3.49 John scored 93 points in this year's AHSME. If the old scoring method were still in effect, he would have scored only 84 points based on the same answers. The number of questions he left unanswered is (the new scoring system awards 5 points for a correct answer, 0 points for a wrong answer, and 2 points for an unanswered question. Under the old scoring system, the base score is 30 points, with 4 points added for each correct answer and 1 point deducted for each wrong answer; unanswered questions neither add nor deduct points)
(A) 6.
(B) 9.
(C) 11.
(D) 14.
(E) Cannot be uniquely determined.
(37th American High School Mathematics Examination, 1986)
［Solution］Let John answer $x$ questions correctly, $y$ questions incorrectly, and leave $z$ questions unanswered. According to the problem, we can set up the following system of equations:
$$
\left\{\begin{array}{l}
5 x+0 \cdot y+2 z=93, \\
30+4 x-y+0 \cdot z=84, \\
x+y+z=30 .
\end{array}\right.
$$
Adding (2) and (3) gives $5 x+z=84$.
Subtracting (4) from (1) gives $z=9$.

Therefore, the answer is $(B)$.