Given Information:
- $p, q, r$ are positive integers.
- $q$ is a multiple of $p$ : This means $q=k p$ for some positive integer $k$.
- $r$ is a factor of $p$ : This means $p=m r$ for some positive integer $m$. (Equivalently, $p / r$ is an integer).

From these two conditions, we can also deduce a relationship between $q$ and $r$ :
Since $q=k p$ and $p=m r$, substituting $p$ into the first equation gives $\(q=k(m r)=(k m) r\)$.
Since $k$ and $m$ are integers, $k m$ is also an integer. So $q$ is a multiple of $r$.
This means $q / r$ is an integer.
Now let's examine each option:
(A) $\(\frac{q-p}{p}\)$
- Since $q=k p$, substitute this into the expression:

$$
\(\frac{k p-p}{p}=\frac{p(k-1)}{p}=k-1 .\)
$$

- Since $k$ is an integer (from $q=k p$ ), $k-1$ is also an integer.
- So, (A) is always an integer.

(B) $\(\frac{q-p}{r}\)$
- We know $q=k p$ and $p=m r$.
- So, $\(\frac{k p-p}{r}=\frac{p(k-1)}{r}\)$.
- Since $\(p / r=m\)$ (an integer), we can rewrite this as $\((k-1) \cdot \frac{p}{r}=(k-1) m\)$.
- Since $k$ and $m$ are integers, $(k-1) m$ is always an integer.
- So, (B) is always an integer.

(C) $\(\frac{q-r}{p}\)$
- We know $q=k p$. We also know $r$ is a factor of $p$, so $r$ is a factor of $k p$.
- We need to determine if $r / p$ is always an integer. We know $p / r$ is an integer, but $r / p$ is generally not.
- Let's test with an example:

Let $p=4$.
Then $r$ can be 1,2 , or 4 .
Let $r=2$. (So $p=2 r, m=2$ )
Let $q$ be a multiple of $p$. Let $q=1 \cdot p=4$. (So $q=k p, k=1$ )
Then $\(\frac{q-r}{p}=\frac{4-2}{4}=\frac{2}{4}=\frac{1}{2}\)$.
- Since $\frac{1}{2}$ is not an integer, this expression is not necessarily an integer.
(D) $\frac{q p^2}{r}$
- We know $q=k p$ and $p=m r$.
- Substitute $q$ and $p$ :

$$
\(\frac{(k p)(m r)^2}{r}=\frac{k p \cdot m^2 r^2}{r}=k p m^2 r\) .
$$

- Since $k, p, m, r$ are all integers, their product $k p m^2 r$ is always an integer.
- So, (D) is always an integer.

(E) $\(\frac{q+p r}{p}\)$
- Since $q=k p$, substitute this into the expression:

$$
\(\frac{k p+p r}{p}=\frac{p(k+r)}{p}=k+r \).
$$

- Since $k$ is an integer (from $q=k p$ ) and $r$ is an integer (given), $k+r$ is always an integer.
- So, (E) is always an integer.

Based on the analysis, option (C) is the one that is not necessarily an integer.

The final answer is (C)