Is a

Question

Is a < b?

(1) a^b < b^a
(2) a/b > 1

OptimusPrepJanielle · Answer

Statement 1: a^b < b^a Considering values a = 1, b = 2 a^b = 1 and b^a = 2 In this case a < b a = 2, b = -1 a^b = 1/2 and b^a = 1 Here a > b INSUFFICIENT Statement 2: a/b > 1 If a > 0, then a > b If a < 0, then a < b INSUFFICIENT Combining: a^b < b^a and a/b > 1 a > 0 Assume a = 2, b = 1 a < 0 Assume a = - 2, b = -1 Still nothing can be said about the relation between a and b INSUFFICIENT Option E

stne · Answer

When taking the combined case, if a=2 and b=1 then the first statement is violated. Please can you show , how you arrived at E

ArtyomYom · Answer

1. a^b < b^a Let's structure possible variants: 1.1 ab, let's prove that this variant exists a=4 b=3 than a^b < b^a hold true (64<81) So, this condition is INSUFFICIENT 2. a/b > 1 Let's structure possible variants: 2.1 a>0, b > 0 => a>b 2.2 a<0, b < 0 => ab exists because of logic in point 1.2 and 2.1 BUT we don't know about ab or a

CrackverbalGMAT · Answer

This is a DS question where, although the question stem is quite straightforward in what it asks, the statements are tricky to interpret. The best approach, therefore, is to plug in simple values and check. From statement I alone, we know a^b < b^a . If a = 2 and b = 3, 2^3 < 3^2 is true. Also, ab and we get a NO as an answer to the main question. Statement I alone is insufficient. Answer options A and D can be eliminated. Possible answers are B, C or E. Statement II alone says a/b > 1. A common mistake that some students do when they see this inequality is that they cross multiply b with 1 and conclude a>b. This is wrong because we do not know the signs of a and b. a/b > 1 can mean two things: Both a and b are positive and a>b. Both a and b are negative and hence ab. If a = -2 and b= -1, (-2)^{-1} = - \frac{1}{2} and (-1)^{-2} = 1. Naturally, a^b (negative) < b^a (positive). But, here a

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Is a < b? (1) a^b < b^a (2) a/b > 1

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