Field Trip Bus Problem
The fourth grade is going on a field trip. There are 187 students and 14 teachers/chaperones. Each bus can seat 40 people. The school district requires that every bus has at least one adult, and no bus may carry more than 40 people total. The office manager first says, “187 ÷ 40 = 4 remainder 27, so 4 buses are enough for students; adults can squeeze in.” A teacher says that is not safe or allowed. You are asked to find the least number of buses needed for all 201 people and to explain how the remainder affects the decision. The bus company charges $275 per bus. The principal wants to know whether 5 buses or 6 buses is the better choice, given that unused seats are allowed but every person must have a seat.
1. Calculate the total number of people and the least number of buses needed. Show the division and explain why the remainder forces an extra bus.
Read the worked answer
Total people = 187 + 14 = 201. Then 201 ÷ 40 = 5 remainder 1. Five buses seat 5 × 40 = 200 people, which is one seat short. Since every person must have a seat, the remainder of 1 forces a sixth bus. The least number is 6 buses, with 6 × 40 = 240 seats and 39 empty seats.
2. The office manager says 4 buses are enough for students, with adults fitting in. Explain the error using the total and the capacity constraint.
Read the worked answer
Four buses seat only 160 people, and there are 187 students alone. Even if adults were not counted, 4 buses cannot hold the students; with adults the total is 201. The office manager divided only students and then treated the 27 students in the remainder as if adults could “squeeze in.” That violates the 40-person limit and the requirement that every person has a seat. The correct comparison is 201 people versus 40 seats per bus, not 187 students versus 40 seats per bus.
3. Compare the cost of 5 buses and 6 buses and explain why 5 buses cannot be chosen.
Read the worked answer
Five buses cost 5 × $275 = $1,375 and provide 200 seats, which is less than 201 people, so that plan is not allowed. Six buses cost 6 × $275 = $1,650 and provide 240 seats. Although 5 buses is cheaper, it cannot transport everyone; the least valid choice is 6 buses.