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# Is 5 > x > 0 ?

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Manager
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Is 5 > x > 0 ?  [#permalink]

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Updated on: 23 Feb 2014, 05:17
3
00:00

Difficulty:

65% (hard)

Question Stats:

55% (01:30) correct 45% (01:35) wrong based on 127 sessions

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Is 5 > x > 0 ?

(1) x is an integer greater than 4
(2) 5x^2 - 25x > 0

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The proof of understanding is the ability to explain it.

Originally posted by GMATD11 on 01 Mar 2011, 00:18.
Last edited by Bunuel on 23 Feb 2014, 05:17, edited 3 times in total.
Edited the question
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Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
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01 Mar 2011, 00:21
1) is Sufficient. Easy

2) rephrased as x(x-5) > 0

hence x < 0 or x > 5. Sufficient. Even if you draw the number line the region 0 < x < 5 is excluded. Hence the answer is NO.

D it is.
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01 Mar 2011, 02:24
GMATD11 - your workings are correct but conclusion is not. For statement B, you have proven that x is either less than 0 or more than 5, so we can certainly say that x does not lie between 0 and 5 and this is sufficient to answer the question is 5>x>0, hence sufficient and answer is D.

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01 Mar 2011, 02:36
GMATD11 wrote:
a) x>4

so x does not lie between 0 to 5. Sufficient

b) 5(x^2-5*x)>0

x(x-5)>0
Both x and (x-5) ill have the same sign

x>0 and x-5>0 OR x<0 and x-5<0

x>0 nd x>5 OR x<0 nd x<5

may lie may not lie ---NS

A

Is 5 > x > 0 ?

(1) x is an integer greater than 4 --> x could be 5, 6, 7, ... in any case x is not in the range 0<x<5, so the answer to the question is NO. Sufficient.

(2) 5x^2 - 25x > 0 --> x^2-5x>0 --> x(x-5)>0 -> x<0 or x>5, again x is not in the range 0<x<5, so the answer to the question is NO. Sufficient.

For solving inequalities like x^2-5x>0 check:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
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09 Mar 2011, 11:14
Bunuel, out of curiosity how would you graph (x^2) - 5x?

I understand how you did it here (x2-4x-94661.html#p731476), but graphing it doesn't really seem to be working for this equation.
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09 Mar 2011, 11:32
1
m990540 wrote:
Bunuel, out of curiosity how would you graph (x^2) - 5x?

I understand how you did it here (x2-4x-94661.html#p731476), but graphing it doesn't really seem to be working for this equation.

x^2-5x=0:
Attachment:

MSP216419ef1hab97a0ab1200005h6b9b331di3gh21.gif [ 3.64 KiB | Viewed 2516 times ]
As you can see x^2-5x>0 for x<0 and x>5.
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09 Mar 2011, 11:36
Thank you! That makes it much easier to see the answer.
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09 Mar 2011, 11:38
m990540 wrote:
Bunuel, out of curiosity how would you graph (x^2) - 5x?

I understand how you did it here (x2-4x-94661.html#p731476), but graphing it doesn't really seem to be working for this equation.

You don't really need the ability to draw the actual graph - You just need to know that an equation of this form with likely multiple roots would change signs at those roots.

Lets look at$$x^2-5x$$

This can be rewritten as $$x*(x-5)$$, so points of interest are $$x = 0$$ and $$x=5$$ (where this function equals to zero). So, it can be deduced that this function will change a sign at $$x=0$$ and then again at$$x=5$$.

Whats the sign when$$x<0$$. Lets put$$x=-1$$ and see, it is +ve , so between 0 and 5, function will be negative and again for $$x > 5$$, it will be positive.

So, without drawing exact graph, you can estimate that for x<0, the curve would be in 2nd quadrant (x is -ve and y is positive), between 0 and 5 it will be 4th quadrant (x is positive and y is negative) and again after x=5, it will be 1st quadrant (both x and y are positive)
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09 Mar 2011, 11:44
beyondgmatscore wrote:
m990540 wrote:
Bunuel, out of curiosity how would you graph (x^2) - 5x?

I understand how you did it here (x2-4x-94661.html#p731476), but graphing it doesn't really seem to be working for this equation.

You don't really need the ability to draw the actual graph - You just need to know that an equation of this form with likely multiple roots would change signs at those roots.

Lets look at$$x^2-5x$$

This can be rewritten as $$x*(x-5)$$, so points of interest are $$x = 0$$ and $$x=5$$ (where this function equals to zero). So, it can be deduced that this function will change a sign at $$x=0$$ and then again at$$x=5$$.

Whats the sign when$$x<0$$. Lets put$$x=-1$$ and see, it is +ve , so between 0 and 5, function will be negative and again for $$x > 5$$, it will be positive.

So, without drawing exact graph, you can estimate that for x<0, the curve would be in 2nd quadrant (x is -ve and y is positive), between 0 and 5 it will be 4th quadrant (x is positive and y is negative) and again after x=5, it will be 1st quadrant (both x and y are positive)

All that is explained here: x2-4x-94661.html#p731476
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Re: Is 5 > x > 0 ?  [#permalink]

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23 Feb 2014, 05:18
Bumping for review and further discussion.
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Re: Is 5 > x > 0 ?  [#permalink]

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13 Mar 2016, 07:48
We need to check the range here
Statement 1 is sufficient to say that x will never be in the range (0,5)
and Statement 2 too .
hence D
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Re: Is 5 > x > 0 ?  [#permalink]

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28 Nov 2018, 21:41
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Re: Is 5 > x > 0 ?   [#permalink] 28 Nov 2018, 21:41
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