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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Anurag@Gurome
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Given: y = |x| + xRezinka wrote:If y is an integer and y = |x| + x, is y=0?
(1) x<0
(2) y<1
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 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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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!
Regards,
Gagan
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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
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
[/i]Anurag@Gurome wrote:Given: y = |x| + xRezinka wrote:If y is an integer and y = |x| + x, is y=0?
(1) x<0
(2) y<1
Therefore,Statement 1: x < 0 => Sufficient
- 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 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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How do we know that x > 0 in this case?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.
...
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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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
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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sushantgupta
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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.
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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Sanjay2706
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prashant misra
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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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mourinhogmat1
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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.
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.
