Let the price of the book $= p$. Total paid $= 1.04p$.

She paid $10 and received change less than $3.00, meaning total paid was more than $7.00:

$1.04p > 7.00 \Rightarrow p > 6.73$

She also received some change, so total paid $< \$10.00$:

$1.04p < 10.00 \Rightarrow p < 9.615$

So $6.73 < p < 9.615$. Now check each statement:

• Statement A: $p < \$9.50$ — Not necessarily, since $p$ can be up to $9.615$. ✗
• Statement B: $p > \$6.90$ — Not necessarily, $p$ can be $6.80$ (giving change $\approx \$2.93 < \$3.00$). ✗
• Statement C: Sales tax $= 0.04p < 0.04 \times 9.615 \approx \$0.385 < \$0.45$. This is always true within the valid range. ✓
• Statement D: Tax $> \$0.50$ would require $p > 12.50$, which is impossible here. ✗

Only Statement C must be true.

Notice $P = I + N$ where $N=\begin{pmatrix}0&0&0\\3&0&0\\9&3&0\end{pmatrix}$ (strictly lower triangular, so $N^3=0$).

$P^5 = (I+N)^5 = I + 5N + 10N^2$ (since $N^3=0$).

$N^2 = \begin{pmatrix}0&0&0\\0&0&0\\9&0&0\end{pmatrix}$... computing: $(N^2)_{31}=3\cdot3=9$, others 0.

$P^5 = I + 5N + 10N^2$. So $Q = P^5 + I = 2I + 5N + 10N^2$.

$q_{21} = 5\cdot3 = 15$, $q_{31} = 5\cdot9+10\cdot9=45+90=135$... wait: $q_{31}=(5N)_{31}+(10N^2)_{31}=5\cdot9+10\cdot9=45+90=135$. Wait $N_{31}=9, N^2_{31}=9$: $q_{31}=5(9)+10(9)=135$. $q_{32}=(5N)_{32}=5\cdot3=15$.

$\dfrac{q_{21}+q_{31}}{q_{32}} = \dfrac{15+135}{15} = \mathbf{10}$