14.15 Randomly select a real number $x$ 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 .
(40th American High School Mathematics Examination, 1989)
[Solution] Given $[\sqrt{x}]=12$, then we know

that is $\square$
$$
\begin{array}{l}
12 \leqslant \sqrt{x}<13, \\
12^{2} \leqslant x<13^{2},
\end{array}
$$

Also, $[\sqrt{100 x}]=120$, then $120 \leqslant \sqrt{100 x}<121$, which means $120 \leqslant 10 \sqrt{x}<121$, or $12 \leqslant \sqrt{x}<12.1$,
thus
$$
12^{2} \leqslant x<12.1^{2} \text {. }
$$

Therefore, the required probability $\quad p=\frac{12.1^{2}-12^{2}}{13^{2}-12^{2}}=\frac{2.41}{25}=\frac{241}{2500}$, hence the answer is $(B)$.