31-14 Let $a, b$ be natural numbers not exceeding 10. Then, the number of pairs $(a, b)$ that make the solution of the equation $a x=b$ greater than $\frac{1}{4}$ and less than $\frac{1}{3}$ is
(A) 2.
(B) 3.
(C) 4.
(D) 1.
（1st "Five Sheep Cup" Junior High School Mathematics Competition, 1989)
[Solution] From the problem, we have $x=\frac{b}{a}$, then $\frac{1}{4}<\frac{b}{a}<\frac{1}{3}$, which means $3 b<a<4 b$. Also, $a, b$ are natural numbers not exceeding 10, so $3 b<10$, hence $b \leqslant 3$.
If $b=1$, then there is no solution for $a$;
If $b=2$, then $6<a<8$, we get $\quad a=7$;
If $b=3$, then $9<a<12$, we have $a=10$.
In summary, $(a, b)=(7,2)$ or $(10,3)$.
Therefore, the answer is $(A)$.