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If y is an integer and y=|x|+x, is y=0?

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If y is an integer and y=|x|+x, is y=0? [#permalink] New post 14 Dec 2010, 21:03
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If y is an integer and y=|x|+x, is y=0?

(1) x<0
(2) y<1
[Reveal] Spoiler: OA
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Re: tricky q from GMATPrep2 [#permalink] New post 15 Dec 2010, 00:50
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Fijisurf wrote:
I am sure it is posted somewhere on the forum already , I just can't find it.


If y is an integer and y=|x|+x, is y=0?

(1)x<0
(2)y<1


If y is an integer and y = |x| + x, is y = 0?
(1) x < 0
(2) y < 1

Note: as y=|x|+x then y is never negative. If x>{0} then y=x+x=2x>0 and if x\leq{0} (when x is negative or zero) then y=-x+x=0.

(1) x<0 --> y=|x|+x=-x+x=0. Sufficient.

(2) y<1, as we concluded y is never negative, and we are given that y is an integer, hence y=0. Sufficient.

Answer: D.

Also discussed in Inequality and absolute value questions from my collection: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope it's clear.
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Re: tricky q from GMATPrep2 [#permalink] New post 14 Dec 2010, 21:39
Each of (1) and (2) are sufficient to answer the question. Thus the answer should be "D".
From(1), any value of X<0, whether it's integer or fraction will lead to Y = 0 because the mode function will result in positive value and positive value added to same negative value will result in Zero.
From (2), We can have Y as negative Integer ( -1,-2....) if and only if the Right hand side of equation is different. It's just adding the same X and it can't lead to the negative value.

Yes it can definitely lead to the Zero.

Thus Answer should be D.

You can plug and play certain values ( numbers ) in order to verify the above.
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Re: tricky q from GMATPrep2 [#permalink] New post 14 Dec 2010, 22:07
I got it. thanks
I just forgot that "y" is an integer. For some reason I thought that "y" can be 0<y<1.
Thanks. I need to be more careful.
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Re: tricky q from GMATPrep2 [#permalink] New post 15 Dec 2010, 06:33
I orginially had this as A, because I didn't read the question thoroughly to see y was an integer.

just goes to show how easy it is to miss something if you are going too fast.
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Re: tricky q from GMATPrep2 [#permalink] New post 16 Dec 2010, 00:33
tfincham86 wrote:
I orginially had this as A, because I didn't read the question thoroughly to see y was an integer.

just goes to show how easy it is to miss something if you are going too fast.

Same mistake... It kills me !!!!!!!!! :beat
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Re: tricky q from GMATPrep2 [#permalink] New post 16 Dec 2010, 05:51
Expert's post
Fijisurf wrote:
I am sure it is posted somewhere on the forum already , I just can't find it.


If y is an integer and y=|x|+x, is y=0?

(1)x<0
(2)y<1


A word of caution: When you read "If y is an integer and y=|x|+x", analyze it there and then. y can be a positive integer when x is positive, y will be 0 when x is negative and y will be 0 when x is 0."
Another important point to note here: When the author puts in extra effort to write "y is an integer" rather than "x and y are integers" , take special note that x may not be an integer. Not that it matters very much here but in many questions such a statement will have special significance.
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please explain, why not D why only B [#permalink] New post 25 Feb 2011, 07:26
Q19:
If y is an integer and y = !x! + x, is y = 0?
(1) x < 0
(2) y < 1
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Re: please explain, why not D why only B [#permalink] New post 25 Feb 2011, 08:02
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Re: If y is an integer and y=|x|+x, is y=0? [#permalink] New post 06 Feb 2014, 03:45
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Re: If y is an integer and y=|x|+x, is y=0? [#permalink] New post 07 Feb 2014, 16:23
If y is an integer and y = |x| + x, is y = 0?
y = 0, when x < 0; y = 2x, when x >=0
(1) x < 0 => y = 0
(2) y < 1 -> Because y is an integer, y has to be zero (y cannot be a negative integer because the least value of y is zero).

D
Re: If y is an integer and y=|x|+x, is y=0?   [#permalink] 07 Feb 2014, 16:23
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