In which one of the following choices must p be greater than q?

This question is tricky. You might end up losing lots of valuable time trying to plug in numbers in each option and getting lost mid-way. The trick is to use exponents theory. I like to use number plugging only where it is apparent that it is useful (e.g. in options A and B above to rule them out in 10 sec) or where I can't think of anything else. The specific theory which helps in dealing with this and similar questions:

If the base is positive, no matter what power you give to it, it will not become negative.
e.g. 5^n must be positive no matter what the value of n is.
The reason is that 5^n is just 5 multiplied by itself n number of times, whatever n is. So this number can never be negative.

Also,
If n is positive, 5^n > 1
If n = 0, 5^n = 1
If n is negative, 0 < 5^n < 1

Another thing, the positive base, a, can lie between one of two ranges - 'between 0 and 1' and 'greater than 1'. We saw what happens if the base is greater than 1 above (in 5^n example)

Let's see what happens when the base lies between 0 and 1:
If n is positive, a^n < 1
If n = 0, a^n = 1
If n is negative, a^n > 1

These relations hold the other way around too. If a is between 0 and 1, and a^n > 1, n must be negative etc.
Plug in values to understand them.

Now look at this question:

In these three options, each term is positive so we can take the term on the right to left hand side so that we have to deal with a single base.

(C)\(0.9^p > 0.9^q\)
\((0.9)^{p-q} > 1\)
Here, p-q must be negative i.e. p-q < 0 or p<q (from red above)

(D) \(9^p < 9^q\)
\(9^{p-q} < 1\)
p-q < 0 (from blue above) or p < q

(E)\(9^p > 9^q\)
\(9^{p-q} > 1\)
So p-q > 0 (from green above) or p > q