A car reaches from $A$ to $B$ in $t$ hours. To reach from $A$ to $C$ in the same amount of time, it needs to cover each km $m$ minutes faster, because the distance between $A$ and $C$ is $k \mathrm{~km}$ longer than between $A$ and $B$. How far is $B$ from $A$?
If the distance between $A$ and $B$ is $x \mathrm{~km}$, then $A C=x+k \mathrm{~km}$.

Between $A$ and $B$, the car covers $1 \mathrm{~km}-\mathrm{t}$ in $\frac{60 t}{x}$ minutes. If we subtract $m$ minutes from this, we get the time it took to cover 1 km of the $A C$ distance, which is $\frac{60 t}{x+k}-t$:

$$
\frac{60 t}{x}-m=\frac{60 t}{x+k}
$$

Multiplying through by the product of the denominators and rearranging, we get

$$
m x^{2}+m k x-60 t k=0 \quad \text {, }
$$

the solution to this equation is:

$$
x_{1,2}=\frac{-m k \pm \sqrt{m^{2} k^{2}+240 m t k}}{2 m}
$$

From the perspective of the problem, only the positive root is valid. The sought distance is therefore

$$
A B=\frac{-m k+\sqrt{m^{2} k^{2}+240 m t k}}{2 m}
$$

Zahumenszky Zille (Bp. VIII., Fazekas g. I. o. t.)