jtht wrote:
Bunuel wrote:
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.
(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.
Hello,
In statement 2, don't we have to also assume that x must be odd along with y being negative? It is not necessary to solve this problem here, but I would like to make sure that my logic is sound.
Thanks in advance!!
____________________
Yes, that's correct.
Statistics : Posted by Bunuel • on 07 Dec 2012, 08:48 • Replies 31 • Views 84709
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