A market trader uses the formula \(C=\dfrac{2n+30}{n}\) for the average cost per item, in pounds, when packing \(n\) items into a box including a fixed box ...
Question 1 Report
A market trader uses the formula \(C=\dfrac{2n+30}{n}\) for the average cost per item, in pounds, when packing \(n\) items into a box including a fixed box fee. The table shows some values, with one unknown, labelled \(k\).
| \(n\) | 5 | 10 | 15 | 20 |
|---|---|---|---|---|
| \(C\) | 8 | \(k\) | 4 | 3.5 |
- Find the value of \(k\). (1)
- Find the value of \(n\) for which \(C=2.5\). (3)
Answer Details
Substituting a table value into the cost formula finds the missing entry \(k\); setting the formula equal to a target cost and rearranging (multiplying through by \(n\) to clear the fraction) then solves for \(n\).
(a) Substituting \(n=10\): \(C = \dfrac{2(10)+30}{10} = \dfrac{50}{10} = 5\), so \(k=5\). [1 mark]
(b) Setting \(C=2.5\): \(\dfrac{2n+30}{n}=2.5\). Multiplying both sides by \(n\): \(2n+30=2.5n\). Subtracting \(2n\) from both sides: \(30=0.5n\). Dividing by 0.5: \(n=60\) items. [3 marks]
Download The App On Google Playstore
Everything you need to excel in your exams
