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.
x and y
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- rijul007
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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]
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]
- rijul007
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Statement 1 saysmankey wrote:Someone please explain this.
Thanks.
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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use numbers -- make the question easy, at least in my caseGmatKiss 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.
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