A job can be completed by three workers, with the first taking $p$ more days than the second, and $q$ more days than the third. The first two workers together can complete the job in the same amount of time as the third worker alone. How much time does each worker need individually to complete the job?
Let $x$ be the number of days it takes for the first worker to complete the job alone. Thus, the second and third workers need $x-p$ and $x-q$ days, respectively, to complete the work, and in one day they complete $\frac{1}{x}$, $\frac{1}{x-p}$, and $\frac{1}{x-q}$ of the work, respectively. Since the first two workers together can complete the work in the same amount of time as the third worker alone, the fractions of the work they complete in one day are equal:

$$
\frac{1}{x}+\frac{1}{x-p}=\frac{1}{x-q}
$$

Here, clearly $x > p$, $x > q$, and it must be that $p < q$, otherwise the right side would be smaller than the second term on the left side. From this, by rearranging,

$$
x^{2}-2 q x+p q=0, \quad x_{1,2}=q \pm \sqrt{q(q-p)}
$$

Only

$$
x_{1}=q+\sqrt{q(q-p)}
$$

fits the content of the problem, in which case the number of days required for the second and third workers are

$$
q-p+\sqrt{q(q-p)}, \quad \text { and } \quad \sqrt{q(q-p)}
$$

József Berényi (Debrecen, Ref. Kollégium g. I. o. t.)