If p and q are positive integers, is 21^p/630^q a terminating

We'll simplify the expression so we understand what we need to do.
This is a Precise approach.

We'll extract all common factors between 21 and 630, creating an easier-to-understand expression.
\(\frac{(21)^p}{(630)^q}=\frac{(21)^p}{(21*30)^q}=\frac{(21)^{p-q}}{(30)^q}=\frac{(3*7)^{p-q}}{(3*10)^q}=\frac{(3)^{p-2q}*(7)^{p-q}}{(10)^q}\)
A terminating decimal is an integer divided by some power of 10.
Therefore, for the above to be a terminating decimal, \((3)^{p-2q}*(7)^{p-q}\) must be an integer.
As 3 and 7 are both prime numbers then this occurs only when p-2q and p-q are both positive. This happens only when p>2q.
Looking at our statements (1) lets us answer the question with a NO and (2) does not let us answer the question.

Our answer is (A).

*Note 1 - a terminating decimal is basically one long integer with the decimal point somewhere in the middle.
Therefore you can multiply it by a power of 10 to create an integer. This also means that every terminating decimal is an integer divided by a power of 10.
*Note 2 - A number divided by another number is an integer only if the denominator has the same prime factors as the numerator.
As this can never happen for powers of 3 and 7, they must both be in the numerator meaning that p-2q and p-q need to be positive.