The ferry timetable

The ferry crosses the water every forty minutes. The first boat leaves at six in the morning and the last at ten at night, which makes twenty-five sailings a day.

The ferry crosses every $40$ min. Boat number $k$ leaves at $T_k=6:00+40k$ min. First boat $6:00$, last $22:00$: $(22-6)\times 60=960$ min, $960÷40=24$ gaps, so $25$ sailings a day.

$T_k=6:00+40k$ min, $k=0,\dots ,24$: $T_{24}=22:00$, so $25$ sailings.

One passenger reached the quay at ten to ten. The boat of twenty past nine had left half an hour earlier, so she waited ten minutes for the ten o'clock boat. The crossing takes twenty-five minutes, and she stepped ashore at twenty-five past ten.

A passenger reached the quay at $9:50$. The boat of $9:20$ had left $30$ min before. The next one is at $10:00$, so she waited $10$ min. The crossing takes $25$ min: $10:00+25=10:25$.

$t=9:50$, $T_5=9:20<t<T_6=10:00$. Wait: $10^{\prime }$. $10:00+25^{\prime }=10:25$.

Nobody waits longer than forty minutes. If people arrive at random, the average wait is twenty minutes: half of the gap between two boats.

The wait is never more than $40$ min: $w\le 40$. If people arrive at random, the average wait is $40÷2=20$ min.

wait $w$: $0\le w\le 40^{\prime }$, mean $40÷2=20^{\prime }$.

The harbour master says the lesson is simple: read the timetable at home, not when you reach the water. Looking takes one minute; missing a boat can cost forty.

Lesson: read the timetable at home. Looking takes $1$ min, missing a boat up to $40$ min: $1\ll 40$.

look $1^{\prime }$, miss $\le 40^{\prime }$: $1\ll 40$.