1. The problem asks for the maximum number of iterations needed to find an item using binary search on a sorted list of 7 numbers: [1, 5, 20, 50, 51, 80, 99]. 2. Binary search works by repeatedly dividing the search interval in half. The maximum number of iterations corresponds to the height of a binary search tree or the number of times you can halve the list until you reach one element. 3. The formula for the maximum number of iterations in binary search on a list of size $n$ is: $$\text{max iterations} = \lceil \log_2(n) \rceil$$ where $\lceil x \rceil$ is the ceiling function, which rounds $x$ up to the nearest integer. 4. For $n=7$, calculate: $$\log_2(7) = \frac{\log_{10}(7)}{\log_{10}(2)} \approx \frac{0.8451}{0.3010} \approx 2.807$$ 5. Taking the ceiling: $$\lceil 2.807 \rceil = 3$$ 6. Therefore, the maximum number of iterations needed is 3. 7. The correct answer is B. 3.