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Re: Which is larger, x^x or y^y? [#permalink]
18 Feb 2012, 10:38
2
This post received KUDOS
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 _________________
"Nowadays, people know the price of everything, and the value of nothing."Oscar Wilde
Re: Which is larger, x^x or y^y? [#permalink]
18 Feb 2012, 11:08
Expert's post
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 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
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.
Re: If x and y are both integers, which is larger, x^x or y^y? [#permalink]
18 Feb 2012, 20:05
Had already corrected that before you wrote bunuel. While writing the solution I mistakenly took \(x^y-x\) for \(xy-x\) but corrected it when i was reading my solution after I had posted. Thankyou anyways! _________________
"Nowadays, people know the price of everything, and the value of nothing."Oscar Wilde
Re: Which is larger, x^x or y^y? [#permalink]
27 Dec 2013, 09:09
Bunuel wrote:
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 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
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.
Just curious, why not a GMAT style question?
Thanks
Cheers! J
gmatclubot
Re: Which is larger, x^x or y^y?
[#permalink]
27 Dec 2013, 09:09
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