3.43 If three people, A, B, and C, work together on a task, it takes 6 hours less than if A were to do it alone, 1 hour less than if B were to do it alone, and exactly half the time it would take C to do it alone. If A and B working together need $h$ hours, then $h$ equals
(A) $\frac{5}{2}$.
(B) $\frac{3}{2}$.
(C) $\frac{4}{3}$.
(D) $\frac{5}{4}$.
(E) $\frac{3}{4}$.
(17th American High School Mathematics Examination, 1966)
[Solution] Let the time required for A, B, and C to complete the work alone be $a, b, c$ hours, respectively. According to the problem, we have (setting up equations based on the work completed per unit time)
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a-6}=\frac{1}{b-1}=\frac{2}{c},
$$

Also, $\frac{1}{a}+\frac{1}{b}=\frac{1}{h}$, thus $\frac{1}{h}+\frac{1}{c}=\frac{2}{c}$, which gives $h=c$.
From $\frac{1}{a-6}=\frac{2}{h}$, we get $a=\frac{h+12}{2}$,
and from $\frac{1}{b-1}=\frac{2}{h}$, we get $b=\frac{h+2}{2}$.
Therefore, $\frac{2}{h+12}+\frac{2}{h+2}=\frac{1}{h}$,

which simplifies to $3 h^{2}+14 h-24=0$.
Solving this, we get $h=\frac{4}{3}$ (discarding the negative value).
Thus, the answer is $(C)$.