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.
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(-5)^(-2)=\(\frac{1}{25}\) also 1/25>0

Remember when you square a negative number you get a positive number.


(5)^-2=\(\frac{1}{25}\), and again \(\frac{1}{25}\)>0 not less

when you have X^-Y, it's written out as \(\frac{1}{(X^Y)}\)

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?

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)
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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
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If \(y\neq{0}\), then y^0=1. y^x < 0 does not mean that y^x < (1 = y^0).
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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.
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is x> y?

State1:- x+y>0 .. the sum of two numbers is greater than zero does not tell us anything about how they rank relative to each other
state 2:- y^x<0 --> this implies that x must be odd and y<0 .. If y <0 then x can be -1, +1.
let y=-1. x=1 => y^x = -1 this satisfies
y=-1 . x=-1 => 1/-1^1 => -1 this also satisfies. so x < y and x> y can both betrue. So not sufficient
can y>0 .

combine the two statements y<0 and x+y>0 implies x has to be positive. Also note that since y<0 x+y>0 => x+y-y>0 => x>0. When the two opposite inequality, we can subtract, smaller from larger and get positive result.

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.
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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:

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