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# If x and y are integers, is x > y?

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If x and y are integers, is x > y?  [#permalink]

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27 Jun 2017, 00:39
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95% (hard)

Question Stats:

41% (02:05) correct 59% (01:48) wrong based on 137 sessions

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If x and y are integers, is x > y?

(1) $$y^x < −1$$

(2) $$\frac{x}{|y|} > 1$$

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Re: If x and y are integers, is x > y?  [#permalink]

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27 Jun 2017, 01:07
Bunuel wrote:
If x and y are integers, is x > y?

(1) $$y^x < −1$$

(2) $$\frac{x}{|y|} > 1$$

Statement 1: implies that y<0 and "x" is Odd Positive integer because if x is odd negative integer then in that case y^x = 1/y^x which will be greater than -1. For eg. if x = -3 & y = -2 then (-2)^(-3) = 1/(-2)^3 = -1/8 > -1. Hence "x" is odd positive integer so x>y. Sufficient

Statement 2: x/|y|>1, or x>|Y|. therefore "x" is positive and its magnitude is greater than "y". Hence sufficient.

Option D
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Re: If x and y are integers, is x > y?  [#permalink]

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27 Jun 2017, 01:10
Bunuel wrote:
If x and y are integers, is x > y?

(1) $$y^x < −1$$

(2) $$\frac{x}{|y|} > 1$$

Given : x and y are integers
DS: x > y

Option 1 : y^x < -1 , Since x and y are integers y<-1 and x must be odd and +ve. So x>y

Option 2: x/|y| >1
x>|y|
If y>0, x>y
If y<0 , x>-y i.e. x is +ve and y is -ve . So x>y.

Both options individually solves the problem

So Option D.

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If x and y are integers, is x > y?  [#permalink]

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27 Jun 2017, 01:42
Bunuel wrote:
If x and y are integers, is x > y?

(1) $$y^x < −1$$

(2) $$\frac{x}{|y|} > 1$$

(1)
y^x <-1
x has to be odd integer and positive
odd as the solution is negative
positive because if it is negative

if y=-2 and x = -3
-2^-3 = -1/8 not satisfying the equation
if y=-3 and x = -3
-3^-3 = -1/27 not satisfying the equation
if y = -4 and x=-3
-4^-3 = -1/64 not satisfying the equation
Since x is positive and y is -ve..x>y

(2)x/|y| > 1
Numerator and Denominator will have the same sign for RHS to be true

if y is +ve then x is positive and
x>y
if y is -ve then x will also be negative and since x/y>1
x<y

Hence A

Please correct me if i am wrong
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Luckisnoexcuse

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Re: If x and y are integers, is x > y?  [#permalink]

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14 Mar 2018, 09:17
Luckisnoexcuse wrote:
Bunuel wrote:
If x and y are integers, is x > y?

(1) $$y^x < −1$$

(2) $$\frac{x}{|y|} > 1$$

(1)
y^x <-1
x has to be odd integer and positive
odd as the solution is negative
positive because if it is negative

if y=-2 and x = -3
-2^-3 = -1/8 not satisfying the equation
if y=-3 and x = -3
-3^-3 = -1/27 not satisfying the equation
if y = -4 and x=-3
-4^-3 = -1/64 not satisfying the equation
Since x is positive and y is -ve..x>y

(2)x/|y| > 1
Numerator and Denominator will have the same sign for RHS to be true

if y is +ve then x is positive and
x>y
if y is -ve then x will also be negative and since x/y>1
x<y

Hence A

Please correct me if i am wrong

Hi "Luckisnoexcuse"

Pl note the 'red' portion- 'x' has to be ALWAYS +ve becos denominator being Mod is ALWAYS +ve irrespective of the sign of 'y'.
So when y is -ve or y<0, |y|>0 makes x>0
Thus again x>y from Statement 2 also

Hence 'D' not 'A'

Regards
Dinesh
Re: If x and y are integers, is x > y?   [#permalink] 14 Mar 2018, 09:17
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# If x and y are integers, is x > y?

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