## Task B-1.4.

A pedestrian who crosses $1 \mathrm{~km}$ in 12 minutes, will travel the distance from $A$ to $B$ in the same time it takes a cyclist to travel a $10 \mathrm{~km}$ longer route, when the cyclist covers $1 \mathrm{~km}$ in $4 \frac{1}{2}$ minutes. Determine the distance from $A$ to $B$.
## Solution.

Let the distance from $A$ to $B$ be denoted as $x$. The speed at which the pedestrian walks is

$$
v_{P}=\frac{1}{\frac{12}{60}}=5 \mathrm{~km} / \mathrm{h}
$$

and the speed of the cyclist is

$$
v_{B}=\frac{1}{\frac{4.5}{60}}=\frac{40}{3} \mathrm{~km} / \mathrm{h}
$$

Since the times taken by the pedestrian and the cyclist are equal, and because speed is the ratio of distance to time, it follows that

$$
\frac{x}{5}=\frac{x+10}{\frac{40}{3}}
$$

so we have

$$
\begin{aligned}
\frac{40 x}{3} & =5(x+10) \\
8 x & =3 x+30 \\
x & =6 \mathrm{~km} .
\end{aligned}
$$