Q:

Rate of flow of electrons in a simple circuit, given the current?

I have this question:

The correct answer is C.
I understand that the flow of electrons should be from Y to X because electrons flow opposite to "conventional current", which is from positive to negative.
However, I can't figure out how the rate was found out.
The formula is 
$$Q = It$$
where $Q$ is the charge, $I$ the current and $t$ the time for which the current is considered.
I know that the charge of a single electron is $1.6 \times {10^{-19}}$. And that the current is 4.8 A. But where do I go from there? Time in this case will only be 1 (for 1 second). I can see clearly that the answer was obtained by dividing: $\frac{4.8}{1.6 \times {10^{-19}}}$ But why, I don't understand. The formula for rate should simply be $$(\text{Total charge}) / (\text{one second})$$ but to get the given answer I rearrange the charge formula around, which just gives $ \frac{1}{t} = \frac{I}{Q}$ which doesn't make sense.
I know this is extremely simple but I don't know what I'm missing.

A:

Think about what exactly the rate of flow of electrons is? It is the number of electrons per second passing through! The number of electrons is not included in your expression, and that's the problem.
Let start over but this time with the number of electrons $n$ included:
$$Q=It \Leftrightarrow \\
en=It \Leftrightarrow\\
\frac{n}{t}=\frac{I}{e} \Leftrightarrow\\
$$
Total charge $Q$ is the sum of all the $n$ electron charges $e$, so $Q=ne$.
$n/t$ is number of electrons ($n$) per second ($t$), exactly what you are looking for. This is clearly equal to $I/e$ as you guessed.
Of course if you divide through with $n$ you get the $\frac{1}{t}=\frac{I}{ne}=\frac{I}{Q}$ that you found.

