Problem:
Your school is planning a trip for Year 9 students, and there are three options available: a visit to a local amusement park, a trip to a science museum, or a tour of a historical site. The cost for each option is as follows:
- Amusement park: $30 per student, plus a $200 bus rental fee.
- Science museum: $20 per student, plus a $150 bus rental fee.
- Historical site: $25 per student, plus a $180 bus rental fee.
The school has a budget of $1500 for the trip, and there are 60 students in Year 9. The school wants to choose the option that maximizes the number of students who can attend the trip within the budget.
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Write a linear equation to represent the total cost (C) of each option in terms of the number of students (x) attending the trip.
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Solve the linear equation to find the maximum number of students that can attend each option within the budget.
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Determine which option allows the maximum number of students to attend the trip.
Solution:
- Linear equation for the amusement park option: C = 30x + 200
Linear equation for the science museum option: C = 20x + 150
Linear equation for the historical site option: C = 25x + 180
- To find the maximum number of students that can attend each option within the budget, we need to substitute the total cost (C) with the budget ($1500) in each equation and solve for x.
For the amusement park option: 30x + 200 = 1500 30x = 1500 - 200 30x = 1300 x = 1300/30 x ≈ 43.33
For the science museum option: 20x + 150 = 1500 20x = 1500 - 150 20x = 1350 x = 1350/20 x ≈ 67.5
For the historical site option: 25x + 180 = 1500 25x = 1500 - 180 25x = 1320 x = 1320/25 x ≈ 52.8
- The option that allows the maximum number of students to attend the trip within the budget is the science museum option, as it can accommodate approximately 67 students.