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If y is an integer and y = x + x, is y = 0? [#permalink]
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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
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Re: If y is an integer and y = x + x, is y = 0? [#permalink]
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14 Dec 2010, 21:39
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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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If y is an integer and y = x + x, is y = 0? [#permalink]
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15 Dec 2010, 00:50
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Re: If y is an integer and y = x + x, is y = 0? [#permalink]
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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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Re: If y is an integer and y = x + x, is y = 0? [#permalink]
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21 Feb 2012, 21:07
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If y is an integer and y = x + x, is y = 0?
(1) x < 0 (2) y < 1
I rephrased the original question as Is x<0? Statement 1 : SF Statement II : if y<1; x+x<1..on solving we get 2 ranges for x  X<0 or,  0<X<0.5 Basis this II is insufficient.. Where am I going wrong ?



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If y is an integer and y = x + x, is y = 0? [#permalink]
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21 Feb 2012, 21:19
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Re: If y is an integer and y = x + x, is y = 0? [#permalink]
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21 Feb 2012, 21:43
Bunuel,why didn't i get it with the way i solved it? Am unable to understand where my approach is wrong . Thanks



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Re: If y is an integer and y = x + x, is y = 0? [#permalink]
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21 Feb 2012, 22:35



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Re: If y is an integer and y = x + x, is y = 0? [#permalink]
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04 Jul 2013, 00:24



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Re: If y is an integer and y = x + x, is y = 0? [#permalink]
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07 Jul 2013, 18:30
If y is an integer and y = x + x, is y = 0?
(1) x < 0
y = x + x y=x+x y=0 SUFFICIENT
(2) y < 1 TRICKYYYYYY I originally said it was insufficient because it tells us nothing about the sign. However, if y is less than one and is an integer and it is equal to x+x then it must be zero!! SUFFICIENT
(D)



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Re: If y is an integer and y = x + x, is y = 0? [#permalink]
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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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07 Sep 2014, 01:25
devinawilliam83 wrote: If y is an integer and y = x + x, is y = 0?
(1) x < 0 (2) y < 1
I rephrased the original question as Is x<0? Statement 1 : SF Statement II : if y<1; x+x<1..on solving we get 2 ranges for x  X<0 or,  0<X<0.5 Basis this II is insufficient.. Where am I going wrong ? point to remember is y is an integer if y<1 , then y could be 0, 1, 2 and so on .. here according to the second condition if x= 0 then y= 0 , if x = ve then Y = 0 , if x = +ive like 1 then second choice only will fail, if x = 0.5 then y cant be less than one, if x=.2 or .3 then y cant be integer.



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Re: If y is an integer and y = x + x, is y = 0? [#permalink]
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19 Jan 2015, 00:14
stat 1: X<0 => y = x+x =0 suff stat 2:y is in integer & y<1 => y<=0 ...y can be less than zero only when x is less than zero => y = x+x = 0 and for y=0...ntn to calculate suff => Ans D



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Re: If y is an integer and y = x + x, is y = 0? [#permalink]
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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: inequalityandabsolutevaluequestionsfrommycollection86939.htmlHope 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:\(y=xx=2x\)



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Re: If y is an integer and y = x + x, is y = 0? [#permalink]
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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: inequalityandabsolutevaluequestionsfrommycollection86939.htmlHope 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:\(y=xx=2x\) 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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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: inequalityandabsolutevaluequestionsfrommycollection86939.htmlHope 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:\(y=xx=2x\) 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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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=22 => y=22 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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Re: If y is an integer and y = x + x, is y = 0? [#permalink]
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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=22 => y=22 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/gmatmath ... tevalues/
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Re: If y is an integer and y = x + x, is y = 0? [#permalink]
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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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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?
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