One statement asks whether a composed-then-divided figure can be determined, one composes weekly revenue, and one counts a per-vet figure against a strict threshold.
Weekly revenue is weekly visits × average fee: Bayside 600 × $110 = $66,000, Cedar Vale 900 × $80 = $72,000, Foxglen 400 × $150 = $60,000, Harborview 750 × $100 = $75,000, Juniper 500 × $130 = $65,000.
Statement 1 - Yes. The table prints no revenue-per-exam-room column, but it gives weekly visits, the average fee, and the number of exam rooms, so the figure is computable: Bayside $6,600, Cedar Vale $5,143, Foxglen $8,571, Harborview $6,250, Juniper $7,222 per exam room. The highest is Foxglen, so the answer can be determined. The statement asks only whether it can be determined, and it can. The trap is to answer No because no such column is shown, or to pick the most exam rooms or the highest fee without composing revenue first.
Statement 2 - No. Juniper's weekly revenue is 500 × $130 = $65,000, and Bayside's is 600 × $110 = $66,000, so Juniper earns less and the statement is false. The bait is Juniper's higher fee of $130; but Bayside's larger visit count of 600 more than makes up for its lower fee. Fee is not revenue.
Statement 3 - No. Visits per veterinarian is weekly visits ÷ vets: Bayside 100, Cedar Vale 75, Foxglen 100, Harborview 75, Juniper 100. The statement asks for more than 100, a strict inequality, and the three hospitals at exactly 100 do not clear it; none exceed 100. So zero of the five qualify, not more than half, and the statement is false. The trap is to count the hospitals sitting exactly at 100 as satisfying 'more than 100'; the boundary is excluded.
The lesson: a revenue-per-room figure you can compose and divide is determinable, revenue is visits × fee not fee alone, and 'more than' excludes the boundary value. Correct answers: Yes / No / No.