Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 02 Dec 2012
Posts: 178

If x and y are integers, is x > y? [#permalink]
Show Tags
07 Dec 2012, 08:48
74
This post was BOOKMARKED
Question Stats:
59% (01:27) correct 41% (01:15) wrong based on 1846 sessions
HideShow timer Statistics
If x and y are integers, is x > y? (1) x + y > 0 (2) y^x < 0
Official Answer and Stats are available only to registered users. Register/ Login.



Math Expert
Joined: 02 Sep 2009
Posts: 44623

Re: If x and y are integers, is x > y? [#permalink]
Show Tags
07 Dec 2012, 08:55
12
This post received KUDOS
Expert's post
28
This post was BOOKMARKED
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 23 Jan 2013
Posts: 165
Concentration: Technology, Other
GMAT Date: 01142015
WE: Information Technology (Computer Software)

Re: If x and y are integers, is x > y? [#permalink]
Show Tags
08 Oct 2013, 09:29
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 ? )



Manager
Joined: 26 Sep 2013
Posts: 204
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41 GMAT 2: 730 Q49 V41

Re: If x and y are integers, is x > y? [#permalink]
Show Tags
08 Oct 2013, 17:12
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 \(\frac{1}{(X^Y)}\)



Manager
Joined: 30 Sep 2009
Posts: 108

Re: If x and y are integers, is x > y? [#permalink]
Show Tags
26 Oct 2014, 20:57
3
This post received KUDOS
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



Intern
Joined: 12 Aug 2014
Posts: 17

Re: If x and y are integers, is x > y? [#permalink]
Show Tags
28 Nov 2014, 01:21
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?



Math Expert
Joined: 02 Sep 2009
Posts: 44623

Re: If x and y are integers, is x > y? [#permalink]
Show Tags
28 Nov 2014, 05:22
1
This post received KUDOS
Expert's post
1
This post was BOOKMARKED



Director
Joined: 10 Mar 2013
Posts: 563
Location: Germany
Concentration: Finance, Entrepreneurship
GPA: 3.88
WE: Information Technology (Consulting)

If x and y are integers, is x > y? [#permalink]
Show Tags
20 Dec 2014, 02:09
2
This post received KUDOS
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)
_________________
When you’re up, your friends know who you are. When you’re down, you know who your friends are.
Share some Kudos, if my posts help you. Thank you !
800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50 GMAT PREP 670 MGMAT CAT 630 KAPLAN CAT 660



Intern
Joined: 05 Aug 2014
Posts: 30

Re: If x and y are integers, is x > y? [#permalink]
Show Tags
04 May 2015, 09:47
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!



eGMAT Representative
Joined: 04 Jan 2015
Posts: 1019

Re: If x and y are integers, is x > y? [#permalink]
Show Tags
04 May 2015, 21:21
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
_________________
 '4 out of Top 5' Instructors on gmatclub  70 point improvement guarantee  www.egmat.com



Manager
Joined: 13 Feb 2011
Posts: 95

If x and y are integers, is x > y? [#permalink]
Show Tags
26 Dec 2015, 21:52
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!



Math Expert
Joined: 02 Sep 2009
Posts: 44623

Re: If x and y are integers, is x > y? [#permalink]
Show Tags
27 Dec 2015, 04:47
1
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
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).
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 26 Dec 2014
Posts: 4

If x and y are integers, is x > y? [#permalink]
Show Tags
28 Dec 2015, 19:31
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.



Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 5269
GPA: 3.82

Re: If x and y are integers, is x > y? [#permalink]
Show Tags
29 Dec 2015, 23:10
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.
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The oneandonly World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only $79 for 3 month Online Course" "Free Resources30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons  try it yourself"



Manager
Joined: 28 Oct 2015
Posts: 55

Re: If x and y are integers, is x > y? [#permalink]
Show Tags
14 Jun 2016, 11:00
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!



Manager
Joined: 17 Aug 2015
Posts: 102

Re: If x and y are integers, is x > y? [#permalink]
Show Tags
19 Jun 2016, 06:09
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.



Manager
Joined: 11 Jul 2016
Posts: 80

If x and y are integers, is x > y? [#permalink]
Show Tags
11 Oct 2016, 08:02
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 ?



Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 5269
GPA: 3.82

Re: If x and y are integers, is x > y? [#permalink]
Show Tags
13 Oct 2016, 19:03
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.
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The oneandonly World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only $79 for 3 month Online Course" "Free Resources30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons  try it yourself"



Intern
Joined: 25 Feb 2017
Posts: 42
Location: Korea, Republic of
GPA: 3.67

Re: If x and y are integers, is x > y? [#permalink]
Show Tags
30 Apr 2017, 18:08
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



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 11507
Location: United States (CA)
GRE 1: 340 Q170 V170

Re: If x and y are integers, is x > y? [#permalink]
Show Tags
13 Dec 2017, 21:54
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
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************




Re: If x and y are integers, is x > y?
[#permalink]
13 Dec 2017, 21:54






