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

 It is currently 15 Nov 2019, 08:02 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Is xy > 0? (1) x - y > -2 (2) x - 2y < -6

Author Message
TAGS:

### Hide Tags

Senior Manager  S
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 416
Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

12
134 00:00

Difficulty:   65% (hard)

Question Stats: 63% (02:14) correct 37% (02:22) wrong based on 1039 sessions

### HideShow timer Statistics

Is xy > 0?

(1) x - y > -2
(2) x - 2y < -6

_________________

Originally posted by Baten80 on 05 Jun 2011, 05:06.
Last edited by Bunuel on 09 Jul 2014, 10:49, edited 2 times in total.
Edited the question.
Math Expert V
Joined: 02 Sep 2009
Posts: 59074
Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

31
57
Is xy>0?

Note that question basically asks whether $$x$$ and $$y$$ have the same sign.

(1) x-y > -2 --> we can have an YES answer, if for example $$x$$ and $$y$$ are both positive ($$x=10$$ and $$y=1$$) as well as a NO answer, if for example $$x$$ is positive and $$y$$ is negative ($$x=10$$ and $$y=-10$$). Not sufficient.

(2) x-2y <-6 --> again it' easy to get an YES answer, if for example $$x$$ and $$y$$ are both positive ($$x=1$$ and $$y=10$$) as well as a NO answer, if for example $$x$$ is negative and $$y$$ is positive ($$x=-1$$ and $$y=10$$). Not sufficient.

You can get that the the two statement individually are not sufficient in another way too: we have (1) $$y<x+2$$ and (2) $$y>\frac{x}{2}+3$$. We are asked whether $$x$$ and $$y$$ have the same sign or whether the points (x,y) are in the I or III quadrant ONLY. But all (x,y) points below the line $$y=x+2$$ (for 1) and all (x, y) points above the line $$y=\frac{x}{2}+3$$ cannot lie only in I or III quadrant: points above or below some line (not parallel to axis) lie at least in 3 quadrants.

(1)+(2) Now, remember that we can subtract inequalities with the signs in opposite direction --> subtract (2) from (1): $$x-y-(x-2y)>-2-(-6)$$ --> $$y>4$$. As $$y>4$$ and (from 1) $$x>y-2$$ then $$x>2$$ (because we can add inequalities when their signs are in the same direction, so: $$y+x>4+(y-2)$$ --> $$x>2$$) --> we have that $$y>4$$ and $$x>2$$: both $$x$$ and $$y$$ are positive. Sufficient.

_________________
Manager  Status: How easy it is?
Joined: 09 Nov 2012
Posts: 85
Location: India
Concentration: Operations, General Management
GMAT 1: 650 Q50 V27 GMAT 2: 710 Q49 V37 GPA: 3.5
WE: Operations (Other)
Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

26
1
Please find attached graphical solution. I think its easier and simpler to solve such inequality problem using co-ordinate geometry.

I made this graph in a hurry in paint, so please don't mind the poorly drawn lines.

Consider pressing on Kudos if my post helped you in any way!
Attachments Solution.JPG [ 39.4 KiB | Viewed 86007 times ]

##### General Discussion
Retired Moderator B
Joined: 16 Nov 2010
Posts: 1237
Location: United States (IN)
Concentration: Strategy, Technology
Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

12
3
(1)

x = 1, y = 0

xy = 0 although x-y = 1 > -2

x = -3, y = -2

xy = 6 > 0 and x-y = -1 > -2

Insufficient

(2)

x - 2y < -6

x = 1, y = 4

x - 2y = 1 - 8 = -7 < -6 and xy > 0

x = 0, y = 4

x - 2y = 0 - 8 = -8 < -6 and xy = 0

Insufficient

(1) + (2)

x - y > -2

2y - x > 6

y > 4

Now x has to be such that x > y - 2

So x is positive, because y is 4.1, 5 etc.
and x is > 2.1, 3 etc.

So xy = positive

_________________
Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings
Math Expert V
Joined: 02 Sep 2009
Posts: 59074
Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

18
1
6 Is xy > 0?

$$xy>0$$ means that x and y must have the same sign, so point (x, y) must be either in the first or the third quadrant (green regions).

(1) x-y > -2 --> $$y<x+2$$ --> the area below blue line ($$y=x+2$$). (x, y) may or may not be in green region. Not sufficient.

(2) x-2y < -6 --> $$y>\frac{x}{2}+3$$ --> the area above red line ($$y>\frac{x}{2}+3$$). (x, y) may or may not be in green region. Not sufficient.

(1)+(2) Below blue line and above red line, is yellow region, which is entirely in I quadrant (where $$y>4$$ and $$x>2$$) --> $$xy>0$$. Sufficient.

Hope it helps.

Attachment: xy.png [ 8.35 KiB | Viewed 96942 times ]

_________________
Intern  Joined: 23 Jul 2013
Posts: 5
Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

Bunuel wrote:

Graphic approach:

Attachment:
xy.png

Is xy > 0?

$$xy>0$$ means that x and y must have the same sign, so point (x, y) must be either in the first or the third quadrant (green regions).

(1) x-y > -2 --> $$y<x+2$$ --> the area below blue line ($$y=x+2$$). (x, y) may or may not be in green region. Not sufficient.

(2) x-2y < -6 --> $$y>\frac{x}{2}+3$$ --> the area above red line ($$y>\frac{x}{2}+3$$). (x, y) may or may not be in green region. Not sufficient.

(1)+(2) Below blue line and above red line, is yellow region, which is entirely in I quadrant (where $$y>4$$ and $$x>2$$) --> $$xy>0$$. Sufficient.

Hope it helps.

Hi Bunuel,

I have doubts on graphical method -->

why is line y<x+2 passing through x = -2 ?
the value of x should be greater than -2 (the line may pass through x = -1, 0, 1, 2, ...)
Similarly, why is line y > (x/2) + 3 passing through x = -6
the value of x should be less than -6 (the line may pass through x = -7, -8, -9,...)

With the graphical method, can you illustrate solution of following example ?

Is x + y > 0 ?

1) x - y > 0
2) x^2 - y^2 > 0

Regards,
Akshay
Math Expert V
Joined: 02 Sep 2009
Posts: 59074
Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

AkshayDavid wrote:
Bunuel wrote:

Graphic approach: Is xy > 0?

$$xy>0$$ means that x and y must have the same sign, so point (x, y) must be either in the first or the third quadrant (green regions).

(1) x-y > -2 --> $$y<x+2$$ --> the area below blue line ($$y=x+2$$). (x, y) may or may not be in green region. Not sufficient.

(2) x-2y < -6 --> $$y>\frac{x}{2}+3$$ --> the area above red line ($$y>\frac{x}{2}+3$$). (x, y) may or may not be in green region. Not sufficient.

(1)+(2) Below blue line and above red line, is yellow region, which is entirely in I quadrant (where $$y>4$$ and $$x>2$$) --> $$xy>0$$. Sufficient.

Hope it helps.

Hi Bunuel,

I have doubts on graphical method -->

why is line y<x+2 passing through x = -2 ?
the value of x should be greater than -2 (the line may pass through x = -1, 0, 1, 2, ...)
Similarly, why is line y > (x/2) + 3 passing through x = -6
the value of x should be less than -6 (the line may pass through x = -7, -8, -9,...)

With the graphical method, can you illustrate solution of following example ?

Is x + y > 0 ?

1) x - y > 0
2) x^2 - y^2 > 0

Regards,
Akshay

y < x + 2 is the region which lies BELOW (because of < sign) line y = x + 2.

y > x/2 + 3 is the region which lies ABOVE (because of > sign) line y = x/2 + 3.

Check here for more: graphic-approach-to-problems-with-inequalities-68037.html

Hope it helps.
_________________
Intern  Joined: 14 Oct 2013
Posts: 43
Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

Hi,

I approached the 1&2 decision the same way as Bunuel but just had a clarifying question. When we subtract the two inequalities from eachother..how do we know that the sign stays as greater than (i.e. how do we know that it's y>4 instead of y<4). I assumed it was y>4 and got the question correct because i was subtracting a less than equation from a greater than equation but I have a feeling that's not the right logic. Thanks!
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3141
Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

7
1
healthjunkie wrote:
Hi,

I approached the 1&2 decision the same way as Bunuel but just had a clarifying question. When we subtract the two inequalities from eachother..how do we know that the sign stays as greater than (i.e. how do we know that it's y>4 instead of y<4). I assumed it was y>4 and got the question correct because i was subtracting a less than equation from a greater than equation but I have a feeling that's not the right logic. Thanks!

Hi healthjunkie,

Assume two inequalities x + y > 10......(1)
& x - y < 2........(2)

We see here that both the inequalities have opposite signs.

The basic concept of the inequality says to convert the inequalities into same sign before adding or subtracting them.

We can change the sign of the inequality by multiplying -1 on both sides of the inequality. Let's do it on inequality (2)

x - y < 2, multiplying -1 on both sides we get
-x + y > -2.

Since we have inequality (1) & (2) with the same signs we can add them now

x + y -x + y > 10 -2 i.e. y > 4.

Note here that adding inequality (1) to inequality (2) after multiplying inequality (2) by -1 is similar to subtracting inequality (2) from inequality (1). This concept is used when we say that we can subtract two inequalities with opposite signs.

In this question you asked we subtracted inequality in st-II from inequality in st-I which incidentally meant that st-II was multiplied by -1 and then added to st-I. So, the sign of inequality in st-II flipped.

x - 2y < -6, multiplying it with -1 gave the inequality -x + 2y > 6 which was then added to inequality in st-I

Hope its clear Regards
Harsh
_________________
Manager  S
Joined: 26 Jan 2015
Posts: 91
Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

Bunuel wrote:

Is xy>0?

Note that question basically asks whether $$x$$ and $$y$$ have the same sign.

(1) x-y > -2 --> we can have an YES answer, if for example $$x$$ and $$y$$ are both positive ($$x=10$$ and $$y=1$$) as well as a NO answer, if for example $$x$$ is positive and $$y$$ is negative ($$x=10$$ and $$y=-10$$). Not sufficient.

(2) x-2y <-6 --> again it' easy to get an YES answer, if for example $$x$$ and $$y$$ are both positive ($$x=1$$ and $$y=10$$) as well as a NO answer, if for example $$x$$ is negative and $$y$$ is positive ($$x=-1$$ and $$y=10$$). Not sufficient.

You can get that the the two statement individually are not sufficient in another way too: we have (1) $$y<x+2$$ and (2) $$y>\frac{x}{2}+3$$. We are asked whether $$x$$ and $$y$$ have the same sign or whether the points (x,y) are in the I or III quadrant ONLY. But all (x,y) points below the line $$y=x+2$$ (for 1) and all (x, y) points above the line $$y=\frac{x}{2}+3$$ can not lie only I or III quadrant: points above or below some line (not parallel to axis) lie at least in 3 quadrants.

(1)+(2) Now, remember that we can subtract inequalities with the signs in opposite direction --> subtract (2) from (1): $$x-y-(x-2y)>-2-(-6)$$ --> $$y>4$$. As $$y>4$$ and (from 1) $$x>y-2$$ then $$x>2$$ (because we can add inequalities when their signs are in the same direction, so: $$y+x>4+(y-2)$$ --> $$x>2$$) --> we have that $$y>4$$ and $$x>2$$: both $$x$$ and $$y$$ are positive. Sufficient.

Dear Bunuel,

However, the combined solving of eqns gave me, x<-4 or x>-1, a case in which we cannot comment on sign of x. Can you please help me understand where I have gone wrong?
Attachments IMG_20160401_163125.jpg [ 2.2 MiB | Viewed 65237 times ]

_________________
Kudos is the best way to say Thank you! Please give me a kudos if you like my post
Math Expert V
Joined: 02 Aug 2009
Posts: 8168
Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

3
Alok322 wrote:
Bunuel wrote:

Is xy>0?

Note that question basically asks whether $$x$$ and $$y$$ have the same sign.

(1) x-y > -2 --> we can have an YES answer, if for example $$x$$ and $$y$$ are both positive ($$x=10$$ and $$y=1$$) as well as a NO answer, if for example $$x$$ is positive and $$y$$ is negative ($$x=10$$ and $$y=-10$$). Not sufficient.

(2) x-2y <-6 --> again it' easy to get an YES answer, if for example $$x$$ and $$y$$ are both positive ($$x=1$$ and $$y=10$$) as well as a NO answer, if for example $$x$$ is negative and $$y$$ is positive ($$x=-1$$ and $$y=10$$). Not sufficient.

You can get that the the two statement individually are not sufficient in another way too: we have (1) $$y<x+2$$ and (2) $$y>\frac{x}{2}+3$$. We are asked whether $$x$$ and $$y$$ have the same sign or whether the points (x,y) are in the I or III quadrant ONLY. But all (x,y) points below the line $$y=x+2$$ (for 1) and all (x, y) points above the line $$y=\frac{x}{2}+3$$ can not lie only I or III quadrant: points above or below some line (not parallel to axis) lie at least in 3 quadrants.

(1)+(2) Now, remember that we can subtract inequalities with the signs in opposite direction --> subtract (2) from (1): $$x-y-(x-2y)>-2-(-6)$$ --> $$y>4$$. As $$y>4$$ and (from 1) $$x>y-2$$ then $$x>2$$ (because we can add inequalities when their signs are in the same direction, so: $$y+x>4+(y-2)$$ --> $$x>2$$) --> we have that $$y>4$$ and $$x>2$$: both $$x$$ and $$y$$ are positive. Sufficient.

Dear Bunuel,

However, the combined solving of eqns gave me, x<-4 or x>-1, a case in which we cannot comment on sign of x. Can you please help me understand where I have gone wrong?

Hi,
your Equations are further to be simplified..

you got
2y-6 > x> y-2
we can infer 2y-6>y-2...
or 2y-y>6-2 that is y>4..
if y>4 and x+2>y, x will be >2..
thus both x and y are +ive and our answer is YES for xy>0..

Do not substitute value in 2y-6 > x> y-2 and find the signs of x and y..
_________________
Manager  Joined: 20 Mar 2015
Posts: 54
Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

chetan2u Looks like how I came with the solution is wrong. Since statement 1 and 2 are not sufficient I tried to combine the two statements.

1. X- Y>-2 2. X-2Y <-6

FROM 1: X>Y-2-------(WE HAVE A VALUE FOR X) substituting this value in statement 2 yields;

Y-2 -2Y<-6
OR, -Y < -4
OR, Y> 4 (MUTLIPLYING BOTH SIDES WITH A NEGATIVE SIGN.)

SINCE Y> 4, FROM STATEMETN ONE

X-Y>-2 OR, X- 4.1> -2, SOLVING FOR THIS WOULD GIVE X>2.1.

Is my process correct?
Math Expert V
Joined: 02 Aug 2009
Posts: 8168
Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

bimalr9 wrote:
chetan2u Looks like how I came with the solution is wrong. Since statement 1 and 2 are not sufficient I tried to combine the two statements.

1. X- Y>-2 2. X-2Y <-6

FROM 1: X>Y-2-------(WE HAVE A VALUE FOR X) substituting this value in statement 2 yields;

Y-2 -2Y<-6
OR, -Y < -4
OR, Y> 4 (MUTLIPLYING BOTH SIDES WITH A NEGATIVE SIGN.)

SINCE Y> 4, FROM STATEMETN ONE

X-Y>-2 OR, X- 4.1> -2, SOLVING FOR THIS WOULD GIVE X>2.1.

Is my process correct?

HI,
In this Q it is ok..
But a lot depends on '>' and '<' signs.. you cannot substitute without knowing these..
1. X- Y>-2 ...... x>y-2
2. X-2Y <-6..... x<2y-6..
so y-2<x<2y-6....
so here clearly y-2<2y-6 .... or y>4...
so form equations then substitute...
_________________
Manager  Joined: 13 Jun 2016
Posts: 95
Location: United States
Concentration: Finance, Technology
Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

I didn't know that you can subtract opposite sign equations and actually I feel that many will still get the sign mixed up as to whether it would be < or >. Instead you can flip one of the equations and then you will end up with one sign and then just add from there. That way, there is less chance for a mistake to occur.
Target Test Prep Representative G
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2812
Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

3
Baten80 wrote:
Is xy > 0?

(1) x - y > -2
(2) x - 2y < -6

We need to determine whether the product of x and y is positive. We should recall that the product of two numbers is positive only if both the numbers are positive or if both are negative.

Statement One Alone:

x - y > -2

Statement one tells us that the difference between x and y is -2; it does not tell us anything about the signs of x and y. For instance, if x = 2 and y = 1, we have x - y = 1 > -2, and xy is positive. However, if x = 3 and y = -2, 3 - (-2) = 5 > -2, but xy is negative. Statement one alone is not sufficient. We can eliminate answer choices A and D.

Statement Two Alone:

x - 2y < -6

Again, we have a statement that tells us nothing about the signs of x and y. For instance, if x = 3 and y = 5, then x - 2y = 3 - 2(5) = 3 - 10 = -7 < -6, and xy is positive. However, if x = -1 and y = 5, then x - 2y = -1 - 2(5) = -11 < -6, and xy is negative. Statement two alone is not sufficient. We can eliminate answer choice B.

Statements One and Two Together:

Let’s manipulate the first inequality to read: y < x + 2. Similarly, we can manipulate the second inequality to read: y > (1/2)x + 3.

Thus, we can say the following:

(1/2)x + 3 < y < x + 2

(1/2)x + 3 < x + 2

x + 6 < 2x + 4

2 < x

Thus, x is positive.

We also know the following:

y > (1/2)x + 3

Since x is greater than two, let’s see what we can determine about y, if we substitute 2 for x.

y > (1/2)(2) + 3

y > 4

So y is positive as well. Both statements together are sufficient.

_________________

# Jeffrey Miller

Jeff@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Intern  B
Joined: 17 Mar 2017
Posts: 9
GMAT 1: 510 Q47 V15 GMAT 2: 600 Q49 V22 Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

Hello Bunuel,

I arrive to y>4.

When you replace it in (1) it makes x>2: OK

But

When you replace it in (2) it makes x - 2(4) < -6 ; so x <2 and so can be negative or positive.

Could you advise when replacing in (1) X>2 (answer =C) and when replacing in (2) we have answer (E) as x<2.

Math Expert V
Joined: 02 Sep 2009
Posts: 59074
Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

1
Tpral wrote:
Hello Bunuel,

I arrive to y>4.

When you replace it in (1) it makes x>2: OK

But

When you replace it in (2) it makes x - 2(4) < -6 ; so x <2 and so can be negative or positive.

Could you advise when replacing in (1) X>2 (answer =C) and when replacing in (2) we have answer (E) as x<2.

We got y > 4 not y = 4, so you cannot substitute the way you did. In the solution, x > 2, is derived by adding y > 4 and x > y - 2 (we can add inequalities when their signs are in the same direction): $$y+x>4+(y-2)$$ --> $$x>2$$.
_________________
Intern  Joined: 31 Mar 2018
Posts: 2
Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

I had some trouble with this one:
After eliminating A, B, and D I took a look at both equations:
x-y>-2
x-2y<-6

then I multiplied equation 1 by 3 on both sides:
3(x-y)>3(-2)
3x-3y>-6

with that I combined both statements around -6
x-2y<-6<3x-3y

then collapsed -6 out of the inequality
x-2y<3x-3y

then simplified:
x-2y<3x-3y
x+y<3x [+3y to each side]
y<2x [-x to each side]

and got E insufficient (equation is true if y=1 and x=2 and if y=-1 and x=2); can someone let me know where my misstep was?
Math Expert V
Joined: 02 Sep 2009
Posts: 59074
Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

ktjorge wrote:
I had some trouble with this one:
After eliminating A, B, and D I took a look at both equations:
x-y>-2
x-2y<-6

then I multiplied equation 1 by 3 on both sides:
3(x-y)>3(-2)
3x-3y>-6

with that I combined both statements around -6
x-2y<-6<3x-3y

then collapsed -6 out of the inequality
x-2y<3x-3y

then simplified:
x-2y<3x-3y
x+y<3x [+3y to each side]
y<2x [-x to each side]

and got E insufficient (equation is true if y=1 and x=2 and if y=-1 and x=2); can someone let me know where my misstep was?

Your manipulations are not wrong but they are not complete and thus do not lead to the correct answer.
_________________
Intern  B
Joined: 31 Mar 2018
Posts: 9
WE: Marketing (Telecommunications)
Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

### Show Tags

1
Bunuel wrote:

We got y > 4 not y = 4, so you cannot substitute the way you did. In the solution, x > 2, is derived by adding y > 4 and x > y - 2 (we can add inequalities when their signs are in the same direction): $$y+x>4+(y-2)$$ --> $$x>2$$.

Hi Bunuel ,

Kindly clarify -

"In the Solution of X>2" : Instead of adding Y>4 and X > Y-2 (from equation 1) , can we not add Y>4 and -X> 6-2y ? (from equation 2 : x-2y <-6)

If we do this , we will arrive at X < 2 Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6   [#permalink] 02 Jun 2018, 10:29

Go to page    1   2    Next  [ 29 posts ]

Display posts from previous: Sort by

# Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  