13.188. Worker $A$ spends $k$ minutes less on processing one part than worker $B$. How many parts does each of them process in $t$ hours of work, if $A$ processes $n$ more parts than $B$ in this time?
Solution.

Let $B$ process $x$ parts, $A$ processes $x+n$ parts. For processing 1 part, $A$ spends $\frac{t}{x+n}$ hours, $B$ spends $\frac{t}{x}$ hours. According to the condition, $\frac{t}{x}-\frac{t}{x+n}=\frac{k}{60}$, from which we get
$x=\frac{-k n+\sqrt{k^{2} n^{2}+240 t k n}}{2 k}$

$x+n=\frac{-k n+\sqrt{k^{2} n^{2}+240 t k n}}{2 k}+n=\frac{k n+\sqrt{k^{2} n^{2}+240 t k n}}{2 k}$.

Answer: $\left(-k n+\sqrt{k^{2} n^{2}+240 t k n}\right) /(2 k)$ and $\left(k n+\sqrt{k^{2} n^{2}+240 t k n}\right) /(2 k)$.