A2 June 2024 Q3
3.
[The total weight of the network is 413]
Figure 1 represents a network of cycle tracks between ten towns, A, B, C, D, E, F, G, H, J and K. The number on each arc represents the length, in kilometres, of the corresponding track.
Abi needs to travel along every track shown in Figure 1 to check that they are all in good repair. She needs to start her inspection route at town G and finish her route at either town J or town K.
Abi wishes to minimise the total distance required to traverse every track.
- state which tracks she will repeat in her route
- state the total length of her route
The direct track between town B and town C and the direct track between town H and town K are now closed to all users. A second person, Tarig, is asked to check all the remaining tracks starting at G and finishing at H.
Tarig wishes to minimise the total length of his inspection route.
| Scheme | Marks | AO |
|---|---|---|
 | M1 A1 (ABCDE) A1 (GFH) A1ft (JK) | 1.1b 1.1b 1.1b 1.1b |
| Shortest path from A to J is ACBDGFHJ | A1 | 2.2a |
| (5) |
Notes
In (a) it is important that all values at each node are checked very carefully – the order of the working values must be correct for the corresponding A mark to be awarded e.g. at B the working values must be 25 23 in that order (so 23 25 is incorrect)
It is also important that the order of labelling is checked carefully – some candidates start with a label of 0 at A (rather than 1) – which is fine. Also the order of labelling must be a strictly increasing sequence – so 1, 2, 3, 3, 4, … will be penalised once (see notes below) but 1, 2, 3, 5, 6, … is fine. Errors in the final values and working values are penalised before errors in the order of labelling
M1: A larger value replaced by a smaller value in at least two of the working boxes at either B or D or F or H or K or J
A1: All values in A, B, C, D and E correct. Condone lack of 0 in A’s working value
A1: All values G, F and H correct and the working values in the correct order. Penalise order of labelling only once per question (G, F and H must be labelled in that order and G must be labelled after A, B, C, D and E)
A1ft: All values in K and J correct on the follow through and the working values in the correct order. Penalise order of labelling only once per question. To follow through K, check that the working value at K follows from the candidate’s final values from their feeds into K (which will come from nodes E, F and/or H (in the order in which the candidate has labelled them)) and that the final value, and order of labelling, follows through correctly. Repeat this process for J (which will possibly have working values from E, F and H with the order of these values determined by the candidate’s order of labelling at E, F and H)
A1: CAO - correct path from A to J (ACBDGFHJ)
| Scheme | Marks | AO |
|---|---|---|
| If finishing at J then pair A, B, C and K: AB + CK = 23 + 73 = 96 AC + BK = 18 + 68 = 86* AK + BC = 91 + 5 = 96 | M1 A1ft | 3.1b 1.1b |
| If finishing at K then pair A, B, C and J: AB + CJ = 23 + 76 = 99 AC + BJ = 18 + 71 = 89 AJ + BC = 94 + 5 = 99 | M1 dep A1ft | 1.1b 1.1b |
| Finish at J and repeated AC, BD, DG, GF, FH, HK | A1 | 2.2a |
| Total length of route is 413 + 86 = 499 (km) | A1ft | 2.2a |
| (6) |
Notes
(b) Note: FT marks here are only for their values at K and J
M1: One correct set (either ABCK or ABCJ) of three distinct pairings of the correct four odd nodes (so must have AB + CK, AC + BK and AK + BC or AB + CJ, AC + BJ and AJ + BC)
A1ft: Any three rows correct including pairings and totals, from either set ABCK or set ABCJ (this can be 3 correct from one set or 2 correct from one set and 1 correct from the other set)
dM1: All six distinct pairings for nodes ABCK and ABCJ – dependent on first M mark
A1ft: All six rows correct including pairings and totals
A1: cao correct edges clearly stated and not just in their working. Must be edges AC, BD, DG, GF, FH, HK and clearly selecting to finish at J
A1ft: Either 499 from correct working or 413 + their “86” – dependent on first four marks in this part
To check pairings use
| AB | CK | 96 or Final value at K + 5 |
| AC | BK | 86 or Final value at K - 5 |
| AK | BC | 96 or Final value at K + 5 |
| AB | CJ | 99 or Final value at J + 5 |
| AC | BJ | 89 or Final value at J - 5 |
| AJ | BC | 99 or Final value at J + 5 |
| Scheme | Marks | AO |
|---|---|---|
| e.g. Tarig’s route length: (94 + 2) – 5 – 6 = 85 or with the addition of 413 giving 413 + 94 + 2 – 5 – 6 = 498 or 402 (413 – 5 – 6) + 96 (AJ) = 498 | M1 | 3.1b |
| Therefore Tarig’s route is shorter (dependent on 498 and 499 or 85 and 86 seen) | A1 | 2.2a |
| (2) | ||
| (13 marks) |
Notes
M1: Calculating the length of Tarig’s route by considering the length of shortest path from A to J (either 94 or follow through final value at J from (a)) + 2 (as we can no longer take the shortest path from A to D as BC has been removed) – 5 (BC) – 6 (HK) A correct value of 498 or 85 implies this mark (but not from incorrect working)
A1: CAO (dependent on obtaining both correct values of 499 and 498 or 86 and 85)