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DS - Inequalities

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DS - Inequalities

by spepakay » Mon Jan 03, 2011 5:14 am
What is the most efficient way of solving the below problem? Please provide step by step solution. Just to clarify, the ^ (caret) implies "to the power of". Thanks in advance.

Is A^B > B^A?

(1) A^A > A^B

(2) B^A > B^B
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Source: — Data Sufficiency |
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by goyalsau » Mon Jan 03, 2011 6:03 am
spepakay wrote:What is the most efficient way of solving the below problem? Please provide step by step solution. Just to clarify, the ^ (caret) implies "to the power of". Thanks in advance.

Is A^B > B^A?

(1) A^A > A^B

(2) B^A > B^B
We can not take negative fractions for A & B , Because Then it will be imaginary Numbers and we don't deal with those numbers in Gmat.

I would like to proof it with putting in numbers for A & B, Starting with +ve integers , -ve integers , +ve fractions and Zero.

It looks that we are in for a very lengthy solution. :P

(1) A^A > A^B

Let A = 4 , B = 3

A^B > B^A :arrow: 4^3 = 64 , 3^4 = 81 , So answer is No.

Let A = 3 , B = 2

A^B > B^A :arrow: 3^2 = 9 , 2^3 = 8 , Answer is Yes.

So insufficient.

(2) B^A > B^B

Let A = 3 , B = 2

A^B > B^A :arrow: 3^2 = 9 , 2^3 = 8 , Answer is Yes.

Let A = 4 , B = 3

A^B > B^A :arrow: 4^3 = 64 , 3^4 = 81 , So answer is No.

So insufficient. Again.

NOTE: By Putting in same values in both the same statements we get a yes or no. So even by combining the statement we get the same answer. So still insufficient.

MY personal suggestion. Always start with easy number like +ve integers. then go to - ve integers and only if required go to fractions.........
:wink:

My answer is E, What is the OA?
Saurabh Goyal
[email protected]
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EveryBody Wants to Win But Nobody wants to prepare for Win.
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by aleph777 » Mon Jan 03, 2011 11:23 am
Great question, and I'm finding more and more that the best way to deal with this sort of abstract problem is with a firm grasp of number properties.

IS A^B > B^A?


The first thing to notice is that the question doesn't pose any limitations on A or B, such as "both A and B are positive integers." This is a good clue from the start, because it means it will be harder to sufficiently answer the question if we can't confirm one or both are positive.

(1) A^A > A^B

First the plug-in-and-solve answer: Since we're dealing with the same base here, we know this means: A > B. But this doesn't give us any insight into the original question because if A = 3 and B = 2, then A^B > B^A is true, but if A =5 and B = 2, the question is false. Therefore: INSUFFICIENT.

There's another way to approach this, though. Since the question poses no limitation on what kind of numbers A and B are, we could simply think in terms of negative evens and odds.

Here's a great rule to remember for the GMAT: a negative number raised to an EVEN exponent becomes positive, but a negative number raised to an ODD exponent remains negative. That's because when multiplying or dividing, two negatives make a positive, but if you then multiply that positive by a negative, you get another negative. So without even thinking in terms of the numbers, we can't determine if A or B is pos or neg, nor can we determine if they're odd or even? If A = -3 and B = -2, then the statement and question are both true. But if A = -2 and B = -3, then the statement remains true, but the question becomes false. Therefore: INSUFFICIENT.

The second approach requires a bit more memorization in the beginning, but it's a steadfast rule that works in many different questions and cuts down on computation time.


(2) B^A > B^B
This is the same sort of statement as above, so the same rule applies here. Therefore: INSUFFICIENT.


But what happens when we look at the statements together? We know both statements are true, so what does that tell us about the questions?

Well, we still end up with insufficient information.

Let's say A = 3 and B = 2 again. Then the question is true, because 3^2 > 2^3.

But what if A = 2 and B = -3? Then both statement are true, but the question becomes false because 2^-3 < -3^2.

Therefore (E)
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