# 4.3. Condition:

In front of the elevator stand people weighing 150, 62, 63, 66, 70, 75, 79, 84, 95, 96, and 99 kg. The elevator's load capacity is 190 kg. What is the minimum number of trips needed to get everyone up?
# Answer: 6 (or 11)

## Solution.

In one trip, the elevator can move no more than two people, as the minimum possible weight of three people will be no less than $62+63+66=191>190$. Note that no one will be able to go up with the person weighing 150 kg, so a separate trip will be required for his ascent. For the remaining ten people, at least $\frac{10}{2}=5$ trips will be needed, as no more than 2 people can fit in the elevator. Thus, the trips will be: (62, 99), $(63,96),(66,95),(70,84),(75,79)$ and (150).

In this problem, 11 is also considered a correct answer if the solution takes into account not only the elevator's ascents but also its descents to pick up people.