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18 Feb 2012, 09:33
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
64% (02:07) correct
36%
(01:57)
wrong
based on 36
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If x and y are both integers, which is larger, x^x or y^y?
(1) x = y + 1
(2) x^y > x and x is positive.
(1) x = y + 1
(2) x^y > x and x is positive.
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18 Feb 2012, 11:38
Kudos
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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.
Knowing that \(x>0\), \(x^y>x\) is only possible if \(y>1\) . Please note even when \(y=0\) , \(x>0\) and \(x\) is an integer. So now we know that \(y\) is positive and \(x\) is positive but we do not know which is larger. Hence Insufficient
Combined:
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
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.
Knowing that \(x>0\), \(x^y>x\) is only possible if \(y>1\) . Please note even when \(y=0\) , \(x>0\) and \(x\) is an integer. So now we know that \(y\) is positive and \(x\) is positive but we do not know which is larger. Hence Insufficient
Combined:
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
18 Feb 2012, 12:08
Kudos
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omerrauf
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.
