If x and y are integers, is x y?

Question

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

(1) x + y > 0
(2) y^x < 0

Bunuel · Answer

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

(1) x + y > 0. Given that the sum of two numbers is greater than zero, but we cannot determine which one is greater. Not sufficient.

(2) y^x < 0. This statement implies that y is a negative number. Now, if y=-1 and x=1, then x>y BUT if y=-1 and x=-1, then x=y. Not sufficient.

(1)+(2) Since from (2) we have that y is a negative number, then -y is a positive number. Therefore from (1) we have that x>-y=positive, which means that x is a positive number. So, we have that x=positive>y=negative. Sufficient.

Answer: C.

abhi398 · Answer

Further explaining option c

from 2, we know y is a -ve number and to make x+y > 0 , x has to be positive to make the equation greater than zero .

so c is sufficient

KarishmaB · Answer

Question: Is X>Y?

1) x + y > 0 
Sum of x and y is positive. We don't know which one is greater. Not sufficient.

2) y^x < 0
y to the power x is negative. The power cannot change the sign of the expression. If the expression is negative, it is imperative that y is negative (and x is odd). Whether x is negative or positive doesn't matter. 
Here we know that y is negative but x could be positive or negative so we don't know which of x and y is greater. Not sufficient.

Using both together, you know that y is negative but x+y is positive. When will this be possible? Only when x is positive and with absolute value greater than y. 
So x is positive but y is negative. This means x is greater than y.

Answer (C)

mulhinmjavid · Answer

From 2, why we have decuded that y is a -ve number? y^x= 3^-3 is also less than zero and in this case y is positive. Can you please explain?

Bunuel · Answer

3^{(-3)} = \frac{1}{3^3} = \frac{1}{27} > 0 , not less than 0.

BrainLab · Answer

Question is x>y ?

(1) x+y>0 we can not determine wether x>y or not, but this statement says that one of the vaiables must be positive to satisfy it --> Not sufficient
(2) Y^x<0 the only way it to be negative is when Y is negative, but X could be also negative (-2^-3 = -1/8) --> Not Sufficient, as we don't know wether x>y

(1)+(2) Statement 1 says that one of the variables must be positive + Statement 2 says that Y is in all cases negative --> if Y is negative X must be positive and is > Y

Answer (C)

naeln · Answer

Would anybody please explain why we ruled out statement 2. I thought the that the only was for y^x<0 is for y to be negative and X to be an odd power. I ruled out the negative values of X thinking that if X is negative then y would be a negative fraction and the question stem says that y is an integer. Why was I wrong here? Please help!

EgmatQuantExpert · Answer

Hi  naeln ,

From statement-II, we can deduce that  y  is a negative integer but we can't say if  x  is a negative or a positive odd integer. Let's evaluate both the cases:

Case-I:  x  is positive  
If  x  is a positive odd integer, then  y^x < 0 . For example, assuming  y = -2  and  x = 3  would give  y^x = -2^3 = -8 < 0

Case-II:  x  is negative  
When  x  is negative,  y^x  would become a negative fraction and not  y  itself.

For example: if  y = -2  and  x = -3 , then  x  &  y  both are integers and  y^x = -2^{-3} = \frac{-1}{8} < 0 . Here  y^x = \frac{-1}{8}  is a fraction and not  y  itself.

Hope its clear why statement-II does not give us a unique answer.

Regards
Harsh

Dienekes · Answer

Thanks for the explanation. I got the correct answer using the same approach, however at the beginning of my solution I tried a different way to understand statement 2: Since  y^0=1  and  0<1 , so if  y^x<0 , it should mean that  y^x<y^0  and hence  x<0 . I know this is not correct but I am unable to understand what am I doing wrong here.
Can you please help confirm that?

Thanks again!

Bunuel · Answer

If  y
eq{0} , then y^0=1. y^x < 0 does not mean that y^x < (1 = y^0).

kjarrah · Answer

(1) x+y>0 : For this to be true, we will have the following 3 senarios: a) x > 0 and y > 0 b) x > 0 and y <= 0 , which means x > y c) y > 0 and x <= 0 , which means y > x Therefore, this statement is Insufficient (2) y^x < 0 : For this to be true, y must be less than 0 and x must be an odd integer . Therefore, this statement is insufficient. (1) and (2) together: From statement (2), we know y < 0 , therefore the only viable scenario for this in statement (1) is option (b): x > 0 and y < 0 , which means x > y => Sufficient. C

MathRevolution · Answer

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, is x > y?
(1) x + y > 0
(2) y^x < 0

There are two variables (x and y) in the original condition. In order to match the number of variables and the number of equations, we need 2 equations. Since the condition 1) and 2) each has 1 equation, there is high chance that C is going to be the answer.
Using both the condition 1) and 2), we know that the condition 2) states y^x<0, which means y<0 and x=odd. Also, the condition 1) states x+y>0, which means y<0. So, x>0 and it is always the case that x>y. The answer is ‘yes’ and the conditions are sufficient. Therefore, since the condition 1) and 2) are not sufficient alone, the correct answer is C.

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.

Manonamission · Answer

How to determine whether the variable approach will it into a particular Q?
There are cases where at first glance ,it seems two options will convert to two equations but they turn out to be a single equation. 
Is there any pattern which I can look into ?

MathRevolution · Answer

Hi, 
Since x>y? is a question, we figured out the values of x and y.
Also, in a case where 1)=2) and there is 1 equation, it is 95% that D is an answer and 5% that E is an answer.
Thank you.

EMPOWERgmatRichC · Answer

Hi All,

We're told that X and Y are integers. We're asked if X > Y. This is a YES/NO question. We can solve it with a mix of Number Properties and TESTing VALUES.

1) X+Y > 0

With this Fact, we know that at least one of the two variables is positive. 
IF.... X = 1, Y = 0 then the answer to the question is YES. 
IF.... X = 0, Y = 1 then the answer to the question is NO. 
Fact 1 is INSUFFICIENT

2) Y^X < 0

With this Fact, we know that Y CANNOT be positive (since a positive raised to any integer power is a positive). Thus, Y MUST be NEGATIVE and X must be ODD (since raising a negative to an EVEN power will end in a positive result).

IF.... X = 1, Y = -1 then the answer to the question is YES. 
IF.... X = -3, Y = -1 then the answer to the question is NO. 
Fact 2 is INSUFFICIENT

Combined, we know that there must be at least one positive value and that Y must be negative. By extension, that means that X MUST be positive - and the answer to the question is ALWAYS YES. 
Combined, SUFFICIENT

Final Answer:  C

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

fskilnik · Answer

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) \(x \leqslant y\)

The contradiction follows:
\(\left( 1 \right)\,\,\,\, \Rightarrow \,\,\,\,\,y < 0\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( 3 \right)} \,\,\,x < 0\,\,\,\,\,\,\, \Rightarrow \,\,\,x + y < 0\,\,\,\,\, \Rightarrow \,\,\,\,\left( 2 \right)\,\,{\text{contradicted}}\)

proabhinav · Answer

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 ?

??

Bunuel · Answer

x + y < 0

Subtract y from both sides: (x + y) - y < 0 - y;

x < -y.

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EgmatQuantExpert · Answer

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

For more information on Inequalities, go through the following articles

Different methods to solve absolute value equations and inequalities

Solving inequalities- Number Line Method

Wavy Line Method Application - Complex Algebraic Inequalities     
 
  https://e-gmat.com/blogs/wp-content/uploads/2018/08/Ability-Quiz-Algebra-e1533292163516.png

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

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