At a local market's checkout, the waiting time, \(W\) minutes, is directly proportional to the number of customers in the queue, \(n\), and inversely propor...
Question 1 Report
At a local market's checkout, the waiting time, \(W\) minutes, is directly proportional to the number of customers in the queue, \(n\), and inversely proportional to the square of the number of tills open, \(t\). When \(t=2\), the graph in the diagram shows that \(n=20\) gives \(W=15\).
- Write a formula for \(W\) in terms of \(n\), \(t\) and a constant \(k\). (1)
- Find the value of \(k\). (2)
- Find \(W\) when \(n=30\) and \(t=3\). (2)
Answer Details
Combining "directly proportional to the number of customers" and "inversely proportional to the square of the number of tills" gives a single formula for waiting time; the constant is found by reading one point from the graph (with the number of tills fixed) before the formula is used to predict waiting time for a different queue length and a different number of tills.
- \[ W=\frac{kn}{t^2} \] [1 mark]
- Reading the graph with \(t=2\): \(n=20\) gives \(W=15\). Substituting: \[ 15=\frac{k\times20}{2^2}=\frac{20k}{4}=5k \] so \[ k=3 \] [2 marks]
- With \(n=30\), \(t=3\): \[ W=\frac{3\times30}{3^2}=\frac{90}{9}=10 \text{ minutes} \] [2 marks]
Even though the queue in part (c) is longer than the one on the graph (\(30\) customers against \(20\)), the waiting time comes out shorter (\(10\) minutes against \(15\)), because opening an extra till (\(t=3\) instead of \(2\)) divides the waiting time by \(t^2\), and this squared effect outweighs the increase in queue length.
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