## PROBLEM 4. FERRIES

Two ferries shuttle across a river at constant speeds, turning at the banks without losing time. They depart simultaneously from opposite banks and meet for the first time 700 feet from one of the banks, continue to the banks, return, and meet for the second time 400 feet from the opposite bank. As an oral exercise, determine the width of the river. (In the UK and US, a foot is denoted by a prime: 700 feet $-700^{\prime}$.)

![](https://cdn.mathpix.com/cropped/2024_05_21_2f4fa61a46fbb276c744g-12.jpg?height=320&width=1132&top_left_y=1510&top_left_x=470)

Fig. 5
Solution. The total distance covered by the ferries at the time of their first meeting is exactly equal to the width of the river (Fig. 5). However, it may be surprising that by the time they meet again, the total distance they have covered is three times the width of the river. Since the speeds are constant, the second meeting will occur after a time three times longer than the first. Before the first meeting, ferry A, for example, had traveled 700 feet. In three times the duration, it would have traveled 2100 feet. But by the time of the second meeting, it reaches the riverbank and, turning back, travels an additional 400 feet. Thus, the width of the river is $2100-400=1700$ feet.