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Numerator is 1. For the fraction to be >1, denominator must be <1, but it cannot be negative or zero.
So denominator must lie between 0 and 1.
So the question is basically asking whether 0 < (p^q−q^p) < 1

Statement 1. q^p < p^q
Or p^q - q^p > 0... but we don't know whether its less than 1 or not. So Insufficient.

Statement 2. p^q < 1 + q^p
Or p^q - q^p < 1 but we don't know whether is greater than 0 or not. So Insufficient.

Combining the two statements, we get 0 < p^q - q^p < 1. So Sufficient.
Hence C answer
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Official Solution



Steps 1 & 2: Understand Question and Draw Inferences

To find:

    • Is \(\frac{1}{{p^q−q^p}}> 1\)

Let us try to understand what this means:

    • Consider \((p^q - q^p)\) to be \(X\).

So the question becomes:

    • Is \(\frac{1}{X} > 1\)……………………………………………..(Ineq 1)

This is possible only when

    • \(0 < X < 1\)

Therefore,

    • If \(0 < (p^q - q^p ) < 1\) then

    • \(\frac{1}{{p^q - q^p}} > 1\) and the answer to the question will be a definite YES

So we need to tell if

    • \(0 < (p^q - q^p ) < 1\)

We need to get definite YES or a definite NO to answer this question.

Step 3: Analyze Statement 1 independently

    • \(q^p < p^q\)

Subtracting q^p from both sides we get:

    • \(0 < p^q - q^p\)

We do not know if \(p^q - q^p < 1\)

Therefore, Statement 1 alone is NOT sufficient to answer this question.


Step 4: Analyze Statement 2 independently

    • \(p^q < 1 + q^p\)

Subtracting qp from both sides we get:

    • \(p^q - q^p < 1\)

    • We do not know if \(p^q - q^p > 0\)

Therefore, Statement 2 alone is NOT sufficient to answer this question.


Step 5 – Analyse both Statements together:

From Statement 1 we got:

    • \(0 < p^q - q^p\)

From Statement 2 we got:

    • \(p^q - q^p < 1\)

Combining both statements we get:

    • \(0 < (p^q - q^p ) < 1\)

Correct Answer: Option C


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Saquib
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There are two reasons why test takers find inequality questions tricky:

1. The equations given are sometimes too complex to analyze
2. Test takers resort to plugging in values and forget to test for all types of values

Plugging in values in any inequality question should always be the last resort when solving questions. The cardinal rule that every student should keep in mind while solving inequalities is 'ALWAYS BREAKDOWN THE QUESTION STEM'. This rule will serve you really well on test day.

There are two stages to breakdown a question stem when dealing with a question on inequalities, these help in better analysis and should be treated as hygiene factors while solving inequalities:

1. Always make sure you keep the RHS of the inequality 0
2. Simplify the LHS to a product or division of values


These 2 steps of simplification makes our analysis simpler.
For e.g. if xy > 0, then we can clearly say that x and y both have to be of the same sign. We need only to worry about the signs of x and y and nothing else.
But if x + y > 0, then the analysis becomes more trickier, in addition to the signs of x and y, we also need to worry about the magnitudes.

The main point here is just that multiplication and division of values are easier to analyze than addition or subtraction. So always simplify the LHS of the inequation to a product or a division.

With these points in place, lets now get to the question.

Is 1/(p^q - q^p) > 1

Subtracting 1 on both sides we get,

1/(p^q - q^p) - 1 > 0

Simplifying the LHS to a product or division of values we get,

(1 - p^q + q^p)/(p^q - q^p) > 0

For this equation to hold, we need the numerator and denominator to have the same sign.

Statement 1 : q^p < p^q

q^ p < p^q -----> q^p - p^q < 0.

This tells us that the denominator is negative. The numerator can be positive or negative, since q^p - p^q can be a negative integer < -1, in which case we get a NO, or a negative fraction, in which case we get a YES. Insufficient.

Statement 2 : p^q < 1 + q^p

p^q < 1 + q^p -----> 0 < 1 + q^p - p^q ----> 1 + q^p - p^q > 0.

So clearly the entire numerator is positive, but we do not have any information on the sign of the denominator. Insufficient.

Combining the statements, we have the numerator to be positive and the denominator to be negative, this gives us a definite NO. Sufficient.

Hope this helps!

Aditya
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