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Joined: 04 Mar 2018
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Re: If x and y are integers, is x > y?
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07 Jun 2018, 19:54
I'll try to give the simplest solution I could think of:

1. \(x + y > 0\)
Therefore: \(x > y\)
Example: if x = 3 and y = 2 , this is satisfied. This also satisfied x > y (that is, 3 > 2.) Counterexample: if x = 1 and y = 1, this is satisfied. However, this violates the x > y equation. We have found both an example and a counterexample  therefore, 1 by itself is insufficient.

2. \(y^x < 0\)
The only way this is true is if x is odd and y is negative.
But there's an infinite number of possibilities where x is odd and y is negative. Therefore, 2 by itself is insufficient.

What about equation 1 and equation 2 combined?
If we know y is negative, then the equation in 1:
\(x > y\)
becomes
\(x > y\) (does that look familiar?)
Therefore, by definition, both 1 and 2 together are necessary to solve the problem.



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Re: If x and y are integers, is x > y?
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22 Aug 2018, 10:05
Walkabout wrote: If x and y are integers, is x > y?
(1) x + y > 0 (2) y^x < 0 combining two statements Y must be negative and \(x + y > 0\) means that \(y^x < 0\) could be following values \(3^5\), \(3^9\) etc X will always be greater



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Re: If x and y are integers, is x > y?
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22 Aug 2018, 14:52
The interesting issue here may be explicitly presented below: \({y^x} < 0\,\,\,;\,\,\,x + y > 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,x > y\) To prove that, let´s suppose, on the contrary, that we could have SIMULTANEOUSLY the following: (1) \({y^x} < 0\) (2) \(x + y > 0\) (3) \(x \leqslant y\) The contradiction follows: \(\left( 1 \right)\,\,\,\, \Rightarrow \,\,\,\,\,y < 0\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( 3 \right)} \,\,\,x < 0\,\,\,\,\,\,\, \Rightarrow \,\,\,x + y < 0\,\,\,\,\, \Rightarrow \,\,\,\,\left( 2 \right)\,\,{\text{contradicted}}\)
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Re: If x and y are integers, is x > y?
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16 Sep 2018, 11:12
Hi, here are my two cents for this questions Question is IS x>y or we can rephrase this as IS xy>0 Lets understand in which conditions is xy>0 x>0,y>0, and x>y x<0,y<0, and y>x x>0,y<0, and x>y x<0,y>0, this case is not possibleSo Statement A: x+y>0 we can see following cases x>0, y>0 x<0, y>0 and y>x x>0,y<0 and x>y x<0,y<0 this case is not possible We can't conclude anything if x>y Hence A is insufficientFrom Statement B: \(y^x\)<0 so we have two cases y is negative and x is positive and odd y is negative and x is negative and odd So we can't conclude anything from above if x>y Hence B is insufficientNow combining the above two we have there is only one conditions that satisfies both Statement A & B which is x>0 and y<0. So we can conclude x>y Hence C is sufficient.
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Re: If x and y are integers, is x > y?
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08 Oct 2018, 20:38
Dear All Please can you help me with a general query regarding such questions. x + y > 0  Inequalities is my area of learning If we subtract y from both sides , will the sign change ? i.e. x+y<0 when y is subtracted will it become x > y ? ?? Walkabout wrote: 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?
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08 Oct 2018, 21:16
proabhinav wrote: Dear All Please can you help me with a general query regarding such questions. x + y > 0  Inequalities is my area of learning If we subtract y from both sides , will the sign change ? i.e. x+y<0 when y is subtracted will it become x > y ? ?? Walkabout wrote: If x and y are integers, is x > y?
(1) x + y > 0 (2) y^x < 0 x + y < 0 Subtract y from both sides: (x + y)  y < 0  y; x < y. 9. Inequalities For more check Ultimate GMAT Quantitative Megathread
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If x and y are integers, is x > y?
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08 Oct 2018, 21:23
proabhinav wrote: Dear All
Please can you help me with a general query regarding such questions.
x + y > 0 
Inequalities is my area of learning If we subtract y from both sides , will the sign change ?
i.e. x+y<0 when y is subtracted will it become x > y ?
Hi proabhinav, Adding or Subtracting any number will not affect the sign of the inequality. Only, when we multiply or divide an inequality by a negative number, we need to change the sign. So, if x + y < 0, subtracting y from both sides, we get, • x + y  y < y • Thus, x < y For more information on Inequalities, go through the following articlesDifferent methods to solve absolute value equations and inequalitiesSolving inequalities Number Line MethodWavy Line Method Application  Complex Algebraic Inequalities
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If x and y are integers, is x > y?
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16 May 2019, 16:26
I think if you can get used to writing down cases you can get a better understanding than just plugging in numbers to test... my attempt:
(1) x+y > 0 We have 3 cases: 1... x is + & y is + (can be both Yes or No) 2... bigger +x & smaller y (Yes) 3... smaller x & bigger +y (No)
Not sufficient.
(2) y^x < 0 We have 3 cases: 1... y & +Odd number x (Yes) 2... y & Odd# x bigger than y(Yes) 3... y & Odd# x smaller than/equal to y(No)
Against not sufficient.
(1&2) Since y must be negative only Case 2 works in (1), concurrently x must be positive so only Case 1 works in (2). Now there's only 1 way to fulfill both statements so it's sufficient.



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Re: If x and y are integers, is x > y?
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16 May 2019, 19:16
Walkabout wrote: If x and y are integers, is x > y?
(1) x + y > 0 (2) y^x < 0 Statement 1. X Y 2 3 3 2, so not sufficient. Statement 2. y has to be a negative Integer and x has to be an odd to satisfy the condition. Now, X Y 3 2 (X>y) 3 2 (Y>x) Not sufficient. Let's check for condition C (Statement 1+2): Since y is a negative integer X has to be a positive (odd) integer >y to satisfy x+y>0 So, C is answer (Combining Statement 1 & 2).
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Re: If x and y are integers, is x > y?
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19 May 2019, 07:08
This is a moderately difficult question on Inequalities, with the second statement being slightly tricky to analyse. So you will have to be slightly careful in interpreting it. Option B is one of the trap answers here. We know that x and y are integers, but we do not know their signs. We are trying to ascertain if x>y. In other words, we are trying to ascertain if (xy)>0. From statement I alone, we only know that (x+y) > 0. This is hardly sufficient to uniquely answer whether (xy)>0. If x = 3 and y = 2, (x+y) = (3+2) = 5 which is definitely greater than 0; also, for these values of x and y, x>y and the answer to the main question will be a YES. If x = 1 and y = 2, (x+y) = (1+2) = 1 which is definitely greater than 0; but, for these values of x and y, x<y and the answer to the main question will be a NO. Therefore, statement I alone is insufficient. So, answer options A and D can be ruled out. The possible answer options are B, C or E. From statement II alone, we know that \(y^x\) < 0 which means \(y^x\) is negative. So, we can surely say that y has to be negative i.e. y<0. But we cannot say anything about the sign of x. If y = 3 and x = 1, \((3)^1\) <0. Here, x>y; if y = 3 and x = 5, \((3)^5\)<0, here x<y. Therefore, statement II alone is insufficient. So, answer option B can be ruled out. The possible answer options are C or E. Using the data from statements I and II, from the second statement, we know that y is negative, but x can be positive or negative. But, both x and y cannot be negative since x+y>0. Hence, x has to be positive and y has to be negative. Therefore, we can surely say that x>y. So, the correct answer option is C. As mentioned earlier, you have to exercise a fair degree of caution while interpreting statement II, because it can trick you into assuming that x has to be positive. Also, trying out simple values while evaluating the inequalities will help analyse them better.. Hope this helps!
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Re: If x and y are integers, is x > y?
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