than x.

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Re: Is x^2 greater than x ?  [#permalink]

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New post 10 Feb 2014, 14:14
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I'm going with (A) too.

(1) x^2 is greater than 1. Any number greater than 1 will be greater when squared (sign doesn't matter here).
(2) x is greater than - 1. Includes decimals, 0 and 1, so insufficient.
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Re: Is x^2 greater than x ?  [#permalink]

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New post 10 Feb 2014, 16:21
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Is x^2 greater than x ?

(1) x^2 is greater than 1.
(2) x is greater than - 1.


\(x^2>x\). So eather \(x>1\) or \(x<0\)

St(1) Say that \(x^2>1\) so \(x>1\). Sufficient.
St(2) Say that \(x>-1\) So x could be 0 or 2. Not sufficient.

ans A
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Re: Is x^2 greater than x ?  [#permalink]

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New post 29 Mar 2014, 22:46
Bunuel wrote:
SOLUTION

Is x^2 greater than x ?

Is \(x^2 > x\)? --> is \(x(x-1)>0\)? --> is \(x\) in the following ranges: \(x<0\) or \(x>1\)?

(1) x^2 is greater than 1 --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Sufficient.

(2) x is greater than -1 --> \(x>-1\). Not sufficient.

Answer: A.


Hi Bunuel,
If the answer choice satisfy any one of the "Or" inequality does it mean its sufficient?
for eg in 1) x > 1 satisifies ( x >1 or x < 0 ) but x < -1 dosent,
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Re: Is x^2 greater than x ?  [#permalink]

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New post 30 Mar 2014, 11:05
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abid1986 wrote:
Bunuel wrote:
SOLUTION

Is x^2 greater than x ?

Is \(x^2 > x\)? --> is \(x(x-1)>0\)? --> is \(x\) in the following ranges: \(x<0\) or \(x>1\)?

(1) x^2 is greater than 1 --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Sufficient.

(2) x is greater than -1 --> \(x>-1\). Not sufficient.

Answer: A.


Hi Bunuel,
If the answer choice satisfy any one of the "Or" inequality does it mean its sufficient?
for eg in 1) x > 1 satisifies ( x >1 or x < 0 ) but x < -1 dosent,


The question asks whether \(x<0\) or \(x>1\):
Attachment:
MSP101751cei440531bcgfbc0000210f0c958eb6ie40.gif
MSP101751cei440531bcgfbc0000210f0c958eb6ie40.gif [ 997 Bytes | Viewed 4599 times ]
So, whether x is in the blue ranges above.

(1) says that \(x<-1\) or \(x>1\). Now, let me ask you a question: is x in the blue ranges???
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Re: Is x^2 greater than x ?  [#permalink]

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New post 17 Mar 2015, 00:10
Bunuel wrote:
\(x^2 > x\)? --> is \(x(x-1)>0\)? --> is \(x\) in the following ranges: \(x<0\) or \(x>1\)?


I understand the answer to this questions. Kindly help me understand how we have moved x from the RHS to the LHS here. X could either be +ve or -ve. If it is -ve, then we will have to flip the sign! Am I missing something here?

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Re: Is x^2 greater than x ?  [#permalink]

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New post 17 Mar 2015, 01:04
joseph0alexander wrote:
Bunuel wrote:
\(x^2 > x\)? --> is \(x(x-1)>0\)? --> is \(x\) in the following ranges: \(x<0\) or \(x>1\)?


I understand the answer to this questions. Kindly help me understand how we have moved x from the RHS to the LHS here. X could either be +ve or -ve. If it is -ve, then we will have to flip the sign! Am I missing something here?

Thank you!


hi ,
you do not have to change the sign..
say x is a negative number.. then that -ive sign is already there in x...
x=-2.... so wherever x is there you can write -2 and the sign will be there..
but if u change the sign in front of x, it will be mistake...
in this case too.. if u make it -x, this will be equal to -(-2)=+2, which would be wrong...
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Re: Is x^2 greater than x ?  [#permalink]

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New post 20 Mar 2015, 02:55
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joseph0alexander wrote:
Bunuel wrote:
\(x^2 > x\)? --> is \(x(x-1)>0\)? --> is \(x\) in the following ranges: \(x<0\) or \(x>1\)?


I understand the answer to this questions. Kindly help me understand how we have moved x from the RHS to the LHS here. X could either be +ve or -ve. If it is -ve, then we will have to flip the sign! Am I missing something here?

Thank you!


Hi Joseph0alexander!
You can move variables from one side to another without changing signs. You only change signs if you multiply or divide by a negative number. In general if you have a problems with inequalities try the following. Find the numbers that turn inequality into zero. The best way to do it is move all variables to the left or to the right (as you like it) and test intervals on a sign. Here we have x(^2)>x. Hence we can move x to the left and have the following: x(^2)-x>0 ==> x(x-1)>0. Two points that turn this inequality into 0 are 0 and 1. So we have three intervals. If x>1 then inequality is positive, if 0<x<1 then negative and if x<0 again positive. Then start checking statements in the same way. Hope it helps
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Is x^2 greater than x ?  [#permalink]

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New post 01 Jun 2015, 19:00
Hi All,

This DS question is based on a couple of Number Property rules; if you know the rules, then you can answer this question with a more logic-based approach. If you don't know the rules, then you can still discover the patterns involved by TESTing VALUES.

We're asked if X^2 is greater than X. This is a YES/NO question.

By doing a little bit of work up-front, we can make dealing with the two Facts easier. We just have to think about what X COULD be and whether that would make X^2 greater than X (or not).

IF... X = ANY negative value....
Then X^2 = positive and X^2 > X. The answer to the question would be YES.

IF.... X = 0 or X = 1
Then X^2 = X. The answer to the question would be NO.

IF.... X = A positive FRACTION...
Then X^2 < X. The answer to the question would be NO.

IF.... X > 1
Then X^2 > X. The answer to the question would be YES.

Fact 1: X^2 is greater than 1

Here, X COULD be any negative LESS than -1 (eg. -2, -3, -4, -1.5, etc.)....and the answer to the question would be YES.
X COULD also be any positive GREATER than 1 (e.g. 2, 3, 4, 1.5, etc.)...and the answer to the question would also be YES.
The answer to the question is ALWAYS YES.
Fact 1 is SUFFICIENT.

Fact 2: X is greater than -1

X COULD be 0...and the answer to the question would be NO.
X COULD also be any positive GREATER than 1 (e.g. 2, 3, 4, 1.5, etc.)...and the answer to the question would be YES.
Fact 2 is INSUFFICIENT

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Re: Is x^2 greater than x?  [#permalink]

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New post 02 Jun 2015, 09:46
Is x^2 greater than x?

For x^2 to be greater than x, x should be greater than 1 (or less than -1) or negative if it's less than 1

(1) x^2 is greater than 1 if x^2 is greater than 1, so is x is also (or a negative integer). x = 2 or - 2, x^2 = 4, for example. Sufficient
(2) x is greater than -1 If x = 2, x^2= 4. The statement is true. But if x = 1/2 for example, x^2 = 1/4 and this is not true. Not sufficient
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Re: Is x^2 greater than x?  [#permalink]

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New post 29 Dec 2015, 23:03
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution

Is x^2>x?
(1) x^2 is greater than 1
(2) x is greater than -1

In case of inequality questions, it is important to note that the conditions are sufficient if the range of the question includes the range of the conditions.
First, we need to modify the original condition and the question.
The original question asks if x^2-x>0. Then, by modifying the question, we can see that the question is asking if x(x-1)>0. This means, essentially, we need to prove if x<0 or 1<x. There is 1 variable (x), and in order to match the number of variables and the number of equations, we need 1 equation. Since the condition 1) and 2) each has 1 equation, there is high chance D is going to be the answer.
In case of the condition 1), it states x^2>1. From this, we can obtain x^2-1>0. Then, (x-1)(x+1)>0. So, x<-1 or 1<x. Since the range of the question includes the range of the condition, the condition is sufficient.
In case of the condition 2), it states x>-1. Since the range of the question includes the range of the condition, the condition is not sufficient. Hence, the correct answer is A.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: Is x^2 greater than x ? &nbs [#permalink] 18 Jul 2018, 04:57
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