Field Trip Buses, Admission, and the Binding Constraint
A middle school is planning a field trip for 187 students and 12 chaperones. The transportation company charges $425 per bus for the day, and each bus seats 48 passengers. The school has $3,400 budgeted for buses. The museum charges $6.50 admission per person, but its group policy gives one free adult admission for every 10 paying students; free-adult slots cannot be used by students, and the school will not pay for more adults than the 12 already attending. The trip coordinator wants to know whether the combined bus and admission cost fits the budget and how much cushion remains. A parent volunteer argues they should book only 4 buses because 187 students divided by 48 is about 3.9, but the coordinator is unsure whether chaperones change the calculation.
1. How many buses are needed, and what is the total bus cost? Is the $3,400 transportation budget enough before admission is added?
Read the worked answer
Count all passengers, not just students: 187 students + 12 chaperones = 199 passengers. Each bus seats 48, so divide: 199 ÷ 48 = 4.1458. Since a partial bus still must be rented, round up to 5 buses. The parent’s 4-bus estimate uses only students and ignores both the chaperones and the remainder. Bus cost = 5 × $425 = $2,125. The transportation budget alone is $3,400, so yes, the bus cost fits, with $3,400 − $2,125 = $1,275 left before admission.
2. What is the total admission charge under the museum’s free-adult policy?
Read the worked answer
Only students pay. There are 187 paying students. The policy gives one free adult admission for every 10 paying students, so the number of free-adult slots is the greatest whole number less than or equal to 187 ÷ 10, which is 18. The school is bringing only 12 adults, and free slots cannot be used by students, so all 12 adults are free. Admission = 187 × $6.50. Compute 187 × 6 = 1,122 and 187 × 0.50 = 93.50, so the total is $1,215.50.
3. What is the combined cost, and if the school wanted to add students without renting a sixth bus, would bus seats or budget run out first? Show the maximum number of additional students.
Read the worked answer
Combined cost = $2,125 + $1,215.50 = $3,340.50. Cushion = $3,400 − $3,340.50 = $59.50. Bus capacity with 5 buses is 5 × 48 = 240 passengers; current passengers are 199, so seats allow 41 more passengers. Budget allows only $59.50 ÷ $6.50 ≈ 9.15, so at most 9 additional students can be afforded. Budget runs out first: 9 more students cost $58.50, leaving $1.00, while 10 more would cost $65 and exceed the cushion. The binding constraint is money, not bus seats.