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

 It is currently 20 Feb 2019, 15:43

### 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

## Events & Promotions

###### Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### Free GMAT Prep Hour

February 20, 2019

February 20, 2019

08:00 PM EST

09:00 PM EST

Strategies and techniques for approaching featured GMAT topics. Wednesday, February 20th at 8 PM EST

February 21, 2019

February 21, 2019

10:00 PM PST

11:00 PM PST

Kick off your 2019 GMAT prep with a free 7-day boot camp that includes free online lessons, webinars, and a full GMAT course access. Limited for the first 99 registrants! Feb. 21st until the 27th.

# If x and y are integers and xy ≠ 0, is x - y > 0?

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 53020
If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

### Show Tags

24 Mar 2015, 03:46
00:00

Difficulty:

55% (hard)

Question Stats:

60% (01:57) correct 40% (02:13) wrong based on 254 sessions

### HideShow timer Statistics

If x and y are integers and xy ≠ 0, is x - y > 0?

(1) x/y < 1/2

(2) $$\sqrt{x^2}= x$$ and $$\sqrt{y^2} = y$$

Kudos for a correct solution.

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 53020
Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

### Show Tags

30 Mar 2015, 03:32
3
2
Bunuel wrote:
If x and y are integers and xy ≠ 0, is x - y > 0?

(1) x/y < 1/2

(2) $$\sqrt{x}= x$$ and $$\sqrt{y} = y$$

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:
Attachment:

isx-ypositive_explanation.png [ 36.76 KiB | Viewed 2300 times ]

Q: For statement 1, why can't we multiply both sides by y to get x < 1/2y? Wouldn't that be sufficient on its own?

A: You have good intuition, and this might work for an equation, but there's a crucial difference between equations and inequalities: if you divide (or multiply) both sides by a negative number, then you have to change the direction of the inequality sign.

If I have -5x > 10

and divide both sides by -5, I need to flip the inequality so I have:

x < -2

Let's see why with actual numbers:

Equation:

8 = 8

divide both sides by - 2:

-4 = -4 <--- still true!

Inequality:

16 > 8

divide both sides by -2

-8 > -4 ....no! We changed the relationship when dividing by -2. So we need to flip the sign:

-8 < -4

Now this is correct.

In this problem, we have to ask "Is y negative?" If it isn't, then we can rework the equation as you did. If it is negative, then we have to flip the sign. But we don't know if it's negative or not! We can't multiply by y.

So, we cannot multiply across an inequality with variables unless we know for certain whether they're positive or negative.

Q: Can you please explain to me how I know that the √y² = y, is positive? Why can't it be positive and negative? For example, √4 is +2 or -2.

A: The radical sign (more commonly known as the square root sign) actually means the POSITIVE square root of a given number. This positive square root is often referred to as the principal square root. Anytime we see the radical sign, we want the principal (positive) square root.

Also, from Wikipedia

"Every non-negative real number a has a unique non-negative square root, called the principal square root, which is denoted by [square root symbol], where [square root symbol] is called radical sign.

For example, the principal square root of 9 is 3, denoted √9 = 3, because 3² = 3 × 3 = 9 and 3 is non-negative."

So if x²= 4, then x = 2 or -2

But if √4 = x, then x = 2 only

√4 is the positive root of 4 only.

When you see a "√" sign with something inside, there is only one possible output.

So √(x^2) = |x|

_________________
##### General Discussion
Manager
Joined: 25 Mar 2014
Posts: 133
Location: India
Concentration: Operations, Finance
GMAT Date: 05-10-2015
GPA: 3.51
WE: Programming (Computer Software)
Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

### Show Tags

24 Mar 2015, 05:26
Given that non of a and y is 0.
Statement 1 is not enough, as relationship between x and Y are different for +ve and -ve values of x and y.
Statement 2 is enough. With this information x = y =1, as they can not be 0. SUFFICIENT.
_________________

Please give Kudos to the post if you liked.

Director
Joined: 07 Aug 2011
Posts: 529
GMAT 1: 630 Q49 V27
Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

### Show Tags

24 Mar 2015, 05:42
Bunuel wrote:
If x and y are integers and xy ≠ 0, is x - y > 0?

(1) x/y < 1/2

(2) $$\sqrt{x}= x$$ and $$\sqrt{y} = y$$

Kudos for a correct solution.

1) x/y < 1/2
At X=1,Y=3 above inequality holds but is x>y ? NO
At X=-1,Y=-4 above inequality holds but is x>y ? YES
NOT SUFFICIENT.

(2) $$\sqrt{x}= x$$ and $$\sqrt{y} = y$$
As X and Y are integers, we can infer X=Y=1
SUFFICIENT.

Manager
Joined: 18 Sep 2014
Posts: 227
Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

### Show Tags

24 Mar 2015, 11:26
St 1) x/y < 1/2
At X=1,Y=3 above inequality holds but is x>y ? NO
At X=-1,Y=-4 above inequality holds but is x>y ? YES
NOT SUFFICIENT.

St 2) \sqrt{x}= x and \sqrt{y} = y
As X and Y are integers, we can infer X=Y=1
SUFFICIENT.
Ans: B
_________________

Kindly press the Kudos to appreciate my post !!

Senior Manager
Joined: 28 Feb 2014
Posts: 294
Location: United States
Concentration: Strategy, General Management
Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

### Show Tags

24 Mar 2015, 17:00
Bunuel wrote:
If x and y are integers and xy ≠ 0, is x - y > 0?

(1) x/y < 1/2

(2) $$\sqrt{x}= x$$ and $$\sqrt{y} = y$$

Kudos for a correct solution.

Statement 1: x/y < 1/2
test numbers
If x=-3, y=2 No
if x=3 y=-2 yes
Insufficient

Statement 2:
x, y = 1
sufficient

Manager
Joined: 26 Dec 2012
Posts: 146
Location: United States
Concentration: Technology, Social Entrepreneurship
WE: Information Technology (Computer Software)
Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

### Show Tags

24 Mar 2015, 23:21
x &y integers & x*y not=0; is x>y?

1. Resolve the equation and 2x<y; this will hold true for +ve value but not hold true for -ve values; Not sufficient

2. Square of number = number itself only for 1, hence x=y=1; sufficient

Thanks,
Intern
Joined: 11 Apr 2013
Posts: 12
Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

### Show Tags

08 Apr 2015, 04:42
2
Bunuel could you please explain this more briefly as why B is not the answer as the above fluff is bouncing all above the head.
if x^2=x and y^2=y this implies that x=1 and y=1. what am i missing here?????
Math Expert
Joined: 02 Sep 2009
Posts: 53020
Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

### Show Tags

08 Apr 2015, 04:52
anshul2014 wrote:
Bunuel could you please explain this more briefly as why B is not the answer as the above fluff is bouncing all above the head.
if x^2=x and y^2=y this implies that x=1 and y=1. what am i missing here?????

Sorry, there was a typo in (2). It should read: $$\sqrt{x^2}= x$$ and $$\sqrt{y^2} = y$$

Edited. Thank you.
_________________
Manager
Joined: 03 Sep 2014
Posts: 74
Concentration: Marketing, Healthcare
Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

### Show Tags

08 Apr 2015, 08:51
Bunuel wrote:
If x and y are integers and xy ≠ 0, is x - y > 0?

(1) x/y < 1/2

(2) $$\sqrt{x^2}= x$$ and $$\sqrt{y^2} = y$$

Kudos for a correct solution.

Hi,

The way I approached is-

1) x/y< 1/2 => 2x < y

Now if x = -3 and y = -1 => 2x < y => -6 < -1. Then is x - y > 0 - NO.

if x = 1 and y = 3 => 2x < y => 2 < 3. Then is x - y > 0 - NO.

if x = -1/2 and y = -1/6 => 2x < y => -1 < -1/6. Then is x - y > 0 - NO.

if x = 1/6 and y = 1/2 => 2x < y => 1/3 < 1/2. Then is x - y > 0 - NO.

To simply put, shouldn't A be sufficient enough to say that if 2x < y then x - y will never be greater zero??

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13562
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

### Show Tags

11 Apr 2015, 10:19
1
HI King407,

TESTing VALUES is a great approach for this question, but the problem with your approach is that it does not factor in all of the possibilities that can occur. If you look at your work, you chose to make X and Y either BOTH positive OR both NEGATIVE, but you COULD have 1 positive and 1 negative....

We're told that X/Y < 1/2

IF...
X = 1
Y = -3
then -1/3 < 1/2
and the answer to the question is.....(1) - (-3) = 4 > 0 so the answer is YES.
Fact 1 is INSUFFICIENT

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!*****

Manager
Joined: 03 Sep 2014
Posts: 74
Concentration: Marketing, Healthcare
Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

### Show Tags

11 Apr 2015, 10:38
EMPOWERgmatRichC wrote:
HI King407,

TESTing VALUES is a great approach for this question, but the problem with your approach is that it does not factor in all of the possibilities that can occur. If you look at your work, you chose to make X and Y either BOTH positive OR both NEGATIVE, but you COULD have 1 positive and 1 negative....

We're told that X/Y < 1/2

IF...
X = 1
Y = -3
then -1/3 < 1/2
and the answer to the question is.....(1) - (-3) = 4 > 0 so the answer is YES.
Fact 1 is INSUFFICIENT

GMAT assassins aren't born, they're made,
Rich

Thanks it is clear now...

.
Intern
Joined: 23 Sep 2014
Posts: 34
Location: India
Concentration: Marketing, Finance
GMAT 1: 670 Q48 V34
Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

### Show Tags

08 Nov 2016, 05:01
Hi,

Can someone pls explain why stmt 2 by itself is not correct?

The statement deduces to x=1 and y=1, thus x-y=0.

Am I missing something?
Math Expert
Joined: 02 Sep 2009
Posts: 53020
Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

### Show Tags

08 Nov 2016, 05:33
1
believer700 wrote:
Hi,

Can someone pls explain why stmt 2 by itself is not correct?

The statement deduces to x=1 and y=1, thus x-y=0.

Am I missing something?

Any positive x and y satisfy the second statement, not only x = 1 and y = 1.
_________________
Intern
Joined: 23 Sep 2014
Posts: 34
Location: India
Concentration: Marketing, Finance
GMAT 1: 670 Q48 V34
Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

### Show Tags

09 Nov 2016, 00:55
Bunuel wrote:
believer700 wrote:
Hi,

Can someone pls explain why stmt 2 by itself is not correct?

The statement deduces to x=1 and y=1, thus x-y=0.

Am I missing something?

Any positive x and y satisfy the second statement, not only x = 1 and y = 1.

Many thanks!!

I messed up on the square part.
GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 763
Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

### Show Tags

08 Oct 2018, 10:33
Bunuel wrote:
If x and y are integers and xy ≠ 0, is x - y > 0?

(1) x/y < 1/2

(2) $$\sqrt{x^2}= x$$ and $$\sqrt{y^2} = y$$

$$\left( * \right)\,\,\,\,x,y\,\,\, \ne 0\,\,\,\,{\rm{ints}}\,\,\,\,\left( {xy \ne 0} \right)\,\,\,\,$$

$$x\,\,\mathop > \limits^? \,\,y$$

$$\left( 1 \right)\,\,{x \over y} < {1 \over 2}\,\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( { - 1, - 3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$

$$\left( 2 \right)\,\,\,\left\{ \matrix{ \,\left| x \right| = x\,\,\,\, \Rightarrow \,\,\,\,x \ge 0\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,x > 0 \hfill \cr \,\left| y \right| = y\,\,\,\, \Rightarrow \,\,\,\,y \ge 0\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,y > 0 \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {2,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\,$$

$$\left( {1 + 2} \right)\,\,\,\,\,\left\{ \matrix{ \,\,{x \over y} < {1 \over 2}\,\,\,\,\,\mathop \Rightarrow \limits_{y\, > \,\,0}^{ \cdot \,\,2y} \,\,\,\,\,2x < y \hfill \cr \,\,\,1 < 2\,\,\,\,\,\mathop \Rightarrow \limits_{\,x\, > \,\,0}^{ \cdot \,\,x} \,\,\,\,\,\,\,x < 2x \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,x < 2x < y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x < y\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________

Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net

Re: If x and y are integers and xy ≠ 0, is x - y > 0?   [#permalink] 08 Oct 2018, 10:33
Display posts from previous: Sort by