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# If x is an integer, is x|x| < 2x? (1) x < 0 (2) x =

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If x is an integer, is x|x| < 2x? (1) x < 0 (2) x = [#permalink]

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25 Jun 2008, 09:07
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If x is an integer, is x|x| < 2x?

(1) x < 0
(2) x = -10

The OA is D. Statement 2 is obviously sufficient. I don't understand how statement 1 is sufficient?

Take the example of x = -2.

-2(2) = 2(-2) - therefore x|x| = 2x

Take the example of x = -3.

-3(3) < 2(-3) - therefore x|x| < 2x
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25 Jun 2008, 09:36
terp06 wrote:
If x is an integer, is x|x| < 2x?

(1) x < 0
(2) x = -10

The OA is D. Statement 2 is obviously sufficient. I don't understand how statement 1 is sufficient?

Take the example of x = -2.

-2(2) = 2(-2) - therefore x|x| = 2x

Take the example of x = -3.

-3(3) < 2(-3) - therefore x|x| < 2x

You did all the work

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25 Jun 2008, 09:42
brokerbevo wrote:
terp06 wrote:
If x is an integer, is x|x| < 2x?

(1) x < 0
(2) x = -10

The OA is D. Statement 2 is obviously sufficient. I don't understand how statement 1 is sufficient?

Take the example of x = -2.

-2(2) = 2(-2) - therefore x|x| = 2x

Take the example of x = -3.

-3(3) < 2(-3) - therefore x|x| < 2x

You did all the work

With the work I did above, I thought the answer would be B? I found 2 negative integers for statement 1, 1 which satisfies the condition and one which does not, meaning that statement 1 should be insufficient?
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25 Jun 2008, 09:50
terp06 wrote:
If x is an integer, is x|x| < 2x?

(1) x < 0
(2) x = -10

The OA is D. Statement 2 is obviously sufficient. I don't understand how statement 1 is sufficient?

Take the example of x = -2.

-2(2) = 2(-2) - therefore x|x| = 2x

Take the example of x = -3.

-3(3) < 2(-3) - therefore x|x| < 2x

As you said statement 1 is not sufficient.

x = -1 -> x*abs(x) = -1 > 2*(-1)
x = -3 -> x*abs(x) = -9 < -6 = 2*(-3)

D is incorrect, the answer has to be B.
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25 Jun 2008, 10:02
terp06 wrote:
brokerbevo wrote:
terp06 wrote:
If x is an integer, is x|x| < 2x?

(1) x < 0
(2) x = -10

The OA is D. Statement 2 is obviously sufficient. I don't understand how statement 1 is sufficient?

Take the example of x = -2.

-2(2) = 2(-2) - therefore x|x| = 2x

Take the example of x = -3.

-3(3) < 2(-3) - therefore x|x| < 2x

You did all the work

With the work I did above, I thought the answer would be B? I found 2 negative integers for statement 1, 1 which satisfies the condition and one which does not, meaning that statement 1 should be insufficient?

Yes, maybe I should read your entire post. Sorry about that. Yes, the answer should be B Where did you find this problem?
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25 Jun 2008, 10:03
This problem is in the OG11 and the OA is D.
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25 Jun 2008, 10:09
terp06 wrote:
This problem is in the OG11 and the OA is D.

Hmm. What is the explanation in the back of the book for how they arrived at D? Because even moreover, if you try -1 the inequality is > and if you try -3 the inequality is < and if you try -2, it is an equality --> Its all over the charts!!
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25 Jun 2008, 10:46
terp06 wrote:
If x is an integer, is x|x| < 2x?

(1) x < 0
(2) x = -10

The OA is D. Statement 2 is obviously sufficient. I don't understand how statement 1 is sufficient?

You wrote the question incorrectly. It makes a huge difference.

The question is:

If x is an integer, is x |x| < 2^x?

(1) x < 0
(2) x = -10

Statement one, plug in (-1,-2)

-1 > 1/2 yes
-4 > 1/4 yes

Sufficient

Statement 2

-20 < -1/2^10

Sufficient

Last edited by 2010mba on 25 Jun 2008, 11:02, edited 1 time in total.
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25 Jun 2008, 11:33
2010mba wrote:
terp06 wrote:
If x is an integer, is x|x| < 2x?

(1) x < 0
(2) x = -10

The OA is D. Statement 2 is obviously sufficient. I don't understand how statement 1 is sufficient?

You wrote the question incorrectly. It makes a huge difference.

The question is:

If x is an integer, is x |x| < 2^x?

(1) x < 0
(2) x = -10

Statement one, plug in (-1,-2)

-1 > 1/2 yes
-4 > 1/4 yes

Sufficient

Statement 2

-20 < -1/2^10

Sufficient

In the book, it is written as 2x.
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26 Jun 2008, 09:21
terp06 wrote:
In the book, it is written as 2x.

The x is small, and half way up the 2, meaning it is an exponent. I hope that helps. If it were 2x, then the x would be on same line as the 2.
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26 Jun 2008, 10:08
2010mba wrote:
The question is:

If x is an integer, is x |x| < 2^x?

(1) x < 0
(2) x = -10

Statement one, plug in (-1,-2)

-1 > 1/2 yes
-4 > 1/4 yes

Sufficient

Is this a proof (1) is sufficient ?? You just test it for 2 values (why -1 and -2 by the way ?) and therefore it is sufficient ? Are you sure it works for x=-1/8 ?

I would say something like: if x<0, then x |x| <0

And since 2^x is positive, then we have x |x| < 2^x (this is no longer that plugging numbers )
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27 Jun 2008, 13:47
2010mba wrote:
terp06 wrote:
In the book, it is written as 2x.

The x is small, and half way up the 2, meaning it is an exponent. I hope that helps. If it were 2x, then the x would be on same line as the 2.

My understanding is that this is printed incorrectly in earlier printings of the Guide, and was later corrected. In some books, it is actually printed as 2x, and not as 2^x, in the question (though not in the solution section of the book).
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29 Jun 2008, 13:42
2010mba wrote:
It says x is an integer.

Yeah, okay...

So (if you prefer): did you try x= -4679? x= -27?
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01 Jul 2008, 02:24
Oski wrote:
2010mba wrote:
It says x is an integer.

Yeah, okay...

So (if you prefer): did you try x= -4679? x= -27?

I think 2^-27 will still be a positive value !!
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01 Jul 2008, 02:36
saurabhkowley18 wrote:
I think 2^-27 will still be a positive value !!

I know.

But the reason for that IS NOT that it is the case for x=-1 and x=-2
Re: OG DS 128   [#permalink] 01 Jul 2008, 02:36
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