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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.

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.

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.
_________________

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

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21 Feb 2012, 22:07

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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 ?

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

Notice that since \(y=|x|+x\) then \(y\) is never negative. If \(x>{0}\) (so if x is positive) then \(y=x+x=2x\) and for \(x\leq{0}\) then (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.

Bunuel,why didn't i get it with the way i solved it? Am unable to understand where my approach is wrong . Thanks

You forgot that it's given that y=integer. (2) says y<1, thus |x|+x must also be some integer less than 1: 0, -1, ... but if you refer to my solution you'll see that |x|+x can never be negative, so the only valid solution for |x|+x (or which is the same for y) is 0.

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

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07 Jul 2013, 19:30

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

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

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07 Feb 2014, 17:23

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).

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

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07 Sep 2014, 02:25

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.

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

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19 Jan 2015, 01:14

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

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

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08 Oct 2016, 03:35

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 \(y=-x+x=0\).

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 \(y=-x+x=0\).

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

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08 Oct 2016, 04:00

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 \(y=-x+x=0\).

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.
_________________

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

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08 Oct 2016, 04:42

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.

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

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08 Oct 2016, 04:51

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.

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

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10 Nov 2016, 16:38

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.