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Senior Manager
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If y is an integer, y=|x|+x, is y=0? 1)x<0 2)y<1 [#permalink]
 28 Dec 2006, 00:16
If y is an integer, y=|x|+x, is y=0?
1)x<0
2)y<1
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Director
Joined: 06 Feb 2006
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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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Senior Manager
Joined: 23 Jun 2006
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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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Director
Joined: 30 Nov 2006
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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
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x is negative then y = x - x = 0 .. YES y = 0
statement 1 is sufficient
(2) y<1
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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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Director
Joined: 06 Feb 2006
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Damn, again, i somehow tend to misread this "interger" thing.... yes should be D...
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Senior Manager
Joined: 24 Oct 2006
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You are right! OA is D
Mind works better after a good night's sleep!
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VP
Joined: 25 Jun 2006
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D.
y can be only >= 0. so (2) is actually Suff.
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Senior Manager
Joined: 08 Jun 2006
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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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Senior Manager
Joined: 24 Nov 2006
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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.
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