Here is an interesting question

1. a^b < b^a

2. a/b > 1

This is a good question because it helps us consider several possibilities for examples.

Let us start with Statement I alone:

1. a^b < b^a

Now, if a = 1, b = 100 a^b < b^a and a < b

If we can think of one example where a^b < b^a such that a > b, we can say this statement is insufficient.

If we have “b” as a negative number this could be very easy (especially if a were an even number). Let us say b = -2 and a = 2

2^(-2) < (-2)^2, but 2 > -2

So, statement I alone is insufficient.

Another counter example could just be 5^3 < 3^5, but 5> 3

Now, let us move on to statement II.

Statement II alone is clearly insufficient. If a, b are both positive then a/b > 1 => a > b.

If both are negative, it means the opposite.

So, statement II alone is not sufficient.

For instance a = 5, b = 3 satisfies this condition, a = -5, b = -3 also satisfies this condition.

Now, let us take both statements together

If a/b is greater than 1, then either both are positive or both are negative.

If both are positive then a has to be greater than b. a^b > b^a condition need not be combined and tested.

So, we can establish that the conditions together are not sufficient if we can find one example where a, b both are negative, and such that a < b and a^b > b^a.

If we can have b as an even number and a as an odd number, we should be through.

(-3)^(-2) > (-2)^(-3)

In this case, a < b.

So, taking the two statements together is still not sufficient.

Excellent question, purely because it makes one think of many examples. And THAT skill is very essential for DS. Students should be thinking along the lines of “Can I figure out a counter example for this”.

Just to recap.

a = 5, b = 3 satisfies both statements. And a > b

a = -3, b = -2 satisfies both statements. And a < b.

So, satisfying both statements we have two different combinations. So, we cannot answer the question even when both are used together .

Answer Choice E.

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