C1 June 2014 (R) Q10
10. Xin has been given a 14 day training schedule by her coach.
Xin will run for \(A\) minutes on day 1, where \(A\) is a constant.
She will then increase her running time by \((d + 1)\) minutes each day, where \(d\) is a constant.
Yi has also been given a 14 day training schedule by her coach.
Yi will run for \((A - 13)\) minutes on day 1.
She will then increase her running time by \((2d - 1)\) minutes each day.
Given that Yi and Xin will run for the same length of time on day 14,
Given that Xin runs for a total time of 784 minutes over the 14 days,
| Scheme | Marks |
|---|---|
| Attempts to use \(a + (n - 1)\text{"}d\text{"}\) with \(a = A\) and “\(d\)”\(= d + 1\) and \(n = 14\) | M1 |
| \(A + 13(d + 1) = A + 13d + 13\) * | A1* |
| (2) |
Notes
M1: Attempts to use \(a + (n - 1)d\) with \(a = A\) and \(d = d + 1\) AND \(n = 14\)
A1*: cao This is a given answer and there is an expectation that the intermediate answer is seen and that all work is correct with correct brackets.
The expressions \(A + 13(d + 1)\) and \(A + 13d + 13\) should be seen
N.B. If brackets are missing and formula is not stated
e.g. \(A + 13d + 1 \Rightarrow A + 13d + 13\) or \(A + (13)d + 1 \Rightarrow A + 13d + 13\) then this is M0A0
If formula is quoted and \(a = A\) and \(d = d + 1\) is quoted or implied, then M1 A0 may be given
So \(a + (n - 1)d\) followed by \(A + (13)d + 1 = A + 13d + 13\) achieves M1A0
| Scheme | Marks |
|---|---|
| Calculates time for Yi on Day 14\(= (A - 13) + 13(2d - 1)\) | M1 |
| Sets times equal \(A + 13d + 13 = (A - 13) + 13(2d - 1) \Rightarrow d = \ldots\) | M1 |
| \(d = 3\) | A1 cso |
| (3) |
Notes
M1: States a time for Yi on Day 14 = \((A - 13) + 13(2d - 1)\)
M1: Sets their time for Yi, equal to \(A + 13d + 13\) and uses this equation to proceed to \(d =\)
A1: cso \(d = 3\) Needs both M marks and must be simplified to 3 (not 39/13)
[NB Setting each of the times separately equal to 0 leads to \(d = 3\) – this will gain M0A0]
| Scheme | Marks |
|---|---|
| Uses \(\dfrac{n}{2}\left\{2A + (n - 1)(D)\right\}\) with \(n = 14\), and with \(D = d\) or \(d + 1\) | M1 |
| Attempts to solve \(\dfrac{14}{2}\left\{2A + 13\times\text{‘}(d + 1)\text{’}\right\} = 784 \Rightarrow A = \ldots\) | dM1 |
| \(A = 30\) | A1 |
| (3) | |
| (8 marks) |
Notes
M1: Uses the sum formula \(\dfrac{n}{2}\left\{2A + (n - 1)(D)\right\}\) with \(n = 14\) and \(D = d + 1\) or allow \(D = d\) (usually 4 or 3)
NB May use \(\dfrac{n}{2}\left\{A + (A + 13D)\right\}\) with \(n = 14\) and and \(D = d + 1\) or allow \(D = d\) (usually 4 or 3)
dM1: Attempts to solve \(\dfrac{14}{2}\left\{2A + 13\times\text{‘}4\text{’}\right\} = \text{"}784\text{"} \Rightarrow A = \ldots\) (Must use their \(d + 1\) this time)
Allow miscopy of 784
A1: cao \(A = 30\)