Quantity $A$ is greater than Quantity $B$.
Algebraic solution
Given $\(\frac{p+q}{p}=\frac{10}{7}\)$, divide numerator by $p$ :

$$
\(\frac{p+q}{p}=1+\frac{q}{p}=\frac{10}{7} \Rightarrow \frac{q}{p}=\frac{10}{7}-1=\frac{3}{7}\) .
$$


So $\(q=\frac{3}{7} p\)$.
Now compute Quantity A:

$$
\(\frac{p-q}{q}=\frac{p-\frac{3}{7} p}{\frac{3}{7} p}=\frac{\frac{4}{7} p}{\frac{3}{7} p}=\frac{4}{3}\) .
$$


Since $\(\frac{4}{3}>\frac{5}{3}\)$ is false, correct that: actually $\(\frac{4}{3}<\frac{5}{3}\)$, so Quantity $A$ is less than Quantity $B$.