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Modulus

by Rezinka » Thu Dec 30, 2010 12:21 am
If y is an integer and y = |x| + x, is y=0?
(1) x<0
(2) y<1

OA : D
IMO : A

[spoiler]I don't think B gives me a definite value because the question only mentions y is an integer.
Case 1 : x = 0; So y = 0......... SUFFICIENT
Case 2 : x = 0.4; So y = 0.8... INSUFFICIENT[/spoiler]
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by Anurag@Gurome » Thu Dec 30, 2010 12:43 am
Rezinka wrote:If y is an integer and y = |x| + x, is y=0?
(1) x<0
(2) y<1
Given: y = |x| + x
Therefore,
  • 1. If x = 0 => |x| = 0 => y = 0
    2. If x > 0 => |x| = x => y = (x + x) = 2x > 0
    3. If x < 0 => |x| = -x => y = (-x + x) = 0
Statement 1: x < 0 => Sufficient

Statement 2: y < 1
Note that from above analysis y can never be negative.
Now there is only one non-negative integer less than 1, which is zero.
Thus y = 0, because y is declared as an integer.

Sufficient

The correct answer is D.

Note: You cannot take take x = 0.4 as that doesn't satisfies the fact that y is an integer. You should choose the value of x accordingly.
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by Rezinka » Thu Dec 30, 2010 12:44 am
Missed that..!!

Thanks :)
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by Gags » Thu Dec 30, 2010 11:32 am
Hello Anurag,
Could you please explain how |x| + x = -x + x. Shouldnt it be -x + ( -x ) = -x - x = -2x i.e y < 0.

the 3rd condition says that x < 0 then can we add a positive x and get the value of y as '0'.

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by Anurag@Gurome » Thu Dec 30, 2010 12:11 pm
Gags wrote:Hello Anurag,
Could you please explain how |x| + x = -x + x. Shouldnt it be -x + ( -x ) = -x - x = -2x i.e y < 0. |x| = -x for x < 0. But x is always equal to x whatever be the value of x. What is the logic behind x should be -x?

the 3rd condition says that x < 0 then can we add a positive x and get the value of y as '0'. If x is less than zero, that means x is negative. There is no negative x which can be positive!

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by Gags » Thu Dec 30, 2010 1:49 pm
Thanks for the explanation. I guess i got it. However, when you wrote y = -X + X , i went in to a wrong direction ..
actually the value of |X| = -X .. so if X < 0 then -x = - (-x) i.e the negative value will be finally turned in to positive and the value of x, and not |x| , in y = |x| + x will be be - x ,
So y = x + (-x)

I hope my understanding is correct now !

Thanks,
Gagan
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by ragz » Thu Dec 30, 2010 2:06 pm
Anurag@Gurome wrote:
Rezinka wrote:If y is an integer and y = |x| + x, is y=0?
(1) x<0
(2) y<1
Given: y = |x| + x
Therefore,
  • 1. If x = 0 => |x| = 0 => y = 0
    2. If x > 0 => |x| = x => y = (x + x) = 2x > 0
    3. If x < 0 => |x| = -x => y = (-x + x) = 0
Statement 1: x < 0 => Sufficient

Statement 2: y < 1
Note that from above analysis y can never be negative.
Now there is only one non-negative integer less than 1, which is zero.
Thus y = 0, because y is declared as an integer.

But, based on your derivations from the questions, we are trying to solve for #2 which is y=2x. But, we also know since x>0 in this case, 2x>0 and hence y>0. So, doesn't that mean x cannot be 0.

Sufficient

The correct answer is D.

Note: You cannot take take x = 0.4 as that doesn't satisfies the fact that y is an integer. You should choose the value of x accordingly.
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by Anurag@Gurome » Thu Dec 30, 2010 9:22 pm
ragz wrote:...
But, based on your derivations from the questions, we are trying to solve for #2 which is y=2x. But, we also know since x>0 in this case, 2x>0 and hence y>0. So, doesn't that mean x cannot be 0.
...
How do we know that x > 0 in this case?
Statement 2 only says y is less than 1 and nothing about x.
From the question stem we know that only possible values of y are 0 (when x ≤ 0) and 2x (when x > 0) and y is also an integer. Now statement 2 says y is less than 1, which leaves us with only one possible value of y which is zero. And y = 0 is possible for any x ≤ 0.
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by tgou008 » Wed Apr 27, 2011 7:14 am
I thought this was pretty straightforward, but think if you moved through it too quickly you would have picked A over D.

This took me just over a minute by my thought process was as follows:

y = [x] + x AND y = integer
Does y = 0?
I know from the above equation that y will equal zero if x is <0 OR if x=0

(1) x<0. Sufficient; as I point out in the previous sentence y will = 0 if x < 0 (this holds, even if X isn't an integer)

(2) y < 1

I thought about this two ways
a) y = 0. If y = 0 then X was either negative or zero. Either way this is sufficient to answer question
b) But what if y = -1. This is actually not possible due to the fact that one of the x values is an absolute, therefore Y can only ever be positive or zero.
SUFFICIENT

Correct answer is D
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by sushantgupta » Mon Jul 04, 2011 11:06 am
statement 1 says x < 0

hence |x|+ x = 0
hence statement 1 is sufficient.

Statement 2 says y < 1
which means |x| + x < 1 had x been an integer statement 2 wouls have been sufficient but there is no such constraint on x so x can be anything less then 1/2 so value of y is not fixed.


So statement 1 alone is sufficient.
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by Sanjay2706 » Mon Jul 04, 2011 8:46 pm
Yes D is spot on.:)
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by pankajks2010 » Thu Jul 07, 2011 8:24 am
amazing question..totally missed it, till the time saw Anurag's explanation..great!!
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by mirantdon » Thu Jul 07, 2011 9:21 am
+1 for D .
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by prashant misra » Wed Nov 09, 2011 11:53 pm
oh i misread the second statement and thought that y can take any value but it was given that y is an integar value thus plugging in numbers in both statements also gives us an integar value so the answer is option D
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by mourinhogmat1 » Wed Feb 08, 2012 6:13 pm
I believe the approach needs to be different for this question.

Given:
y = |x|+x

If x is positive --> y = 2x
If x is negative --> y = 0
If x is 0 --> y = 0

(1) x<0
Using second equation we know y = 0. SUFFICIENT.

(2) y<1
Taking first equation if y<1, 2x<1 --> x<1/2. So, when x is 0, -1, etc. y = 0 SUFFICIENT.

ANSWER IS D.
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