Science Exhibition Room and Volunteer Plan
A school is holding a science exhibition on Saturday from 9 a.m. to 1 p.m. There are six exhibits: a solar oven, a model wind turbine, a water-filtration demo, a robotics arm, a chemistry volcano, and a sound-wave display. Three rooms are available. Room A has two power sockets, opens onto the courtyard, and holds 30 visitors. Room B has one power socket, a window that opens, and holds 20 visitors. Room C has no power socket, no opening window, has a water tap, and holds 40 visitors. The solar oven and the chemistry volcano need ventilation and cannot be placed in Room C. The wind turbine and the robotics arm need power. The water-filtration demo needs the water tap. Each exhibit needs one volunteer for the whole morning, but only five volunteers are available. Visitor groups of 25 rotate every 20 minutes. At 10 a.m., the robotics arm blows a fuse. The only spare fuse is in the school office, which is locked until 11 a.m.
1. Allocate the six exhibits to the three rooms so that all power, water, and ventilation constraints are met. Then name the one exhibit you would leave without a dedicated volunteer, and justify your choice using safety and visitor value.
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Feasible allocation:
- Room A (2 sockets, ventilation, capacity 30): wind turbine, robotics arm, solar oven.
- Room B (1 socket, ventilation, capacity 20): sound-wave display, chemistry volcano.
- Room C (water tap, no socket, no opening window, capacity 40): water-filtration demo. This satisfies the power needs (two in A, one in B), the water-tap need (C), and the ventilation rule (solar oven and volcano are not in C). Leave without a dedicated volunteer: the sound-wave display. It is the least hazardous exhibit: it has no heat, moving blades, water, or chemicals. It can run on a fixed loop with a clear instruction sign and be checked periodically by a roving volunteer. The other exhibits need continuous supervision because of heat, moving parts, chemicals, or water spillage.
2. At 10 a.m. the robotics arm fails. Describe your immediate response for the next 60 minutes, including volunteer reassignment and visitor management.
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- Stop and secure: switch off the robotics arm and place a 'closed for repair' sign at the door of Room A. Do not attempt a repair before the spare fuse is available.
- Reassign the volunteer: move the robotics volunteer to the unstaffed sound-wave display in Room B, or keep them in Room A to control the crowd around the solar oven and volcano.
- Manage visitors: at the next 20-minute rotation, tell groups that the robotics arm is unavailable and offer extra time at the water-filtration demo or the sound-wave display. If Room A becomes crowded, redirect groups to Room C, which holds 40.
- Repair: at 11 a.m., collect the fuse from the office, replace it, test the arm, and reopen only if it works safely. Keep the sign up until the test is complete.
3. Four groups of 25 visitors are scheduled to rotate every 20 minutes. Using the room capacities, calculate the maximum number of visitors who can be inside the three rooms at one time, and explain whether all four groups can be accommodated in a single rotation. Recommend a practical adjustment.
Read the worked answer
Maximum capacity: 30 + 20 + 40 = 90 visitors at any one time. Four groups: 4 × 25 = 100 visitors, so they cannot all be inside at once; at least 10 must wait. There is a second problem: a 25-person group does not fit entirely into Room B, which holds only 20. If each room takes one whole group, Room B is over capacity by 5. At most three whole groups (75 visitors) can be placed one per room, and even then Room B's group is too large. Recommendation: reduce each rotation group to 20 visitors, or split each 25-person group into 20 + 5 and send the 5 to Room A or C only if space is available. Alternatively, keep one group in a holding area each rotation and rotate four groups through three rooms in a cycle. This keeps every room within capacity and avoids crowding.