Higher June 2022 Paper 6 Q11
11 Amir, Beth and Charlie work in a cafe.
Customers give spare change as tips.
At the end of each week, Amir, Beth and Charlie share the total amount of tips between them in the ratio matching the number of hours they worked that week.
This week:
- Amir’s share of the tips was £25.40.
- Beth worked twice as many hours as Amir.
- Charlie worked 5 more hours than Amir.
- The total hours worked by Amir, Beth and Charlie was 85 hours.
Calculate the total amount of tips received this week.
You must show your working. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 107.95 with correct working | 6 | B1 for \(2a\) or \(a + 5\) or \(4a + 5\) or 25.4[0] + 5\(x\) seen M1 for \(a + 2a + a + 5 = 85\) or better or for a trial correctly evaluated A1 for [\(a\) =] 20 [hours] | “correct working” requires at least M1ANDM1 or M2 If working in pence: • Allow up to 5 part marks for consistent working • Allow full marks if answer is clearly stated as 10795 p[ence] M1 implied by sub into \(a + 2a + a\ [+ 5]\) with evaluation B1 max possible for using \(5a\) instead of \(a + 5\) |
| AND M2 for \(\dfrac{25.4[0]}{\textit{their } 20} \times 85\) or 1.27 × 85 oe or 25.4[0] + 50.8[0] + \(\dfrac{25.4[0]}{\textit{their } 20} \times \textit{their } 25\) oe or 25.4[0] + 1.27 × 40 + 1.27 × 25 oe or M1 for \(\dfrac{25.4[0]}{\textit{their } 20}\) implied by 1.27 or \(\dfrac{25.4[0]}{4}\) implied by 6.35 | e.g. M2 for \(\dfrac{25.4[0]}{4} \times 17\) Method marks may be earned in stages May see equivalent algebraic methods. See Appendix. Non-algebraic methods may earn up to full marks. | ||
| If 0 or 1 scored, instead award SC2 for 107.95 with no or insufficient working If 0 scored, instead award SC1 for 20 [hours] with no or insufficient working | |||
Appendix: algebraic methods for Q11
There are possibly many algebraic methods for this question. Examiners should use the main scheme as a template, matching steps or positions in the solution as best as possible. If in doubt, contact your Team Leader. For example:
| Response | Judgement | Mark |
|---|---|---|
| (tips): Amir : Beth : Charlie are 25.4 : 50.8 : 25.4 + 5\(x\) (where \(x\) is hourly rate of tips) | This is on the scheme at B1 | B1 |
| (total tips): 25.4 + 50.8 + 25.4 + 5\(x\) = 85\(x\) | There is an equation on the scheme, so M1 would be a good judgement | M1 |
| (solving): \(x\) = 1.27 | And then this would be the A1 | A1 |
| (substitution into either side of the equation) eg 85 × 1.27 | This is on the scheme at M2 | M2 |
| (final answer) 107.95 | The answer is correct and the candidate has satisfied the “correct working” requirement and so is awarded full marks |