Is r/s^2 a terminating decimal?

Several questions have been posted about terminating decimals lately. Below is the theory about this issue:

Theory:
Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be the terminating decimal.

We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.

Now:

For (1) \(\frac{r}{s^2}=\frac{r}{225^2}=\frac{r}{9^2*5^4}\), we can not say whether this fraction will be terminating, as 9^2 can be reduced or not.

(2) is clearly insufficient.

(1)+(2) \(\frac{r}{s^2}=\frac{9^2}{9^2*5^4}=\frac{1}{5^4}\), as denominator has only 5 as prime, hence this fraction is terminating decimal.

Answer: C.