It is currently 21 Jan 2018, 18:47

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Manager
Manager
avatar
Joined: 02 Dec 2012
Posts: 178
If x and y are integers, is x > y? [#permalink]

Show Tags

New post 07 Dec 2012, 07:48
71
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

59% (01:25) correct 41% (01:16) wrong based on 1678 sessions

HideShow timer Statistics

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

(1) x + y > 0
(2) y^x < 0
[Reveal] Spoiler: OA
Expert Post
13 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 43348
Re: If x and y are integers, is x > y? [#permalink]

Show Tags

New post 07 Dec 2012, 07:55
13
This post received
KUDOS
Expert's post
26
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?
Extra-hard Quant Tests with Brilliant Analytics

Manager
Manager
avatar
Joined: 23 Jan 2013
Posts: 171
Concentration: Technology, Other
Schools: Haas
GMAT Date: 01-14-2015
WE: Information Technology (Computer Software)
GMAT ToolKit User
Re: If x and y are integers, is x > y? [#permalink]

Show Tags

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 ? )
Manager
Manager
avatar
Joined: 26 Sep 2013
Posts: 217
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

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 \(\frac{1}{(X^Y)}\)
3 KUDOS received
Manager
Manager
avatar
Joined: 30 Sep 2009
Posts: 115
GMAT ToolKit User
Re: If x and y are integers, is x > y? [#permalink]

Show Tags

New post 26 Oct 2014, 19:57
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
Intern
avatar
Joined: 12 Aug 2014
Posts: 17
Re: If x and y are integers, is x > y? [#permalink]

Show Tags

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?
Expert Post
1 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 43348
Re: If x and y are integers, is x > y? [#permalink]

Show Tags

New post 28 Nov 2014, 04:22
1
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
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.
_________________

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?
Extra-hard Quant Tests with Brilliant Analytics

2 KUDOS received
Director
Director
User avatar
Joined: 10 Mar 2013
Posts: 588
Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)
GMAT ToolKit User
If x and y are integers, is x > y? [#permalink]

Show Tags

New post 20 Dec 2014, 01:09
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
Intern
avatar
Joined: 05 Aug 2014
Posts: 30
Reviews Badge
Re: If x and y are integers, is x > y? [#permalink]

Show Tags

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!
Expert Post
1 KUDOS received
e-GMAT Representative
User avatar
S
Joined: 04 Jan 2015
Posts: 786
Re: If x and y are integers, is x > y? [#permalink]

Show Tags

New post 04 May 2015, 20:21
1
This post received
KUDOS
Expert's post
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
Harsh
_________________












| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Current Student
avatar
B
Joined: 12 Feb 2011
Posts: 101
GMAT ToolKit User
If x and y are integers, is x > y? [#permalink]

Show Tags

New post 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!
Expert Post
1 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 43348
Re: If x and y are integers, is x > y? [#permalink]

Show Tags

New post 27 Dec 2015, 03:47
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?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
Joined: 26 Dec 2014
Posts: 4
If x and y are integers, is x > y? [#permalink]

Show Tags

New post 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.

[Reveal] Spoiler:
C
Expert Post
Math Revolution GMAT Instructor
User avatar
D
Joined: 16 Aug 2015
Posts: 4705
GPA: 3.82
Premium Member
Re: If x and y are integers, is x > y? [#permalink]

Show Tags

New post 29 Dec 2015, 22:10
Expert's post
1
This post was
BOOKMARKED
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 one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
Find a 10% off coupon code for GMAT Club members.
“Receive 5 Math Questions & Solutions Daily”
Unlimited Access to over 120 free video lessons - try it yourself
See our Youtube demo

Manager
Manager
avatar
Joined: 28 Oct 2015
Posts: 57
Re: If x and y are integers, is x > y? [#permalink]

Show Tags

New post 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!
Manager
Manager
avatar
B
Joined: 17 Aug 2015
Posts: 105
GMAT ToolKit User
Re: If x and y are integers, is x > y? [#permalink]

Show Tags

New post 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+y-y>0 => x>0. When the two opposite inequality, we can subtract, smaller from larger and get positive result.
Manager
Manager
avatar
Joined: 11 Jul 2016
Posts: 80
If x and y are integers, is x > y? [#permalink]

Show Tags

New post 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 ?
Expert Post
Math Revolution GMAT Instructor
User avatar
D
Joined: 16 Aug 2015
Posts: 4705
GPA: 3.82
Premium Member
Re: If x and y are integers, is x > y? [#permalink]

Show Tags

New post 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.
_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
Find a 10% off coupon code for GMAT Club members.
“Receive 5 Math Questions & Solutions Daily”
Unlimited Access to over 120 free video lessons - try it yourself
See our Youtube demo

Intern
Intern
avatar
B
Joined: 25 Feb 2017
Posts: 47
GMAT 1: 720 Q50 V38
GPA: 3.67
Reviews Badge
Re: If x and y are integers, is x > y? [#permalink]

Show Tags

New post 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
Expert Post
EMPOWERgmat Instructor
User avatar
P
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 10734
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Re: If x and y are integers, is x > y? [#permalink]

Show Tags

New post 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:
[Reveal] Spoiler:
C


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

Co-Founder & 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, 20:54
Display posts from previous: Sort by

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.