$15 \cdot 3$ In a junior high school, there are 13 extracurricular interest groups, with the number of people in each group as shown in the table below:

One afternoon, the school simultaneously held Chinese and mathematics lectures. It is known that 12 groups attended the lectures, where the number of people attending the Chinese lecture is 6 times the number of people attending the mathematics lecture. One group remained in the classroom for a discussion, this group is
(A) Group 4.
(B) Group 7.
(C) Group 9.
(D) Group 12.
(Chinese Jiangsu Province Junior High School Mathematics Competition, 1988)
［Solution］Let the number of people attending the Chinese lecture be $a$ people, the number of people attending the math lecture be $b$ people, and the number of people in the discussion group in the classroom be $x$ people. According to the problem, we have
$$
\left\{\begin{array}{l}
a=6 b, \\
a+b+x=2+3+5+\cdots+22+24=158 .
\end{array}\right.
$$
(1) Substituting into (2) we get
$$
x=158-7 b=154-7 b+4=7(22-b)+4,
$$

which means the number of people in the discussion group in the classroom, when divided by 7, leaves a remainder of 4. From the table, only 11 fits.
Therefore, the answer is $(B)$.