Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 15 Jul 2019, 23:20 ### 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.  # If x and y are integers and xy does not equal 0, is xy < 0?

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 56233
If x and y are integers and xy does not equal 0, is xy < 0?  [#permalink]

### Show Tags

2
12 00:00

Difficulty:   95% (hard)

Question Stats: 56% (02:56) correct 44% (02:55) wrong based on 234 sessions

### HideShow timer Statistics Tough and Tricky questions: Inequalities.

If x and y are integers and xy does not equal 0, is xy < 0?

(1) $$y = x^4 – x^3$$

(2) $$-12y^2 – y^2x + x^2y^2 > 0$$

_________________
Intern  Joined: 18 Dec 2013
Posts: 25
GPA: 2.84
WE: Project Management (Telecommunications)
Re: If x and y are integers and xy does not equal 0, is xy < 0?  [#permalink]

### Show Tags

2
Bunuel wrote:

Tough and Tricky questions: Statistics.

If x and y are integers and xy does not equal 0, is xy < 0?

(1) y = x^4 – x^3
(2) -12y^2 – y^2x + x^2y^2 > 0 I got the answer correct in 2 Mins 20 Sec, however, I don't know if my method is correct. Feedback's are welcome if I have missed or omitted anything from my method
***************************************************************************************************************************************************************************************
Now Statement 1 --> $$Y=X^4-X^3$$

From this statement, you can also gather that $$Y's$$ value depends upon $$X$$. So, let take into considerations all possible signs & values of $$X$$.

Assume $$X$$ is positive (+ve), then...

$$Y=X^4-X^3$$
$$Y=X^3*(X-1)$$
$$Y=(+ve)*(+ve - 1) = +ve * +ve$$
But, according to the main statement $$X$$ is not equal to zero, therefore $$X$$ could be equal to 1.
If $$X=1$$, then, $$Y=0$$ --> Substitute $$X=1$$ in statement 1 $$Y=X^4-X^3 = (1)^4-(1)^3 =1-1= Zero$$

Therefore, $$XY = +ve$$ when $$X>1$$ and $$XY = 0$$ when $$X=1$$

Hence, Statement 1 is insufficient, Eliminate A, D

Now, Statement 2 --> $$-12Y^2-Y^2X+X^2Y^2 > 0$$

Rearranging & Simplifying -->
$$Y^2(-12-X+X^2) > 0$$
$$Y^2(X^2-X-12) > 0$$
$$Y^2((X-4)(X+3)) > 0$$

Now, since main stem of the question mentions $$X$$ is not equal to zero
Lets again assume $$X$$ is positive $$(+ve)$$ --> (It doesnt matter whether Y is +ve or -ve since eqn 2 contains $$Y^2$$),

Then... in the simplified statement 2, we again fall into the same scenario where $$XY = +ve$$ when $$X>4$$, $$XY = 0$$ when $$X=4$$ and $$XY = -ve$$ when $$X < 4$$

Hence, Statement 2 is insufficient, Eliminate B

Now, Combining both the statements doesn't provide us any additional information.

Therefore, both statements together are in-sufficient, Eliminate option C

Quote:

Only option E remains, select option E! Manager  Status: I am not a product of my circumstances. I am a product of my decisions
Joined: 20 Jan 2013
Posts: 108
Location: India
Concentration: Operations, General Management
GPA: 3.92
WE: Operations (Energy and Utilities)
Re: If x and y are integers and xy does not equal 0, is xy < 0?  [#permalink]

### Show Tags

1
1
Bunuel wrote:

Tough and Tricky questions: Statistics.

If x and y are integers and xy does not equal 0, is xy < 0?

(1) y = x^4 – x^3
(2) -12y^2 – y^2x + x^2y^2 > 0

Ans is E.

Statement 1 : y = x^4 – x^3

if X = 1, then y = x^4 – x^3 X^4 -X^3 = 0 and since XY does not equal zero X cannot be equal to 1.
if X= -1, then y = x^4 – x^3 X^4 -X^3 = 2 ; X is -ve and Y is +ve therefore XY < 0
If X = 2, then y = x^4 – x^3 X^4 -X^3 = 8 ; X is +ve and Y is +ve therefore XY > 0

Therefore Statement 1 is insufficient.

Statement 2 : -12y^2 – y^2x + x^2y^2 > 0

= y^2 ( -12 -X + X^2 ) > 0
= y^2 ( X^2 -X -12 ) > 0
= y^2 { (X-4) (X+3) } > 0

When X=4 then XY = 0...... Therefore X cannot be 4
When X = 5 then y^2 { (X-4) (X+3) } = y^2 * 8 ; X is +ve and Y can be +ve or - ve
When X = -4 then y^2 { (X-4) (X+3) } = y^2 * 8 ; X is -ve and Y can be +ve or - ve

Statement 2 is insufficient.

Taken together both are still insufficient.

Manager  S
Joined: 22 Jan 2014
Posts: 173
WE: Project Management (Computer Hardware)
Re: If x and y are integers and xy does not equal 0, is xy < 0?  [#permalink]

### Show Tags

Bunuel wrote:

Tough and Tricky questions: Statistics.

If x and y are integers and xy does not equal 0, is xy < 0?

(1) y = x^4 – x^3
(2) -12y^2 – y^2x + x^2y^2 > 0

E.

1) y = x^3(x-1)
x=2, y=8 ; xy=16 (+ve)
x=-1, y=2 ; xy=-2 (-ve)
so inconclusive.

2) y^2(x^2-x-12) > 0
y^2(x-4)(x+3) > 0
y^2 is always positive
so (x-4)(x+3) > 0
x>4 and x<-3
in one case xy is +ve and in other xy is -ve

(1)+(2) --> still inconclusive
_________________
Illegitimi non carborundum.
Intern  Joined: 06 Apr 2015
Posts: 4
Re: If x and y are integers and xy does not equal 0, is xy < 0?  [#permalink]

### Show Tags

thefibonacci wrote:
Bunuel wrote:

Tough and Tricky questions: Statistics.

If x and y are integers and xy does not equal 0, is xy < 0?

(1) y = x^4 – x^3
(2) -12y^2 – y^2x + x^2y^2 > 0

E.

1) y = x^3(x-1)
x=2, y=8 ; xy=16 (+ve)
x=-1, y=2 ; xy=-2 (-ve)
so inconclusive.

2) y^2(x^2-x-12) > 0
y^2(x-4)(x+3) > 0
y^2 is always positive
so (x-4)(x+3) > 0
x>4 and x<-3
in one case xy is +ve and in other xy is -ve

(1)+(2) --> still inconclusive

Hey "Thefibonacci" shouldnt it be x>4 and x>-3? instead of x<-3 ?
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7585
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: If x and y are integers and xy does not equal 0, is xy < 0?  [#permalink]

### Show Tags

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 and xy does not equal 0, is xy < 0?

(1) y = x^4 – x^3
(2) -12y^2 – y^2x + x^2y^2 > 0

In the original condition, there are 2 variables(x,y), which should match with the number equations. So, you need 2 equations. But, for 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. In 1) & 2), from y=x^3(x-1) and y^2(x^2-x-12)>0, y^2(x-4)(x+3)>0, y^2 is always a positive numver. When you divide the 2 equations, (x-4)(x+3)>0 becomes x<-3, 4<x. When x=-4, y=x^3(x-1)>0 --> xy<0, which is yes. When x=5, y=x^3(x-1)>0 -->xy>0, which is no and not sufficient. There fore the answer is E.

-> 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.
_________________
Math Expert V
Joined: 02 Aug 2009
Posts: 7764
Re: If x and y are integers and xy does not equal 0, is xy < 0?  [#permalink]

### Show Tags

nikdiablo129 wrote:
thefibonacci wrote:
Bunuel wrote:

Tough and Tricky questions: Statistics.

If x and y are integers and xy does not equal 0, is xy < 0?

(1) y = x^4 – x^3
(2) -12y^2 – y^2x + x^2y^2 > 0

E.

1) y = x^3(x-1)
x=2, y=8 ; xy=16 (+ve)
x=-1, y=2 ; xy=-2 (-ve)
so inconclusive.

2) y^2(x^2-x-12) > 0
y^2(x-4)(x+3) > 0
y^2 is always positive
so (x-4)(x+3) > 0
x>4 and x<-3
in one case xy is +ve and in other xy is -ve

(1)+(2) --> still inconclusive

Hey "Thefibonacci" shouldnt it be x>4 and x>-3? instead of x<-3 ?

Hi,
it will be x<-3 that is it takes the values -4,-5,-6,-7 and so on..
anything greater than -3, say -2 will make the eq y^2(-2-4)(-2+3)=y^2* -6*1 thus ,making it <0..
_________________
Non-Human User Joined: 09 Sep 2013
Posts: 11649
Re: If x and y are integers and xy does not equal 0, is xy < 0?  [#permalink]

### Show Tags

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________ Re: If x and y are integers and xy does not equal 0, is xy < 0?   [#permalink] 05 Aug 2018, 21:55
Display posts from previous: Sort by

# If x and y are integers and xy does not equal 0, is xy < 0?  