12. John ( $J o h n$ ) scored 93 points in this $A H S M E$. If the old scoring method were still in effect, he would have scored only 84 points with the same answers. How many questions did he leave unanswered? (The new scoring rules are explained on the cover; under the old scoring method, the base score was 30 points, with 4 points added for each correct answer and 1 point deducted for each wrong answer; unanswered questions neither added nor deducted points)
(A) 6; (B) 9 ; (C) 11; (D) 14;
(E) Cannot be uniquely determined.
12
$\mathrm{B}$
12. Let John answer $c$ questions correctly, $w$ questions incorrectly, and leave $u$ questions unanswered. According to the old scoring method, he scores 84 points, i.e., $30+4 c-w=84 \cdots$ (1). According to the new scoring method, he scores 93 points, i.e., $5 c+2 u=93 \cdots$ (2). Also, $c+w+u=30 \cdots(3)$. Solving the system of three linear equations, we get $c=15, w=6$, and $u=9$.