18. (1989 American High School Mathematics Examination) A real number $x$ is chosen at random between 100 and 200. If $[\sqrt{x}]=12$, then the probability that $[\sqrt{100 x}]=120$ is ( ).
A. $\frac{2}{25}$
B. $\frac{241}{2500}$
C. $\frac{1}{10}$
D. $\frac{96}{625}$
E. 1
18. B. Reason: Given $[\sqrt{x}]=12$, we know $12 \leqslant \sqrt{x}<13$,

i.e., $12^{2} \leqslant x<13^{2}$.
Also, $[\sqrt{100 x}]=120$, then $120 \leqslant \sqrt{100 x}<121$, i.e., $120 \leqslant 10 \sqrt{x}<121$, or $12 \leqslant \sqrt{x}<12.1$, thus $12^{2} \leqslant x<12.1^{2}$.
Therefore, the required probability $p=\frac{12.1^{2}-12^{2}}{13^{2}-12^{2}}=\frac{2.41}{25}=\frac{241}{2500}$.