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If x and y are integers, is x > y?
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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?
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07 Dec 2012, 07:55
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?
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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?
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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?
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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?
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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?
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28 Nov 2014, 04:22
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?
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20 Dec 2014, 01:09
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?
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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?
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04 May 2015, 20:21
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 statementII, 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: CaseI: \(x\) is positiveIf \(x\) is a positive odd integer, then \(y^x < 0\). For example, assuming \(y = 2\) and \(x = 3\) would give \(y^x = 2^3 = 8 < 0\) CaseII: \(x\) is negativeWhen \(x\) 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 statementII does not give us a unique answer. Regards Harsh
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If x and y are integers, is x > y?
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26 Dec 2015, 20:52
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. Thanks for the explanation. I got the correct answer using the same approach, however at the beginning of my solution I tried a different way to understand statement 2: Since \(y^0=1\) and \(0<1\), so if \(y^x<0\), it should mean that \(y^x<y^0\) and hence \(x<0\). I know this is not correct but I am unable to understand what am I doing wrong here. Can you please help confirm that? Thanks again!



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Re: If x and y are integers, is x > y?
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27 Dec 2015, 03:47
Dienekes 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. Thanks for the explanation. I got the correct answer using the same approach, however at the beginning of my solution I tried a different way to understand statement 2: Since \(y^0=1\) and \(0<1\), so if \(y^x<0\), it should mean that \(y^x<y^0\) and hence \(x<0\). I know this is not correct but I am unable to understand what am I doing wrong here. Can you please help confirm that? Thanks again! If \(y\neq{0}\), then y^0=1. y^x < 0 does not mean that y^x < (1 = y^0).
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If x and y are integers, is x > y?
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28 Dec 2015, 18:31
Walkabout wrote: If x and y are integers, is x > y?
(1) x + y > 0 (2) y^x < 0 (1) \(x+y>0\) : For this to be true, we will have the following 3 senarios: a) \(x > 0\) and \(y > 0\) b) \(x > 0\) and \(y <= 0\), which means \(x > y\) c) \(y > 0\) and \(x <= 0\), which means \(y > x\) Therefore, this statement is Insufficient (2) \(y^x < 0\): For this to be true, \(y\) must be less than 0 and \(x\) must be an odd integer. Therefore, this statement is insufficient. (1) and (2) together:From statement (2), we know \(y < 0\), therefore the only viable scenario for this in statement (1) is option (b): \(x > 0\) and \(y < 0\), which means \(x > y\) => Sufficient.



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Re: If x and y are integers, is x > y?
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29 Dec 2015, 22:10
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution If x and y are integers, is x > y? (1) x + y > 0 (2) y^x < 0 There are two variables (x and y) in the original condition. In order to match the number of variables and the number of equations, we need 2 equations. Since the condition 1) and 2) each has 1 equation, there is high chance that C is going to be the answer. Using both the condition 1) and 2), we know that the condition 2) states y^x<0, which means y<0 and x=odd. Also, the condition 1) states x+y>0, which means y<0. So, x>0 and it is always the case that x>y. The answer is ‘yes’ and the conditions are sufficient. Therefore, since the condition 1) and 2) are not sufficient alone, the correct answer is C. For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: If x and y are integers, is x > y?
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14 Jun 2016, 10:00
EgmatQuantExpert MathRevolution Can you please explain if y<0 and x+y>0 and we create the following situations x=5,y=3 => 5+(3)>0 but what if x=1,y=12 =>1+(12)<0 How can we not consider the second case above? Thanks in advance!



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Re: If x and y are integers, is x > y?
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19 Jun 2016, 05:09
is x> y?
State1: x+y>0 .. the sum of two numbers is greater than zero does not tell us anything about how they rank relative to each other state 2: y^x<0 > this implies that x must be odd and y<0 .. If y <0 then x can be 1, +1. let y=1. x=1 => y^x = 1 this satisfies y=1 . x=1 => 1/1^1 => 1 this also satisfies. so x < y and x> y can both betrue. So not sufficient can y>0 .
combine the two statements y<0 and x+y>0 implies x has to be positive. Also note that since y<0 x+y>0 => x+yy>0 => x>0. When the two opposite inequality, we can subtract, smaller from larger and get positive result.



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If x and y are integers, is x > y?
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11 Oct 2016, 07:02
MathRevolution wrote: Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution
If x and y are integers, is x > y? (1) x + y > 0 (2) y^x < 0
There are two variables (x and y) in the original condition. In order to match the number of variables and the number of equations, we need 2 equations. Since the condition 1) and 2) each has 1 equation, there is high chance that C is going to be the answer. Using both the condition 1) and 2), we know that the condition 2) states y^x<0, which means y<0 and x=odd. Also, the condition 1) states x+y>0, which means y<0. So, x>0 and it is always the case that x>y. The answer is ‘yes’ and the conditions are sufficient. Therefore, since the condition 1) and 2) are not sufficient alone, the correct answer is C.
For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E. How to determine whether the variable approach will it into a particular Q? There are cases where at first glance ,it seems two options will convert to two equations but they turn out to be a single equation. Is there any pattern which I can look into ?



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Re: If x and y are integers, is x > y?
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13 Oct 2016, 18:03
Manonamission wrote: MathRevolution wrote: Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution
If x and y are integers, is x > y? (1) x + y > 0 (2) y^x < 0
There are two variables (x and y) in the original condition. In order to match the number of variables and the number of equations, we need 2 equations. Since the condition 1) and 2) each has 1 equation, there is high chance that C is going to be the answer. Using both the condition 1) and 2), we know that the condition 2) states y^x<0, which means y<0 and x=odd. Also, the condition 1) states x+y>0, which means y<0. So, x>0 and it is always the case that x>y. The answer is ‘yes’ and the conditions are sufficient. Therefore, since the condition 1) and 2) are not sufficient alone, the correct answer is C.
For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E. How to determine whether the variable approach will it into a particular Q? There are cases where at first glance ,it seems two options will convert to two equations but they turn out to be a single equation. Is there any pattern which I can look into ?Hi, Since x>y? is a question, we figured out the values of x and y. Also, in a case where 1)=2) and there is 1 equation, it is 95% that D is an answer and 5% that E is an answer. Thank you.
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Re: If x and y are integers, is x > y?
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30 Apr 2017, 17:08
If x and y are integers, is x > y?
(1) x + y > 0 (2) y^x < 0
My 2 cents. It is important to realize that on statement 2,
Y = 1, x = 1
OR
Y = 1, x =1
Both values work, thus insufficient. Combining with statm 1, you see that C is the answer



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Re: If x and y are integers, is x > y?
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13 Dec 2017, 20:54
Hi All, We're told that X and Y are integers. We're asked if X > Y. This is a YES/NO question. We can solve it with a mix of Number Properties and TESTing VALUES. 1) X+Y > 0 With this Fact, we know that at least one of the two variables is positive. IF.... X = 1, Y = 0 then the answer to the question is YES. IF.... X = 0, Y = 1 then the answer to the question is NO. Fact 1 is INSUFFICIENT 2) Y^X < 0 With this Fact, we know that Y CANNOT be positive (since a positive raised to any integer power is a positive). Thus, Y MUST be NEGATIVE and X must be ODD (since raising a negative to an EVEN power will end in a positive result). IF.... X = 1, Y = 1 then the answer to the question is YES. IF.... X = 3, Y = 1 then the answer to the question is NO. Fact 2 is INSUFFICIENT Combined, we know that there must be at least one positive value and that Y must be negative. By extension, that means that X MUST be positive  and the answer to the question is ALWAYS YES. Combined, SUFFICIENT Final Answer: GMAT assassins aren't born, they're made, Rich
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