Calculating numbers of moles at equilibrium given the volume of mixture and initial concentration of reactants
Following a discussion that took place in the comments:

If given answers are correct, as Poutnik pointed out, your equation should be :
$$\require{cancel}\ce{A(g) + B(g)&lt;=&gt;2 C(g)}$$ - Mathew Mahindratne
Could you explain how you arrived at that conclusion? - Ollie


This answer solves this question for a general case:
$$\ce{a A(g) + b B(g) &lt;=&gt; c C(g)}$$
Let's assume that initial concentrations of A, B, C are $[A_0]$, $[B_0]$ and $[C_0]$ and the final concentration be $[A]$, $[B]$ and $[C]$.
Now, the given variables are $[A_0]$, $[B_0]$, $[C_0]$ and $[C]$, assuming similar data to the question. Let $\mu$ be change for every mole of C
\begin{align}
\therefore\ &amp;\text{Change}=[C]-[C_0] \tag{1}\\
\implies &amp;\text{Change/mol of C} = \frac{[C]-[C_0]}{c}\tag{2}
\end{align}
Now, building a RICE table we get:
\begin{array}{lc} \hline
 &amp;\ce{a A &amp;+ &amp;b B &amp;&lt;=&gt; &amp;c C} \\ \hline
I &amp; [A_0] &amp;&amp; [B_0] &amp;&amp; [C_0] \\
C &amp; -a\cdot\frac{[C]-[C_0]}{c} &amp;&amp; -b\cdot\frac{[C]-[C_0]}{c} &amp;&amp; +c\cdot\frac{[C]-[C_0]}{c} \\
E &amp; [A_0] - a\cdot\frac{[C]-[C_0]}{c} &amp;&amp; [B_0] - b\cdot\frac{[C]-[C_0]}{c} &amp;&amp; [C_0] + [C]-[C_0] \\ \hline
\end{array}
Therefore, the final concentrations of A and B would be:
\begin{align}
[A] &amp;= [A_0] - a\cdot\frac{[C]-[C_0]}{c} \\
[B] &amp;= [B_0] - b\cdot\frac{[C]-[C_0]}{c} 
\end{align}

Now, onto the specific solution, how to figure out $a$,$b$ and $c$ given the above convoluted generalisation?
Point 1: The change in $[A]$ and $[B]$ are both equal. Looking at change in the above RICE table, we deduce:
\begin{align}
a\cdot\cancel{\frac{[C]-[C_0]}{c}}&amp;=b\cdot\cancel{\frac{[C]-[C_0]}{c}} \\
\implies a&amp;=b
\end{align}
Point 2: We notice that for the change of $\pu{0.04 mol}$ of C, we observe a change of $\pu{0.02 mol}$ in A (or B, both work). Similar to point one, we equate the equations of change of A and C.
\begin{align}
2a\cdot\cancel{\frac{[C]-[C_0]}{c}}&amp;=c\cdot\cancel{\frac{[C]-[C_0]}{c}} \\
\implies 2a&amp;=c
\end{align}
Now, we substitute the values of $a$,$b$ and $c$ in terms of $a$, we get:
$$\ce{\cancel{a} A + \cancel{a} B &lt;=&gt; 2\cancel{a} C}$$
which implies:
$$\ce{A + B &lt;=&gt; 2C}$$
Therefore, your textbook is clearly in the wrong, since it states $\ce{A + B&lt;=&gt;C}$.