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Is x > y?

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Is x > y? [#permalink]

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New post 19 Jun 2011, 03:51
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A
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  25% (medium)

Question Stats:

71% (00:54) correct 29% (00:52) wrong based on 141 sessions

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Is x > y?

(1) x^2 > y

(2) x – |y| > 0

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/is-x-y-1-x-2 ... 15485.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 04 Dec 2017, 00:00, edited 3 times in total.
Edited the question.

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Re: Is x > y? [#permalink]

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New post 13 Feb 2013, 08:42
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Is x > y?

(1) x^2 > y

(2) x – |y| > 0

[Reveal] Spoiler:
My understtanding: All statement 1 tells us for sure is x is not 0
Statement 2 we can break into two staments

x>y
OR
x>-y

What I dont get is only one of those statements being true makes B the correct choice...so how is B the correct choice when there are two possibilities for a solution?

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Last edited by Bunuel on 13 Feb 2013, 08:50, edited 1 time in total.
RENAMED THE TOPIC.

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Re: Is x > y? [#permalink]

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New post 20 Jun 2011, 22:23
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siddhans wrote:
subhashghosh wrote:
|y| is always positive, but y may or may not be positive. If y < 0, then |y| = -y. Also, x > |y| means x is > a positive number, so x is always positive.

Read this :

math-absolute-value-modulus-86462.html



Ok got it ...but in sudhir's ex

st 2: x> |y|
when Y is postive
x> y example X = 6 y = -5 6>-5
when y is negative
x>-y 6>-(-5) 6>5 ( we cant assume Y to be -7 ...)


How come he says Y is +ve and takes y =-5??? Is there some error?


Yes there is an error
Y has to be 5 and not -5

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Re: Is x > y? [#permalink]

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New post 19 Jun 2011, 06:13
siddhans wrote:
Is x > y?

(1) x2 > y

(2) x – |y| > 0


st1: x^2>y
x can be positive or negative ex:
(-5)^2 > 6 but
-5 is not > 6................. in sufficient

st 2: x> |y|
when Y is postive
x> y example X = 6 y = -5 6>-5
when y is negative
x>-y 6>-(-5) 6>5 ( we cant assume Y to be -7 ...)

hence B

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Re: Is x > y? [#permalink]

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New post 19 Jun 2011, 12:59
sudhir18n wrote:
siddhans wrote:
Is x > y?

(1) x2 > y

(2) x – |y| > 0


st1: x^2>y
x can be positive or negative ex:
(-5)^2 > 6 but
-5 is not > 6................. in sufficient

st 2: x> |y|
when Y is postive
x> y example X = 6 y = -5 6>-5
when y is negative
x>-y 6>-(-5) 6>5 ( we cant assume Y to be -7 ...)

hence B


I did not get it...

St 2 :
when Y is positive
x >|y| ex. x=6 and you are taking y =-5????

Also, when y is -ve

why are you writing
x > -y???

how does mod work in this case?

if its given x > |y|???

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Re: Is x > y? [#permalink]

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New post 20 Jun 2011, 21:34
Can someone please reply? Also, how can we assume x is positive in statement 2?

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Re: Is x > y? [#permalink]

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New post 20 Jun 2011, 21:40
|y| is always positive, but y may or may not be positive. If y < 0, then |y| = -y. Also, x > |y| means x is > a positive number, so x is always positive.

Read this :

math-absolute-value-modulus-86462.html
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Re: Is x > y? [#permalink]

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New post 20 Jun 2011, 22:21
subhashghosh wrote:
|y| is always positive, but y may or may not be positive. If y < 0, then |y| = -y. Also, x > |y| means x is > a positive number, so x is always positive.

Read this :

math-absolute-value-modulus-86462.html



Ok got it ...but in sudhir's ex

st 2: x> |y|
when Y is postive
x> y example X = 6 y = -5 6>-5
when y is negative
x>-y 6>-(-5) 6>5 ( we cant assume Y to be -7 ...)


How come he says Y is +ve and takes y =-5??? Is there some error?

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Re: Is x > y? [#permalink]

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New post 13 Feb 2013, 08:58
manimgoindowndown wrote:
Is x > y?

(1) x^2 > y

(2) x – |y| > 0

[Reveal] Spoiler:
My understtanding: All statement 1 tells us for sure is x is not 0
Statement 2 we can break into two staments

x>y
OR
x>-y

What I dont get is only one of those statements being true makes B the correct choice...so how is B the correct choice when there are two possibilities for a solution?



Yes u got it right way! but u gotta complete it buddy :)

ya definitely S1 is not sufficent

But S2 is a trap.
x>y
OR
x>-y

assume numbers for x and y in this case
let x=4 and y=3 so 4>3
like wise x=4 and y=-3 this also 4>-3

So x>|y| always will be x>y

Hope i helped u :)
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Re: Is x > y? [#permalink]

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New post 15 Feb 2013, 01:35
From F.S 1, we can have x=-5,y=-3. Thus , x^2>y, but x<y. Hence the answer is NO. Now, for x=5,y=3, x^2>y and x>y. Here the answer is a YES. Thus not sufficient.

Fro F.S 2, x-mod(y)>0. Thus, x>mod(y). Whatever be the sign of y, x>y(always). Sufficient.

B.
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Re: Is x > y? [#permalink]

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New post 15 Feb 2013, 02:16
1
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BOOKMARKED
manimgoindowndown wrote:
Is x > y?

(1) x^2 > y

(2) x – |y| > 0

[Reveal] Spoiler:
My understtanding: All statement 1 tells us for sure is x is not 0
Statement 2 we can break into two staments

x>y
OR
x>-y

What I dont get is only one of those statements being true makes B the correct choice...so how is B the correct choice when there are two possibilities for a solution?


From 2, we get that x - |y| is positive.

Since |y| is postive, this means that x is also positive.

So, if "y" is negative, it automatically means that x is greater than y.

If "y" is positive,

x - y still gives a postive number. Hence x is again greater than y.

Answer is hence B.
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Re: Is x > y? [#permalink]

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Re: Is x > y? [#permalink]

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New post 03 Dec 2017, 23:59
siddhans wrote:
Is x > y?

(1) x^2 > y

(2) x – |y| > 0


Is x > y?

(1) x^2 > y:

If x = 1 and y = 0, then the answer will be YES.
If x = -1 and y = 0, then the answer will be NO.
Not sufficient.

(2) x – |y| > 0:

x > |y|. Now, if x is greater than the absolute value of y, then it must be greater than y itself. Sufficient.

Answer: B.

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/is-x-y-1-x-2 ... 15485.html
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Re: Is x > y?   [#permalink] 03 Dec 2017, 23:59
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