# 4.1. Condition:

In front of the elevator stand people weighing 50, 51, 55, 57, 58, 59, 60, 63, 75, and 140 kg. The elevator's load capacity is 180 kg. What is the minimum number of trips needed to get everyone up?
Answer: 4 (or 7)

## Solution.

In one trip, the elevator can move no more than three people, as the minimum possible weight of four people will be no less than $50+51+55+57=213>180$. Note that no one will be able to go up with the person weighing 140 kg, so a separate trip will be required for his ascent. For the remaining nine people, at least $\frac{9}{3}=3$ trips will be needed, as no more than 3 people can fit in the elevator. The trips will be as follows: $(50,51,75),(55,57,63),(58,59,60)$ and (140).

In this problem, 7 is also accepted as a correct answer if the solution takes into account not only the elevator's ascents but also its descents to pick up people.