GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 16 Jun 2019, 22:02

### 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

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

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 04 Mar 2018
Posts: 3
Re: If x and y are integers, is x > y?  [#permalink]

### Show Tags

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.
VP
Joined: 09 Mar 2016
Posts: 1279
Re: If x and y are integers, is x > y?  [#permalink]

### Show Tags

22 Aug 2018, 10:05
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
GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 936
Re: If x and y are integers, is x > y?  [#permalink]

### Show Tags

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) $$x \leqslant y$$

$$\left( 1 \right)\,\,\,\, \Rightarrow \,\,\,\,\,y < 0\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( 3 \right)} \,\,\,x < 0\,\,\,\,\,\,\, \Rightarrow \,\,\,x + y < 0\,\,\,\,\, \Rightarrow \,\,\,\,\left( 2 \right)\,\,{\text{contradicted}}$$
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Senior Manager
Joined: 10 Apr 2018
Posts: 265
Location: United States (NC)
Re: If x and y are integers, is x > y?  [#permalink]

### Show Tags

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 x-y>0

Lets understand in which conditions is x-y>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 possible

So 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 insufficient

From 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 insufficient

Now 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.
_________________
Probus

~You Just Can't beat the person who never gives up~ Babe Ruth
Manager
Joined: 03 Aug 2018
Posts: 69
Location: India
Concentration: Strategy, Operations
GMAT 1: 590 Q45 V26
GPA: 3.5
Re: If x and y are integers, is x > y?  [#permalink]

### Show Tags

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 ?

??

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

(1) x + y > 0
(2) y^x < 0
Math Expert
Joined: 02 Sep 2009
Posts: 55618
Re: If x and y are integers, is x > y?  [#permalink]

### Show Tags

08 Oct 2018, 21:16
1
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 ?

??

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

_________________
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2888
If x and y are integers, is x > y?  [#permalink]

### Show Tags

08 Oct 2018, 21:23
1
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

Different methods to solve absolute value equations and inequalities

Solving inequalities- Number Line Method

Wavy Line Method Application - Complex Algebraic Inequalities

_________________
Senior Manager
Status: Gathering chakra
Joined: 05 Feb 2018
Posts: 305
If x and y are integers, is x > y?  [#permalink]

### Show Tags

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.
ISB School Moderator
Joined: 08 Dec 2013
Posts: 419
Location: India
Concentration: Nonprofit, Sustainability
GMAT 1: 630 Q47 V30
WE: Operations (Non-Profit and Government)
Re: If x and y are integers, is x > y?  [#permalink]

### Show Tags

16 May 2019, 19:16
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).
_________________
Kindly drop a '+1 Kudos' if you find this post helpful.

GMAT Math Book

-I never wanted what I gave up
I never gave up what I wanted-
CrackVerbal Quant Expert
Joined: 12 Apr 2019
Posts: 103
Re: If x and y are integers, is x > y?  [#permalink]

### Show Tags

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 (x-y)>0.
From statement I alone, we only know that (x+y) > 0. This is hardly sufficient to uniquely answer whether (x-y)>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!
_________________
Re: If x and y are integers, is x > y?   [#permalink] 19 May 2019, 07:08

Go to page   Previous    1   2   [ 30 posts ]

Display posts from previous: Sort by