C1 June 2010 Q9
9. A farmer has a pay scheme to keep fruit pickers working throughout the 30 day season. He pays £\(a\) for their first day, £\((a + d)\) for their second day, £\((a + 2d)\) for their third day, and so on, thus increasing the daily payment by £\(d\) for each extra day they work.
A picker who works for all 30 days will earn £40.75 on the final day.
A picker who works for all 30 days will earn a total of £1005
| Scheme | Marks |
|---|---|
| \(a + 29d = 40.75 \qquad\) or \(\qquad a = 40.75 - 29d \qquad\) or \(\qquad 29d = 40.75 - a\) | M1 A1 |
| (2) |
Notes
M1: for attempt to use \(a + (n - 1)d\) with \(n = 30\) to form an equation.
So \(a + (30 - 1)d =\) any number is OK
A1: as written. Must see \(29d\) not just \((30 - 1)d\).
Ignore any floating £ signs e.g. \(a + 29d = \)£40.75 is OK for M1A1
These two marks must be scored in (a). Some may omit (a) but get correct equation in (c) [or (b)] but we do not give the marks retrospectively.
| Scheme | Marks |
|---|---|
| \(\left(S_{30}\right) = \dfrac{30}{2}(a + l)\) or \(\dfrac{30}{2}(a + 40.75)\) or \(\dfrac{30}{2}(2a + (30 - 1)d)\) or \(15(2a + 29d)\) | M1 |
| So \(\quad 1005 = 15[a + 40.75] \quad\) * | A1 cso |
| (2) |
Notes
Parts (b) and (c) may run together
M1: for an attempt to use an \(S_n\) formula with \(n = 30\).
Must see one of the printed forms. (\(S_{30} =\) is not required)
A1cso: for forming an equation with 1005 and \(S_n\) and simplifying to printed answer.
Condone £ signs e.g. \(15[a + \)£\(40.75] = 1005\) is OK for A1
| Scheme | Marks |
|---|---|
| \(67 = a + 40.75 \qquad\) so \(\quad \underline{a = (\text{£})\ 26.25 \text{ or } 2625\text{p} \text{ or } 26\tfrac{1}{4} \quad \text{NOT } \tfrac{105}{4}}\) | M1 A1 |
| \(29d = 40.75 - 26.25\) \(\quad = 14.5 \qquad\) so \(\quad \underline{d = (\text{£})0.50 \text{ or } 0.5 \text{ or } 50\text{p}}\) or \(\dfrac{1}{2}\) | M1 A1 |
| (4) | |
| (8 marks) |
Notes
Parts (b) and (c) may run together
1st M1: for an attempt to simplify the given linear equation for \(a\). Correct processes.
Must get to \(ka = \ldots\) or \(k = a + m\) i.e. one step (division or subtraction) from \(a = \ldots\)
Commonly: \(15a = 1005 - 611.25\ (= 393.75)\)
1st A1: For \(a = 26.25\) or 2625p or \(26\tfrac{1}{4}\) NOT \(\tfrac{105}{4}\) or any other fraction
2nd M1: for correct attempt at a linear equation for \(d\), follow through their \(a\) or equation in (a)
Equation just has to be linear in \(d\), they don’t have to simplify to \(d = \ldots\)
2nd A1: depends upon 2nd M1 and use of correct \(a\). Do not penalise a second time if there were minor arithmetic errors in finding \(a\) provided \(a = 26.25\) (o.e.) is used.
Do not accept other fractions other than \(\dfrac{1}{2}\)
If answer is in pence a “p” must be seen.
Sim Equ Use this scheme: 1st M1A1 for \(a\) and 2nd M1A1 for \(d\).
Typically solving: \(1005 = 30a + 435d\) and \(40.75 = a + 29d\).
If they find \(d\) first then follow through use of their \(d\) when finding \(a\).