5. Given that $[x]$ represents the greatest integer not exceeding $x$, if
$$
[x+0.1]+[x+0.2]+[x+0.3]+[x+0.4]+[x+0.5]+[x+0.6]+[x+0.7]+[x+0.8]+[x+0.9]=104
$$

then the minimum value of $x$ is ( ).
A: 11.5
B: 10.5
C: 9.5
D: 8.5
Rongtu A.
Analysis Since $0<0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9<1,104 \div 9=11 \frac{5}{9}$, it follows that $11<x<12$. Because $104=11 \times 4+12 \times 5$, we have
$$
\begin{array}{l}
{[x+0.1]=[x+0.2]=[x+0.3]=[x+0.4]=11} \\
{[x+0.5]=[x+0.6]=[x+0.7]=[x+0.8]=[x+0.9]=12}
\end{array}
$$

From $[x+0.5]=12$, we can deduce that $12 \leq x+0.5<13$, which means $11.5 \leq x<12.5$; from $[x+0.9]=12$, we can deduce that $12 \leq x+0.9<13$, which means $11.1 \leq x<12.1$. Therefore, $11.5 \leq x<12.1$. Thus, the minimum value of $x$ is 11.5.