$31 \cdot 12$ If $a, b$ are natural numbers not exceeding 10, then the number of pairs $(a, b)$ that make the solution of the equation $a x=b$ less than $\frac{1}{2}$ but greater than $\frac{1}{3}$ is
(A) five pairs.
(B) four pairs.
(C) three pairs.
(D) two pairs.
(China Shanghai Junior High School Mathematics Competition, 1983)
［Solution］Let $x=\frac{b}{a}$, and according to the problem, we have $\frac{1}{3}<\frac{b}{a}<\frac{1}{2}, \quad 1 \leqslant a, \quad b \leqslant 10, \quad$ and $a, b \in N$.

Solving, we get $\left\{\begin{array}{l}a=10, \\ b=4 ;\end{array} \quad\left\{\begin{array}{l}a=9, \\ b=4 ;\end{array}\left\{\begin{array}{l}a=8, \\ b=3 ;\end{array}\left\{\begin{array}{l}a=7, \\ b=3 ;\end{array}\left\{\begin{array}{l}a=5, \\ b=2 .\end{array}\right.\right.\right.\right.\right.$ Therefore, the answer is $(A)$.