Question

Option A. The Museum of Pop offers a discount to schools, $18 per ticket, but they require 4 adult chaperones in addition to the number of students.

Option B. Tickets to the Mariners cost $20 per ticket, and the only chaperone required is the teacher.

1. Determine which field trip is cheapest for our class (26 or 27 students).
2. Determine which field trip is cheapest if only 75% of our class can attend.
3. Model each field trip as a linear equation.
4. Detemine:
4a. When is it cheaper to use Option A
4b. When is it cheaper to use Option B
4c. Sketch the graph that helps you make this call.

Answer 1

Summary answers
- For 26 students: Option A and B cost the same ($540).
- For 27 students: Option A is cheapest (A = $558, B = $560).
- If only 75% can attend (see note below about rounding), Option B (Mariners) is cheaper in all reasonable integer cases.

Work and models

Let s = number of students who attend.

Option A (Museum): total cost CA(s) = 18(s + 4) = 18s + 72.
Option B (Mariners): total cost CB(s) = 20(s + 1) = 20s + 20.

1) For s = 26:
- CA(26) = 18(26+4) = 18·30 = $540
- CB(26) = 20(26+1) = 20·27 = $540
=> tie.

For s = 27:
- CA(27) = 18(27+4) = 18·31 = $558
- CB(27) = 20(27+1) = 20·28 = $560
=> Option A is cheaper by $2.

2) If only 75% can attend:
- 75% of 26 = 19.5 → realistically 19 or 20 students.
- If 19 attend: CA = 18(23) = $414, CB = 20(20) = $400 → Option B cheaper.
- If 20 attend: CA = 18(24) = $432, CB = 20(21) = $420 → Option B cheaper.
- 75% of 27 = 20.25 → realistically 20 (or 21).
- If 20 attend: CA = $432, CB = $420 → Option B cheaper.
- If 21 attend: CA = 18(25) = $450, CB = 20(22) = $440 → Option B cheaper.

So with 75% attendance, Mariners (Option B) is the cheaper choice in all integer cases.

4) Comparison / inequality
- When is A cheaper? CA < CB:
18s + 72 < 20s + 20 → 72 − 20 < 20s − 18s → 52 < 2s → s > 26.
So Option A is cheaper when more than 26 students attend.
- When is B cheaper? CA > CB → s < 26 (Option B cheaper if fewer than 26 students).
- If s = 26, costs are equal.

Intersection point
- Solve 18s + 72 = 20s + 20 → s = 26. Cost at intersection: $540. So the two lines cross at (26, 540).

4c) Sketch guidance
- Axes: horizontal = number of students s (≥0), vertical = total cost in dollars.
- Plot the two lines:
- CA(s) = 18s + 72 (slope 18, y-intercept 72)
- CB(s) = 20s + 20 (slope 20, y-intercept 20)
- They intersect at (26, 540). For s < 26 the CB line is below CA (Mariners cheaper); for s > 26 the CA line is below CB (Museum cheaper).
- Example points to plot:
- CA: (0,72), (10,252), (20,432), (26,540), (30,612)
- CB: (0,20), (10,220), (20,420), (26,540), (30,620)

(If you want, I can draw a small plotted sketch or provide a printable graph.)