## Task 4 - 120934

Two pedestrians $A$ and $B$ covered the same distance. They started at the same time. An observer noted:

$A$ walked half the distance at an average speed of $4 \frac{\mathrm{km}}{\mathrm{h}}$, and the rest at $5 \frac{\mathrm{km}}{\mathrm{h}}$. $B$ walked for half of the total time at an average speed of $4 \frac{\mathrm{km}}{\mathrm{h}}$, and the remaining time at $5 \frac{\mathrm{km}}{\mathrm{h}}$.

Who of the two reached the destination first?
We use the relationship speed $=$ distance/time or time $=$ distance/speed. Let $d$ be the total length of the distance.

A requires the time $t_{A}=\frac{d / 2}{4}+\frac{d / 2}{5}=\frac{9}{40} d$ for the distance $d$.

Let $d_{1}, d_{2}$ be the distances over which B travels at $4 \frac{km}{h}$ and $5 \frac{km}{h}$, respectively. Then $d_{1}+d_{2}=d$. For these distances, B requires the times $t_{1}=\frac{d_{1}}{4}$ and $t_{2}=\frac{d_{2}}{5}$.

By assumption, $t_{1}=t_{2}$, so $\frac{d_{1}}{4}=\frac{d_{2}}{5}$. Together with $d_{1}+d_{2}=d$, this yields $d_{1}=\frac{4}{9} d, d_{2}=\frac{5}{9} d$ and thus $t_{1}=t_{2}=\frac{1}{9} d$. Therefore, B requires the total time $t_{B}=t_{1}+t_{2}=\frac{2}{9} d$.

Since $\frac{9}{40}>\frac{2}{9}$, B arrives at the finish line first.