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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] New post 07 Dec 2012, 07:48
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If x and y are integers, is x > y?

(1) x + y > 0
(2) y^x < 0
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Re: If x and y are integers, is x > y? [#permalink] New post 07 Dec 2012, 07:55
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If x and y are integers, is x > y?

(1) x + y > 0. Given that the sum of two numbers is greater than zero, but we cannot determine which one is greater. Not sufficient.

(2) y^x < 0. This statement implies that y is a negative number. Now, if y=-1 and x=1, then x>y BUT if y=-1 and x=-1, then x=y. Not sufficient.

(1)+(2) Since from (2) we have that y is a negative number, then -y is a positive number. Therefore from (1) we have that x>-y=positive, which means that x is a positive number. So, we have that x=positive>y=negative. Sufficient.

Answer: C.
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Re: If x and y are integers, is x > y? [#permalink] New post 08 Oct 2013, 08:29
Bunuel wrote:
If x and y are integers, is x > y?

(1) x + y > 0. Given that the sum of two numbers is greater than zero, but we cannot determine which one is greater. Not sufficient.

(2) y^x < 0. This statement implies that y is a negative number. Now, if y=-1 and x=1, then x>y BUT if y=-1 and x=-1, then x=y. Not sufficient.

(1)+(2) Since from (2) we have that y is a negative number, then -y is a positive number. Therefore from (1) we have that x>-y=positive, which means that x is a positive number. So, we have that x=positive>y=negative. Sufficient.

Answer: C.



Should the answer not be E ?

My explanation as followa

Case 1 :
-2>-5 Stmt 1 correct

(-5)^(-2) < 0

(1/25) < 0 Stmt 2 correct

-2 > 5 No ( Is x > y ? )

Case 2 :

5 > -2 Stmt 1 correct

(5) ^ (-2 )< 0

(1/25 ) < 0 Stmt 2 correct

5 > 2 Yes ( Is x > y ? )
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Re: If x and y are integers, is x > y? [#permalink] New post 08 Oct 2013, 16:12
shelrod007 wrote:
Bunuel wrote:
If x and y are integers, is x > y?

(1) x + y > 0. Given that the sum of two numbers is greater than zero, but we cannot determine which one is greater. Not sufficient.

(2) y^x < 0. This statement implies that y is a negative number. Now, if y=-1 and x=1, then x>y BUT if y=-1 and x=-1, then x=y. Not sufficient.

(1)+(2) Since from (2) we have that y is a negative number, then -y is a positive number. Therefore from (1) we have that x>-y=positive, which means that x is a positive number. So, we have that x=positive>y=negative. Sufficient.

Answer: C.



Should the answer not be E ?

My explanation as followa

Case 1 :
-2>-5 Stmt 1 correct

(-5)^(-2) < 0

(1/25) < 0 Stmt 2 correct

-2 > 5 No ( Is x > y ? )

Case 2 :

5 > -2 Stmt 1 correct

(5) ^ (-2 )< 0


(1/25 ) < 0 Stmt 2 correct

5 > 2 Yes ( Is x > y ? )


(-5)^(-2)=\(\frac{1}{25}\) also 1/25>0

Remember when you square a negative number you get a positive number.


(5)^-2=\(\frac{1}{25}\), and again \(\frac{1}{25}\)>0 not less

when you have X^-Y, it's written out as \(\frac{1}{(X^Y)}\)
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Re: If x and y are integers, is x > y? [#permalink] New post 26 Oct 2014, 12:04
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Re: If x and y are integers, is x > y? [#permalink] New post 26 Oct 2014, 19:57
Further explaining option c

from 2, we know y is a -ve number and to make x+y > 0 , x has to be positive to make the equation greater than zero .

so c is sufficient
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Re: If x and y are integers, is x > y? [#permalink] New post 28 Nov 2014, 00:21
abhi398 wrote:
Further explaining option c

from 2, we know y is a -ve number and to make x+y > 0 , x has to be positive to make the equation greater than zero .

so c is sufficient



From 2, why we have decuded that y is a -ve number? y^x= 3^-3 is also less than zero and in this case y is positive. Can you please explain?
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Re: If x and y are integers, is x > y? [#permalink] New post 28 Nov 2014, 04:22
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mulhinmjavid wrote:
abhi398 wrote:
Further explaining option c

from 2, we know y is a -ve number and to make x+y > 0 , x has to be positive to make the equation greater than zero .

so c is sufficient



From 2, why we have decuded that y is a -ve number? y^x= 3^-3 is also less than zero and in this case y is positive. Can you please explain?


\(3^{(-3)} = \frac{1}{3^3} = \frac{1}{27} > 0\), not less than 0.
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If x and y are integers, is x > y? [#permalink] New post 20 Dec 2014, 01:09
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Question is x>y ?

(1) x+y>0 we can not determine wether x>y or not, but this statement says that one of the vaiables must be positive to satisfy it --> Not sufficient
(2) Y^x<0 the only way it to be negative is when Y is negative, but X could be also negative (-2^-3 = -1/8) --> Not Sufficient, as we don't know wether x>y

(1)+(2) Statement 1 says that one of the variables must be positive + Statement 2 says that Y is in all cases negative --> if Y is negative X must be positive and is > Y

Answer (C)
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Re: If x and y are integers, is x > y? [#permalink] New post 04 May 2015, 08:47
Would anybody please explain why we ruled out statement 2. I thought the that the only was for y^x<0 is for y to be negative and X to be an odd power. I ruled out the negative values of X thinking that if X is negative then y would be a negative fraction and the question stem says that y is an integer. Why was I wrong here? Please help!
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Re: If x and y are integers, is x > y? [#permalink] New post 04 May 2015, 20:21
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naeln wrote:
Would anybody please explain why we ruled out statement 2. I thought the that the only was for y^x<0 is for y to be negative and X to be an odd power. I ruled out the negative values of X thinking that if X is negative then y would be a negative fraction and the question stem says that y is an integer. Why was I wrong here? Please help!


Hi naeln,

From statement-II, we can deduce that \(y\) is a negative integer but we can't say if \(x\) is a negative or a positive odd integer. Let's evaluate both the cases:

Case-I: \(x\) is positive
If \(x\) is a positive odd integer, then \(y^x < 0\). For example, assuming \(y = -2\) and \(x = 3\) would give \(y^x = -2^3 = -8 < 0\)


Case-II: \(x\) is negative

When \(x\) is negative, \(y^x\) would become a negative fraction and not \(y\) itself.

For example: if \(y = -2\) and \(x = -3\), then \(x\) & \(y\) both are integers and \(y^x = -2^{-3} = \frac{-1}{8} < 0\). Here \(y^x = \frac{-1}{8}\) is a fraction and not \(y\) itself.

Hope its clear why statement-II does not give us a unique answer.

Regards
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Re: If x and y are integers, is x > y? [#permalink] New post 05 May 2015, 10:39
Thank you Harsh for the detailed explanation. I got it now :)
Re: If x and y are integers, is x > y?   [#permalink] 05 May 2015, 10:39
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