2. Given that $[x]$ represents the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$. Then the number of positive numbers $x$ that satisfy
$$
20\{x\}+1.1[x]=2011
$$

is ( ).
(A) 17
(B) 18
(C) 19
(D) 20
2. B.

Since $0 \leqslant\{x\}<1$, we have
$$
\begin{array}{l}
0 \leqslant 20\{x\}<20, \\
0 \leqslant 2011-1.1[x]<20, \\
1810<[x]<1829 .
\end{array}
$$

Therefore, $[x]$ can take $1810+k(k=1,2, \cdots, 18)$, and accordingly $\{x\}$ takes $1-\frac{11 k}{200}$.