Museum Trip Budget
A class of 28 students and 3 teachers plans a museum trip. The bus rental is a flat $240. Museum admission is $7 per student and $12 per adult. The school has $450 for the trip. A fundraiser added $85, but the principal rules that fundraiser money may cover admission only, not bus rental. The teacher requires at least one adult per 10 students. Two parents volunteer to chaperone, but the museum charges admission for every adult who enters. The teacher wants every student to attend if the budget allows.
1. Using the principal’s rule, calculate whether all 28 students and the 3 teachers can attend. Show the budget breakdown.
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Bus rental must come from the school’s $450: $450 − $240 = $210 left. Fundraiser money can be added to admission: $210 + $85 = $295 available for admission.
Admission cost:
- Students: 28 × $7 = $196
- Teachers: 3 × $12 = $36
- Total admission: $232
$232 ≤ $295, so all 28 students and 3 teachers can attend. Total trip cost is $240 + $232 = $472 against $535 available, leaving $63. The adult ratio is also satisfied: 28 students require at least 3 adults, and 3 teachers attend.
2. The two parents still want to come. Does including them change the answer? Show the new total and decide whether they should be allowed.
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Including the two parents adds 2 × $12 = $24 in admission. New admission total: $232 + $24 = $256. Total cost: $240 + $256 = $496. Available funds remain $450 + $85 = $535, leaving $39.
So the answer does not change: the trip is still affordable. Because the required adult ratio is already met by the 3 teachers, the parents are optional. Allowing them is reasonable if the $39 cushion is not needed for another required cost; if the school requires a contingency reserve, the teacher could decline the parents without affecting student attendance.
3. Suppose the bus company raises the flat rate to $310. Recalculate and state the maximum number of students who can attend with the 3 teachers, assuming the two parents do not come.
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With a $310 bus rate, school funds left for admission are $450 − $310 = $140. Adding the fundraiser: $140 + $85 = $225 for admission. The 3 teachers cost 3 × $12 = $36, leaving $225 − $36 = $189 for students. At $7 per student, $189 ÷ $7 = 27 students exactly.
So only 27 students can attend; 1 student would be left out. The teacher would need to raise at least $7 more (or find another $7 in the budget) to include the 28th student. This shows the plan is sensitive to the bus rate, even though the original $240 rate worked.