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Is integer x greater than integer y?

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Is integer x greater than integer y? [#permalink]

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New post 12 Sep 2017, 08:02
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Is integer x greater than integer y? [#permalink]

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New post 12 Sep 2017, 08:43
Bunuel wrote:
Is integer x greater than integer y?

(1) x^2 + 2xy + y^2 = 0
(2) x^2 – 2xy + y^2 = 0


Hi Bunuel,

Need some clarity here. From statement 1 it is clear that \(x = -y\) and from statement 2 it is clear that \(x=y\). But in a DS question aren't the two statements supposed to lead to the same value of \(x\) & \(y\)?
Here both the statements are individually sufficient to answer the question stem, but ideally \(x=-y\) and \(x=y\) is only possible when we combine these two to get \(y=0\) & \(x=0\)

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Re: Is integer x greater than integer y? [#permalink]

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New post 12 Sep 2017, 08:48
niks18 wrote:
Bunuel wrote:
Is integer x greater than integer y?

(1) x^2 + 2xy + y^2 = 0
(2) x^2 – 2xy + y^2 = 0


Hi Bunuel,

Need some clarity here. From statement 1 it is clear that \(x = -y\) and from statement 2 it is clear that \(x=y\). But in a DS question aren't the two statements supposed to lead to the same value of \(x\) & \(y\)?
Here both the statements are individually sufficient to answer the question stem, but ideally \(x=-y\) and \(x=y\) is only possible when we combine these two to get \(y=0\) & \(x=0\)


Hints:
1. The statements do not contradict.
2. The answer is not D.
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Re: Is integer x greater than integer y? [#permalink]

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New post 12 Sep 2017, 08:49
Find : is x>y ?

(1) x^2 + 2xy + y^2 = 0
=> (x+y)^2 = 0
=> x+y=0
=> x=-y
Now we dont know if y is negative or positive
if y= +ve , x is -ve and x<y
if y =-ve , x is +ve and x>y
Not sufficient

(2) x^2 – 2xy + y^2 = 0
=> (x-y)^2=0
=> x-y=0
=> x=y
So This implies x is 'not' greater to y but is equal to y.
Sufficient

Answer: B

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Re: Is integer x greater than integer y? [#permalink]

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New post 12 Sep 2017, 09:00
Bunuel wrote:
niks18 wrote:
Bunuel wrote:
Is integer x greater than integer y?

(1) x^2 + 2xy + y^2 = 0
(2) x^2 – 2xy + y^2 = 0


Hi Bunuel,

Need some clarity here. From statement 1 it is clear that \(x = -y\) and from statement 2 it is clear that \(x=y\). But in a DS question aren't the two statements supposed to lead to the same value of \(x\) & \(y\)?
Here both the statements are individually sufficient to answer the question stem, but ideally \(x=-y\) and \(x=y\) is only possible when we combine these two to get \(y=0\) & \(x=0\)


Hints:
1. The statements do not contradict.
2. The answer is not D.


Thanks got it. I mis-read the question thought that it is asking for the value of x ;)

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Re: Is integer x greater than integer y? [#permalink]

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New post 12 Sep 2017, 09:25
(1) x^2 + 2xy + y^2 = 0
(x+y)^2=0
Two different cases can satisfy above equation
Case I: x=y=0
Case II: x=-y
Hence We can not firmly predict whether x is greater than y. Statement 1 is insufficient.

(2) x^2 – 2xy + y^2 = 0
(x-y)^2=0
x=y
as x and y are equal; we can firmly say that x is not greater than y.
Hence statement II is sufficient.

Answer is Option B.

Kudos if it helps.

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Re: Is integer x greater than integer y? [#permalink]

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New post 12 Sep 2017, 12:43
1) (x+y)^2= 0
x=-y

Can be same or can't be.
x=-2 and y=2
or
x= 0 and y=0 (same value)


2) (x-y)^2= 0
x=y

Answer is B
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Re: Is integer x greater than integer y?   [#permalink] 12 Sep 2017, 12:43
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