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Source: — Data Sufficiency |
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by rijul007 » Sat Oct 22, 2011 11:30 am
Statement 1
for y>0
x^x is greater

for y<0
y^y is greater

Statement 1is not sufficient


Statement 2
x^y > x

x>1
y>1

but from this statement alone dsnt tell us whether x is greater or y is greater.
Statement 2is also not sufficient


Combining Statement 1 and 2
x=y+1
x>1
y>1

x^x > y^y

Therefore, C is the correct option.[Both the statements are required together]
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by mankey » Mon Oct 24, 2011 11:12 am
Someone please explain this.

Thanks.
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by rijul007 » Mon Oct 24, 2011 11:34 am
mankey wrote:Someone please explain this.

Thanks.
Statement 1 says
x = y + 1


let us say y = 0
then x = 1

x^x = 1^1
y^y = 0^0
nothing can be said about this as 0^0 is indefinite
not sufficient

statement 2
x^y > x and x is +ve

x^y > x [this is true only when x > 1]
but we need value of y true tell whether x^x is greater or y^y is greater..
not sufficient


combining both the statements
we know that
x > 1 [from statement 2]
y > 0 and x > y [because x=y+1]

from this we can conclude that x^x > y^y

Hence both the statements together are required to solve the ques.



I hope this is clear enough now :)
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by saketk » Mon Oct 24, 2011 12:49 pm
GmatKiss wrote: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.
use numbers -- make the question easy, at least in my case


Stmt 1 -- X= y+1

let X =2
then, Y = 1

here X^X = 2^2 = 4
Y = 1^1= 1

x^x > y^y

now, let X = -2

then, y = -3

X^X = -2^-2 = - (1/4) = - 0.25

y^y = -3^-3 = - (1/27) = -0.037

here x^x < y^y


Hence, statement 1 is insufficient


Statement 2 --

X^y> X and X is positive

Let X = 3
y = 2

X^y = 3^2 = 9>3

x^x > y^y

Let X = 2
and y = 3

X^y = 2^3 = 8>2

x^x < y^y

Hence, Statement 2 is insufficient

Combine the two statement --

We know that x= y+1
and x^y > x

NOw, the important thing here is X cannot be 1

because in that Case y = 0 and 1^0 = 1 which is equal to 1 . This violates the second statement

also, X cannot be = 2 because in that case y = 1

and 2^1 = 2 which again is not in agreement with the second statement

Now, if X = 3, Y will be 2 .. likewise -- we now know that the value of Y will be 1 less that X. Therefore

X^X> Y^Y (always)

Hence, the answer should be OPTION C
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