Let $f$ be a function defined on the set of positive rational numbers with the property that $f(a\cdot b)=f(a)+f(b)$ for all positive rational numbers $a$ and $b$. Suppose that $f$ also has the property that $f(p)=p$ for every prime number $p$. For which of the following numbers $x$ is $f(x)<0$?
$\textbf{(A) }\frac{17}{32} \qquad \textbf{(B) }\frac{11}{16} \qquad \textbf{(C) }\frac79 \qquad \textbf{(D) }\frac76\qquad \textbf{(E) }\frac{25}{11}$
From the answer choices, note that
\begin{align*} f(25)&=f\left(\frac{25}{11}\cdot11\right) \\ &=f\left(\frac{25}{11}\right)+f(11) \\ &=f\left(\frac{25}{11}\right)+11. \end{align*}
On the other hand, we have
\begin{align*} f(25)&=f(5\cdot5) \\ &=f(5)+f(5) \\ &=5+5 \\ &=10. \end{align*}
Equating the expressions for $f(25)$ produces \[f\left(\frac{25}{11}\right)+11=10,\] from which $f\left(\frac{25}{11}\right)=-1.$ Therefore, the answer is $\boxed{\textbf{(E) }\frac{25}{11}}.$
Remark
Similarly, we can find the outputs of $f$ at the inputs of the other answer choices:
\begin{alignat*}{10} &\textbf{(A)} \qquad && f\left(\frac{17}{32}\right) \quad && = \quad && 7 \\  &\textbf{(B)} \qquad && f\left(\frac{11}{16}\right) \quad && = \quad && 3 \\  &\textbf{(C)} \qquad && f\left(\frac{7}{9}\right) \quad && = \quad && 1 \\  &\textbf{(D)} \qquad && f\left(\frac{7}{6}\right) \quad && = \quad && 2 \end{alignat*}
Alternatively, refer to Solutions 2 and 4 for the full processes.
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