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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 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 ?
[Reveal] Spoiler: OA
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Re: If y is an integer and y = |x| + x, is y = 0? (1) x < 0 [#permalink] New post 21 Feb 2012, 22:19
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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.

Answer: D.

For hard inequality and absolute value questions with detailed solutions check this: inequality-and-absolute-value-questions-from-my-collection-86939-40.html

Hope it helps.
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Manager
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Re: If y is an integer and y = |x| + x, is y = 0? (1) x < 0 [#permalink] New post 21 Feb 2012, 22:43
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? (1) x < 0 [#permalink] New post 21 Feb 2012, 23:35
devinawilliam83 wrote:
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.

Hope it's clear.
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is y = 0? [#permalink] New post 21 Mar 2012, 15:07
If y is an integer and y = |x| + x, is y = 0?

(1) x < 0
(2) y < 1
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Re: is y = 0? [#permalink] New post 21 Mar 2012, 15:10
Re: is y = 0?   [#permalink] 21 Mar 2012, 15:10
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