To tackle this problem, we need to analyze the capacity of each bus and the associated costs to determine the most efficient way to transport all 220 students. Let's break it down step by step.
Understanding Bus Capacities and Costs
We have four buses with different capacities and costs:
- Bus A: Capacity = 60 students, Cost = 160
- Bus B: Capacity = 50 students, Cost = 140
- Bus C: Capacity = 40 students, Cost = 125
- Bus D: Capacity = 35 students, Cost = 95
Calculating the Cost per Student
To find the most economical option, we can calculate the cost per student for each bus:
- Bus A: 160 / 60 = 2.67
- Bus B: 140 / 50 = 2.80
- Bus C: 125 / 40 = 3.13
- Bus D: 95 / 35 = 2.71
From this, we can see that Bus A has the lowest cost per student, followed closely by Bus D.
Finding Combinations to Carry All Students
Now, we need to find combinations of buses that can carry all 220 students at the minimum cost. We can use a systematic approach to explore different combinations:
Combination Analysis
Let's start with Bus A, as it has the highest capacity:
- If we use 3 buses of A: 3 x 60 = 180 students (cost = 3 x 160 = 480)
- We still need to carry 40 students. We can use 1 bus of C (cost = 125) or 1 bus of B (cost = 140).
- Using Bus C: Total cost = 480 + 125 = 605.
- Using Bus B: Total cost = 480 + 140 = 620.
Thus, the combination of 3 buses of A and 1 bus of C is optimal.
Exploring Other Combinations
Next, let’s check if using fewer buses of A and more of the others could yield a better cost:
- Using 2 buses of A: 2 x 60 = 120 students (cost = 320)
- We need to carry 100 more students. Possible combinations could be:
- 2 buses of B (100 students, cost = 280) → Total cost = 320 + 280 = 600.
- 2 buses of C (80 students, cost = 250) + 1 bus of D (35 students, cost = 95) → Total = 250 + 95 = 345 → Total = 320 + 345 = 665.
- Using 1 bus of A: 60 students (cost = 160) + 1 bus of B (50 students, cost = 140) + 1 bus of C (40 students, cost = 125) + 1 bus of D (35 students, cost = 95) → Total = 160 + 140 + 125 + 95 = 520.
- However, this combination does not meet the requirement of 220 students.
Minimum Number of Buses Required
From our analysis, the combination of 3 buses of A and 1 bus of C is the most cost-effective way to transport all students, totaling 4 buses:
- 3 buses of A (180 students)
- 1 bus of C (40 students)
This combination allows us to transport all 220 students at the minimum cost of 605.
Summary of Findings
To summarize:
- The optimal bus combination is 3 buses of A and 1 bus of C.
- The minimum cost to transport all students is 605.
- The total number of buses required is 4.
By carefully analyzing the capacities and costs, we can effectively determine the best way to transport all students while minimizing expenses. If you have any further questions or need clarification on any part of this process, feel free to ask!
