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Is x negative ?

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Is x negative ?

by Param800 » Thu Nov 08, 2012 3:15 pm
Can someone please show me how to do this question ?

Is x negative ?

(1) 2x > x^2
(2) x< 1

I thought the answer should be E according to the below strategy -

(1) x^2-2x < 0 = x(x-2) <0
So, that means either x < 0 or x < 2
which is insufficient as x could be 1 or any decimal like 1.5 ( as it doesn't say that x is integer )

(2) x <1
clearly x can be 0.5 or any other decimal so insufficient.

But, the answer is A. Can someone explain me ? Thank You
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by cllong7 » Thu Nov 08, 2012 6:22 pm
Here is how I see it:

(1) 2x > x^2

The only number that satisfies this inequality is 1 (2 > 1). A negative number will not satisfy. For example: 2(-1) > (-1)^2 = -2 > 1...which is incorrect. Because the only number that satisfies is 1, then the answer is sufficient.

(2) x could equal 0 or a negative number, therefore this is insufficient.

Hope this helps!
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by GMATGuruNY » Thu Nov 08, 2012 9:46 pm
Param800 wrote:Can someone please show me how to do this question ?

Is x negative ?

(1) 2x > x^2
(2) x< 1

I thought the answer should be E according to the below strategy -

(1) x^2-2x < 0 = x(x-2) <0
So, that means either x < 0 or x < 2
which is insufficient as x could be 1 or any decimal like 1.5 ( as it doesn't say that x is integer )
The portion in red is not quite correct.
Here's the CRITICAL POINT approach:

2x > x²
x² - 2x < 0
x(x-2) < 0.
The CRITICAL POINTS are x=0 and x=2.
These are the only values where x(x-2) = 0.
To determine the range(s) where x(x-2) < 0, test one value to the left and one value to the right of each critical point.

x<0
Plug x=-1 into 2x > x²:
2(-1) > (-1)²
-2 > 1.
Doesn't work.
x<0 is not a viable range.

0<x<2
Plug x=1 into 2x > x²:
2(1) > 1²
2 > 1.
This works.
0<x<2 is a viable range.

x>2
Plug x=3 into 2x > x²:
2(3) > 3²
6 > 9.
Doesn't work.
x>2 is not a viable range.

0<x<2 is the only range that satisfies statement 1 -- sufficient information to determine that x is not negative.

The correct answer is A.
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by Param800 » Fri Nov 09, 2012 1:40 am
Thanks a lot. I totally get it now :)
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