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

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Joined: 02 Sep 2009
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Is a < b? (1) a^b < b^a (2) a/b > 1  [#permalink]

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16 Dec 2019, 01:26
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Difficulty:

75% (hard)

Question Stats:

41% (02:24) correct 59% (01:52) wrong based on 37 sessions

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Is a < b?

(1) $$a^b < b^a$$

(2) $$\frac{a}{b} > 1$$

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Is a < b? (1) a^b < b^a (2) a/b > 1  [#permalink]

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16 Dec 2019, 13:03
Statement 1:
We know nothing on a or b yet, we cannot simplify this to a < b so cannot prove a < b. Some examples we can use here:
$$2^3 < 3^2$$ => $$a < b$$ . Yes.
$$5^2 < 2^5$$ => $$a > b$$. No.

Statement 2:
This tells us a and b have the same sign. However, we cannot simplify this expression. Insufficient.

Combined:
Positive a and b give a > b from (2). Negative a and b give a < b from (2). We just need to find a case for each scenario that satisfies (1) to say insufficient.
For the positive case, we can recycle $$5^2 < 2^5$$. For the negative case, we can let $$b$$ be a negative odd number and $$a$$ be a negative even number.
$$(-4)^{-3} < (-3)^{-4}$$ for example will satisfy (1) and also give us a < b. Therefore combined it is still insufficient.

Ans: E

Bunuel wrote:
Is a < b?

(1) $$a^b < b^a$$

(2) $$\frac{a}{b} > 1$$

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Re: Is a < b? (1) a^b < b^a (2) a/b > 1  [#permalink]

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27 Dec 2019, 06:08
Option B tells us that a>b as a/b >1, hence answer should be b.
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Re: Is a < b? (1) a^b < b^a (2) a/b > 1  [#permalink]

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27 Dec 2019, 10:49
1
Sejal16 wrote:
Option B tells us that a>b as a/b >1, hence answer should be b.

a = -6 and b = -3
then a/b >1 but a is actually less than b
whereas
a = 6 and b =3
here a is greater than b
hence insufficient
Hope it helps
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Re: Is a < b? (1) a^b < b^a (2) a/b > 1   [#permalink] 27 Dec 2019, 10:49
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