1989 AHSME Problems/Problem 20

Let $x$ be a real number selected uniformly at random between 100 and 200. If $\lfloor {\sqrt{x}} \rfloor = 12$ , find the probability that $\lfloor {\sqrt{100x}} \rfloor = 120$ . ( $\lfloor {v} \rfloor$ means the greatest integer less than or equal to $v$ .)

$\text{(A)} \ \frac{2}{25} \qquad \text{(B)} \ \frac{241}{2500} \qquad \text{(C)} \ \frac{1}{10} \qquad \text{(D)} \ \frac{96}{625} \qquad \text{(E)} \ 1$

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1989 AHSME Problems/Problem 20