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

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.

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..

@bunuel hi Bunuel can you please advise on below

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[/quote]

E.

1) y = x^4-x^3
If x is positive integers x^4>x^3 hence y = positive
If x is negative x^4 will be positive and greater than x^3

=> Y = +ve

x^3 will be negative thus it will be y= value of x^4 + value of x^3
=> Y = +ve

But we don't know x

Not sufficient

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

y^2 is always positive y can be +ve or -ve it alone should be enough to say statement not sufficient to say if XY<0

so from statement 1 and 2 we are left with (x-4)(x+3) > 0 and y= +ve

X>4 or x>-3

If x = -2 then statement 2 become y^2*-6*1< 0

But this is against the given statement 2

Infact x cannot be anything between -3 to 4 x has to be greater than 4 for statement to be true

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


(1) \(y = x^4 – x^3\):

The question becomes:

Is \((x^4 – x^3)x < 0\)?
Is \((x – 1)x^4 < 0\)?

Reduce by x^4 (the square of a number is nonnegative and since we are told that x does not equal to 0, then in this case it's positive):

Is \((x – 1) < 0\)?
Is \(x < 1\)? (\(x \neq 0\))

Since we cannot answer the question whether x < 1, then this statement is not sufficient.

(2) \(-12y^2 – y^2x + x^2y^2 > 0\):

Reduce by y^2 (the square of a number is nonnegative and since we are told that y does not equal to 0, then in this case it's positive):

\(-12 – x + x^2 > 0\);

\(x^2 - x - 12 > 0\);

Factor:

\((x + 3)(x - 4) >0\)

The roots are x= -3 and x = 4. ">" sign indicates that the solution is to the left of the smaller root and to the right of the larger root. Thus x < -3 or x > 4. Check Solving Quadratic Inequalities - Graphic Approach for more on this: https://gmatclub.com/forum/solving-quad ... 70528.html

We cannot answer the question whether xy < 0.

Not sufficient.

(1)+(2) From (1) the question became: is x < 1. From (2) we got that x < -3 or x > 4. So, x may or may not be less than 1. Not sufficient.

Answer: E.

Bunuel KarishmaB

Statement 1: \(y = x^4 – x^3\)

This equation is insufficient to be solved by itself. We would need at least two equations. Plus, we cannot look to solve for XY without increasing the power of X further. Is plugging in a number even necessary? I want to understand a more logical way (without getting bogged down by 4th power of x) to deem stmt 1 Insufficient.

I spent time on statement 1 and later decided this is insufficient for above reason. Thanks for your response.

Question: Is xy < 0?
which means 'Is xy negative?' which means do they have opposite signs or same signs? (They can't be 0).

(1) \(y = x^4 – x^3\)

Easy to give them same signs. For all numbers greater than 1 on the number line, x^4 > x^3. So if x is greater than 1, both x and y are positive.

For opposite signs, I want x to be positive and x^4 < x^3. I know this happens when 0 < x < 1. So here x is positive and y is negative.

Statement not sufficient alone.

(2) \(-12y^2 – y^2x + x^2y^2 > 0\)

y^2 is eliminated and we are left with a quadratic in x. It will give us ranges for x but we don't know anything about the sign of y.
Not sufficient alone.

Using both statements,
Second statement gives us x < -3 or x > 4.
We know that when x > 4, x and y both will be positive (we discussed this in statement 1).
When x = -4, y is positive (recall that numbers between 0 to 1 often behave similarly to numbers less than -1; not always though)
Hence both x and y can have opposite signs too.

Answer (E)


Check this post for how to use number line for DS questions: https://anaprep.com/data-sufficiency-us ... roperties/