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

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SVP
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If y is an integer and y = |x| + x, is y = 0? (1) x < 0 [#permalink]

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New post 27 Oct 2008, 10:51
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D
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If y is an integer and y = |x| + x, is y = 0?

(1) x < 0

(2) y < 1




Please explain your answer.
Thanks

Kudos [?]: 1008 [0], given: 1

Director
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Joined: 23 May 2008
Posts: 801

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Re: DS: absolute value [#permalink]

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New post 27 Oct 2008, 11:11
1)if x less than 0 then lxl + x has to be 0 ..suff
2) if y<1 then then lxl + x is less than one, which means -tve numbers since y is an integer so same results as 1)
..suff

D

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SVP
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Joined: 29 Aug 2007
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Kudos [?]: 843 [0], given: 19

Re: DS: absolute value [#permalink]

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New post 27 Oct 2008, 11:12
tarek99 wrote:
If y is an integer and y = |x| + x, is y = 0?

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

Please explain your answer.
Thanks


Did not you get D? :wink:
Yeah, it should be D.

St 1 is sufficient - no need to expalin.
St 2 is also sufficient - since Y is an integer and <1, Y must be 0 cuz at any cost Y cannot be -ve.

so D.
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Kudos [?]: 843 [0], given: 19

SVP
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Kudos [?]: 1008 [0], given: 1

Re: DS: absolute value [#permalink]

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New post 27 Oct 2008, 11:33
GMAT TIGER wrote:
tarek99 wrote:
If y is an integer and y = |x| + x, is y = 0?

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

Please explain your answer.
Thanks


Did not you get D? :wink:
Yeah, it should be D.

St 1 is sufficient - no need to expalin.
St 2 is also sufficient - since Y is an integer and <1, Y must be 0 cuz at any cost Y cannot be -ve.

so D.



OA is D. I got A when I worked on this problem. Stupid mistake :lol:

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Re: DS: absolute value [#permalink]

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New post 27 Oct 2008, 13:03
key is Y is an integer..so if y<1 then the only option is that its 0..

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Re: DS: absolute value [#permalink]

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New post 27 Oct 2008, 17:32
tarek99 wrote:
If y is an integer and y = |x| + x, is y = 0?

(1) x < 0

(2) y < 1

Please explain your answer.
Thanks


(1) x < 0. Pick any simple -ve number for x. say x = -2

now y = |-2| + (-2) = 2-2 = 0. Statement 1 is sufficient.

(2) y < 1. y could be 0,-1,-2,-3 .........

but we have an extra info in the stem, y = |x| + x. if y<1 then |x| + x < 1 or |x| + x </= 0............ [</= --- lesser than or equal to]

|x| </= -x .... now this can happen only when x </= 0

so y has to be 0.. Statement 2 is sufficient.

D.
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Kudos [?]: 708 [0], given: 1

Re: DS: absolute value   [#permalink] 27 Oct 2008, 17:32
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