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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]

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

(1) x<0
(2) y<1
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Re: tricky q from GMATPrep2 [#permalink]

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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]

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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]

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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]

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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]

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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]

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New post 16 Dec 2010, 05:51
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]

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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]

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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]

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Re: If y is an integer and y=|x|+x, is y=0? [#permalink]

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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
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Re: If y is an integer and y=|x|+x, is y=0? [#permalink]

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Re: If y is an integer and y=|x|+x, is y=0? [#permalink]

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New post 08 Oct 2016, 02:35
Bunuel wrote:
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 \(y=-x+x=0\).

please help!
(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.



the only place i'm stuck is
if \(x\leq{0}\) (when x is negative or zero) then \(y=-x+x=0\)

if x is negative why arent we taking the other x as negative:\(y=-x-x=-2x\)
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Re: If y is an integer and y=|x|+x, is y=0? [#permalink]

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New post 08 Oct 2016, 02:48
nishantdoshi wrote:
Bunuel wrote:
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 \(y=-x+x=0\).

please help!
(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.



the only place i'm stuck is
if \(x\leq{0}\) (when x is negative or zero) then \(y=-x+x=0\)

if x is negative why arent we taking the other x as negative:\(y=-x-x=-2x\)


Knowing that x is a negative number does not mean that you should replace it with -x, this just does not make any sense.
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Re: If y is an integer and y=|x|+x, is y=0? [#permalink]

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New post 08 Oct 2016, 03:00
nishantdoshi wrote:
Bunuel wrote:
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 \(y=-x+x=0\).

please help!
(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.



the only place i'm stuck is
if \(x\leq{0}\) (when x is negative or zero) then \(y=-x+x=0\)

if x is negative why arent we taking the other x as negative:\(y=-x-x=-2x\)


No, you are missing something here.

We are given y = |x| + x, take the value of x as -2 and see the result.

Note that |x| is always positive,

When we say |x| = -x, we mean x will hold a negative value which when multiplied by -ve sign before will give a positive result. NEVER EVER take the values of mod like the way you have taken. I would suggest go through the mod concepts again.
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Re: If y is an integer and y=|x|+x, is y=0? [#permalink]

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New post 08 Oct 2016, 03:42
hey t hanks for the reply

When we say |x| = -x, we mean x will hold a negative value which when multiplied by -ve sign before will give a positive result. NEVER EVER take the values of mod like the way you have taken. I would suggest go through the mod concepts again.

i couldnt understand the above sentence

but my point is when we take x as -2,we get, y=|-2|-2 => y=-2-2 just like we do in the "Critical Points method" (if we get -ve value inside the mod we mult. the mod with the -ve sign.)

please give me detailed explanation if i'm wrong.please.
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If y is an integer and y=|x|+x, is y=0? [#permalink]

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New post 08 Oct 2016, 03:51
nishantdoshi wrote:
hey t hanks for the reply

When we say |x| = -x, we mean x will hold a negative value which when multiplied by -ve sign before will give a positive result. NEVER EVER take the values of mod like the way you have taken. I would suggest go through the mod concepts again.

i couldnt understand the above sentence

but my point is when we take x as -2,we get, y=|-2|-2 => y=-2-2 just like we do in the "Critical Points method" (if we get -ve value inside the mod we mult. the mod with the -ve sign.)

please give me detailed explanation if i'm wrong.please.


No Dude, your reasoning is 100% incorrect. As I said above |x| is always positive, so |-2| is always 2. Note that MOD means MAGNITUTE irrespective of sign.

Now, I can confidently say you are not aware of Mod concept used in mathematics.

Please go through the below link and try solving as many questions as you could.

http://magoosh.com/gmat/2012/gmat-math- ... te-values/
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Re: If y is an integer and y=|x|+x, is y=0? [#permalink]

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New post 08 Oct 2016, 04:00
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abhimahna wrote:
nishantdoshi wrote:
hey t hanks for the reply

When we say |x| = -x, we mean x will hold a negative value which when multiplied by -ve sign before will give a positive result. NEVER EVER take the values of mod like the way you have taken. I would suggest go through the mod concepts again.

i couldnt understand the above sentence

but my point is when we take x as -2,we get, y=|-2|-2 => y=-2-2 just like we do in the "Critical Points method" (if we get -ve value inside the mod we mult. the mod with the -ve sign.)

please give me detailed explanation if i'm wrong.please.


No Dude, your reasoning is 100% incorrect. As I said above |x| is always positive, so |-2| is always 2. Note that MOD means MAGNITUTE irrespective of sign.

Now, I can confidently say you are not aware of Mod concept used in mathematics.

Please go through the below link and try solving as many questions as you could.

http://magoosh.com/gmat/2012/gmat-math- ... te-values/


thanks a lot. and can i pm you for doubts in other topics?
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Re: If y is an integer and y=|x|+x, is y=0? [#permalink]

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New post 08 Oct 2016, 04:16
Break into 2 cases
a) x>0
then
y = 2x,
and y >=1 as y is an integer

b) x<= 0
then
y = 0

therefore
1) is sufficient as for x <0, y is 0
2) is sufficient as only possible value of y <1 is 0, coming from the case b).

Hence, an is D
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Re: If y is an integer and y=|x|+x, is y=0? [#permalink]

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New post 10 Nov 2016, 15:38
Fijisurf wrote:
If y is an integer and y=|x|+x, is y=0?

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



haha, what a classic trap!
i knew it when i saw 50% correct rate...
1. sufficient. x is negative, therefore y=0
2. y<1. sufficient. y must be zero. y can't be a decimal, since we are given the fact that y is an integer.

answer is D.
Re: If y is an integer and y=|x|+x, is y=0?   [#permalink] 10 Nov 2016, 15:38
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