Marathon Riddle

Let the distance between city $A$ and $B$ be $D$ , and the speed of $A$ runner be $v$ km/h; then the speed of runner $B$ is $v - 8$ km/h. Let the time from start till the two runners met be $t$ .

Then, we have $vt + (v-8) t = D$ $\color{#3D99F6}\implies t = \dfrac D{2v-8}$ . We note that $B$ runner run the speed hence took the same time $t$ to return back to city $B$ . Then the average speed of the whole trip is:

$\begin{aligned} v_{\text{avg}} & = \frac D{2\color{#3D99F6}t} & \small \color{#3D99F6} \text{Recall }t = \frac D{2v-8} \\ \implies \frac D{2 \times \frac D{2v-8}} & = 37 \\ v - 4 & = 37 \\ \implies v & = \boxed{41} \end{aligned}$

Suppose the speed of $A$ runner be $x$ and that of $B$ be $y$ . Also, let $S$ be the total distance from city $A$ to $B$ and $T$ be the total time taken for this whole delivery trip.

Then it is clear from the question that $37 = \dfrac{S}{T}$ .

Now let us consider the time used by the $B$ runner, who first ran to meet $A$ runner at some point en route and then later ran back the same distance. Thus, the first leg trip and the second one travelled by $B$ is of the same time and distance, for he used the same speed. In other words, the time spent in the first leg equals $\dfrac{T}{2}$ .

Then in the first leg trip, both runners sprinted to each other. Hence, the speed occurred was the sum of their speeds combined, with the distance $S$ and time $\dfrac{T}{2}$ .

Therefore, we could set up another equation:

$x+y = \dfrac{S}{T/2} = \dfrac{2S}{T}$

Substituting $\dfrac{S}{T} = 37$ from the initial equation, we will obtain:

$x+y = 2\times 37 = 74$

Since their difference in speed is $x-y = 8$ , we can conclude that $2x = 82$ . Finally, $x = \boxed{41}$ .