Is A^B>B^A?
(1)A^A>A^B
(2)B^A>B^B
OA is E. Detailed explanations would be appreciated. Thanks
Confusing Problem
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Statements 1 and 2 tell us the exact same thing, so we can automatically narrow our choices to D and E.
In statement 1:
Plugging 2 in for A we get 2^2>2^B, so B<2 or simply B<A.
However, plugging in -2 for A we get (-2)^(-2)>(-2)^B, in this case B = -1 and B = -3 both satisfy the equation.
We cannot discern a relation between A and B from statement 1 alone.
Statement 2 yields similar results. Both statements are insufficient alone, and provide no new information together.
Answer is E
In statement 1:
Plugging 2 in for A we get 2^2>2^B, so B<2 or simply B<A.
However, plugging in -2 for A we get (-2)^(-2)>(-2)^B, in this case B = -1 and B = -3 both satisfy the equation.
We cannot discern a relation between A and B from statement 1 alone.
Statement 2 yields similar results. Both statements are insufficient alone, and provide no new information together.
Answer is E
knight247 wrote:Is A^B>B^A?
(1)A^A>A^B
(2)B^A>B^B
OA is E. Detailed explanations would be appreciated. Thanks
- navami
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Is A^B>B^A?
(1)A^A>A^B
(2)B^A>B^B
1)
a>b
2)
a>b
so both the statement yeilds the same information a>b
now consider a = 4 b = 2
A^B>B^A ???
A^B = 16
B^A = 16
so A^B>B^A = FALSE
then a = 3 b = 2
A^B>B^A ???
A^B = 9
B^A = 8
so A^B>B^A = TRUE
so we can not conclude even after combining these two.
hence E
(1)A^A>A^B
(2)B^A>B^B
1)
a>b
2)
a>b
so both the statement yeilds the same information a>b
now consider a = 4 b = 2
A^B>B^A ???
A^B = 16
B^A = 16
so A^B>B^A = FALSE
then a = 3 b = 2
A^B>B^A ???
A^B = 9
B^A = 8
so A^B>B^A = TRUE
so we can not conclude even after combining these two.
hence E
This time no looking back!!!
Navami
Navami
- prateek_guy2004
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Well yes this was little bit confusing... i spended around 4 mins.....Is better spend little time rather than confusing yourself.....Finally found answer should be E.