108(977). A person, living in a cottage near a railway station, usually manages to walk from the cottage to the station in time for the train departure in 18 minutes. Once, this person was delayed at home before leaving by several minutes. Although after that, he walked 1.2 times faster than usual, he still missed the train by 2 minutes. How many minutes was he delayed at home before leaving?
Solution. If the usual speed of the gardener is $v$ m/min, then the distance from the garden to the station is $18 v$ meters. When the gardener walked at a speed of $1.2 v$ m/min, it took 15 minutes to travel from home to the garden $(18 v: 1.2 v=15)$. Since the gardener was 2 minutes late, the train left 13 minutes after the gardener left home $(15-2=13)$. Therefore, he was delayed at home for 5 minutes $(18-13=5)$.

Answer: 5 minutes.

$\triangle 109(978)$. At 13:00, water started to be filled into a swimming pool from one pipe to fill it by 16:00 the next day. After some time, another identical pipe was turned on because the pool needed to be filled by 12:00 the next day. At what time was the second pipe turned on?

Solution. Since the pool needed to be filled 4 hours earlier $(16-12=4)$, the second pipe (identical to the first) was turned on 4 hours earlier. Therefore, the second pipe was turned on at 8:00 AM $(12-4=8)$.

Answer: at 8:00 AM.

$\triangle 110(979)$. An electric train passed a signal light in 5 seconds and a platform 150 meters long in 15 seconds. What is the length of the train and its speed?

Solution. Method I. Let the speed of the train be $x$ m/s. Then the length of the train is $5 x$ meters. In 15 seconds, the train travels a distance of $15 x$ meters, or $(150+5 x)$ meters. We have the equation $15 x=150+5 x$, from which $x=15$, i.e., the speed of the train is $15 \mathrm{~m} / \mathrm{s}$, and its length is 75 meters.

Method II. Since the time it takes for any point on the train to pass the platform is 10 seconds $(15-5=10)$, the speed of the train is 15 m/s $(150: 10=15)$, and its length is 75 meters $(15 \cdot 5=75)$.

Answer: $75 \mathrm{~m} ; 15 \mathrm{~m} / \mathrm{s}$.