omerrauf wrote:
Answer should be
C. Here is how:
If \(x\) and \(y\) are both integers, which is larger, \(x^x\) or \(y^y\) ?
Statement A: \(x=y+1\)
So \(x\) and \(y\) are consecutive integers. Remember they can be positive, negative or \(0\) (from the question stem).
Suppose \(y=1\) and \(x=2\) , then \(x^x\) is larger , but suppose \(y=-2\) and \(x =-1\) then \(y^y\) is larger.
Hence Insufficient.
Statement B: \(x^y>x\) and \(x\) is positive.
Let's re-arrage this a bit.
\(x^y>x\) so \(x^y-x>0\) so
x*(y-1)>0 and we know that x is +ve
Now in order for \(x*(y-1)>0\) the factor \((y-1)\) has to be \(>0\) so we know that \(y>1\) but we do not know which is bigger, \(x\) or \(y\).
Hence InsufficientCombined:
From Statement 2 we know that \(x\) and \(y\) are both \(>0\) and from Statement 1 we know that x is bigger. So YES. \(x^x\) is bigger than \(y^y\)
Sufficient. Hence Answer
C Seriously Kudos Hungry
The red part is not correct.
If x and y are both integers, which is larger, x^x or y^y? (1) x = y + 1 --> if \(y\) is positive integer then \(x^x=(y+1)^{y+1}>y^y\) but if \(y=-2\) then \(x=-1\) and \(x^x=-1<\frac{1}{4}=y^y\)
(2) x^y > x and x is positive --> since \(x\) is positive then \(x^{y-1}>1\) --> since \(x\) and \(y\) are integers then \(y>1\). If \(x=1\) and \(y=2\) then \(x^x<y^y\) but if \(x=3\) and \(y=2\) then \(x^x>y^y\). Not sufficient.
(1)+(2) From (2) \(y>1\), so it's a positive integer then from (1) \(x^x=(y+1)^{y+1}>y^y\). Sufficient.
Answer: C.
P.S. Not a GMAT style question.