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

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New post 28 Dec 2006, 00:16
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

1)x<0
2)y<1
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New post 28 Dec 2006, 01:51
I must pick A....

If x is less than 0 then you take any value and you will get y=0

x=-0.5
0.5-0.5=0

x=-1.8
1.8-1.8=0.....
So on.....

B is obviously not enougth....
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New post 28 Dec 2006, 03:06
in fact, st2 is sufficient as well so answer is D.

st1 is sufficient as noted above:
if x<0 then |x|=-x
so y=|x|+x = -x+x = 0
suff.

st2 is sufficient as well:
from stem only we can deduce that y is non-negative:
if x<0 then y=0 as we showed above,
if x=0 then obviously y=0.
if x>0 then |x|=x and y = 2x > 0

also note that y is an integer (!!!!!!!)

st2 says y<1. the only non-negative integer less than 1 is 0.
so y must be 0.

hence sufficient.


answer is D.
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New post 28 Dec 2006, 04:18
I'd go with D

Given: y is integer, y = |x| + x
asked : does y = 0 ?


from the question, 0=< y =< infinity; that is, y can range between zero and any positive integer.

because |x| is always positive.
if x is positive then y = 2x
if x is negative then y = 0


(1) x<0
--------------
x is negative then y = x - x = 0 .. YES y = 0

statement 1 is sufficient

(2) y<1
--------------
the only possible value for y < 1 is zero because y can never be negative and y is an integer so it can't be some fraction less than 1 like 1/2 or so.

So y in this case has to equal zero [ y = 0 ]

statement 2 is sufficient


Answer is : D
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New post 28 Dec 2006, 04:58
Damn, again, i somehow tend to misread this "interger" thing.... yes should be D...
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New post 28 Dec 2006, 07:57
You are right! OA is D

Mind works better after a good night's sleep!
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New post 28 Dec 2006, 22:22
D.

y can be only >= 0. so (2) is actually Suff.
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New post 30 Dec 2006, 19:22
D for me too

From 1) Obvious

From 2)
Possible values of Y is 0 or negative..but negative is not possible because y = |x| + x

|x| + x could be either positive or 0.
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New post 31 Dec 2006, 00:52
y=|x|+x = {0 for x<0; 0 for x=0; 2x for x>0}

Both (1) and (2) are suff to arrive at a single solution, then D.
  [#permalink] 31 Dec 2006, 00:52
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