| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{132}{42} = 3.14...$ so lower bound is 4 | M1 A1 | a1M1: Attempt to find the lower bound ($132 ÷ 22$) / 42 (a value of 3.14 (or better) seen with no working can imply this mark). a1A1: CSO - correct calculation seen or 3.14 followed by 4 – accept 3.1 if correct calculation seen. An answer of 4 with no working scores M0A0 |
| Group 1: **8** 17 9 7; Group 2: 14 **18** 10; Group 3: 12 **22**; Group 4: **15** | M1 A1 | b1M1: First six items placed correctly and at least eight items placed in bins – condone cumulative totals for M1 only (the values in bold). b1A1: CSO (so no additional/repeated values) |
| e.g. middle right: 8 17 9 14 18 12 22 10 15 7; 17 14 18 22 15 **12** 8 9 10 7; 22 **18** 17 14 15 **12** 10 8 9 7; 22 **18** 17 15 14 **12** 10 9 8 7; 22 **18** 17 15 14 **12** 10 9 8 7; 22 **18** 17 15 14 **12** 10 9 8 7 | M1 A1 A1ft A1 | c1M1: Quick sort, pivot, p, chosen (must be choosing middle left or right – choosing first/last item as the pivot is M0). After the first pass the list must read (values greater than the pivot), pivot, (values less than the pivot). **If only choosing one pivot per iteration then M1 only**. c1A1: First pass correct and next pivots chosen correctly for the second pass (but the second pass does not need to be correct) – so they must be choosing (if middle right) a pivot values of 18 and 10 for the second pass or (if middle left) a pivot value of 14. a2A1ft: Second and third passes correct (follow through from their first pass and choice of pivots). They do not need to be choosing a pivot for the fourth pass for this mark. c3A1: CSO (correct solution only – all previous marks in this part must have been awarded) including if middle right a fourth pass with the 15 and 7 used as pivots or if middle left a fifth pass with the 8 used as a pivot |
| Group 1: **22 18**; Group 2: **17 15** 10; Group 3: 14 12 9 7; Group 4: 8 | M1 A1 | d1M1: First six items placed correctly and at least eight items placed in bins – condone cumulative totals for M1 only (the values in bold). d1A1: CSO (so no additional/repeated values) |
| B(E)C + G(I)H = (11.2 + 14.5) + (8.3 + 17.2) = 51.2*; B(F)G + C(E)JH = (10.3 + 15.2) + (14.5 + 7.5 + 16.2) = 63.7; B(E)JH + C(I)FG = (11.2 + 7.5 + 16.2) + (14.5 + 4.3 + 15.2) = 68.9; Repeat as: BE, CE, GI, HI | M1 A1 A1 A1 | e1M1: Correct three pairings of the correct four odd nodes (B, C, G and H). e1A1: Any one row correct including pairings and totals. e2A1: All three rows correct including pairings and totals. e3A1: CAO correct arcs clearly stated: BE, CE, GI and HI – must be these arcs and not e.g. BEC, GIH, or BC via E, etc. |
| Route e.g. ABEBFECECIJFGIHIJDCA; Length $= 227.2 + 51.2 = 278.4$ (m) | B1 B1ft | f1B1: Any correct route (checks: 21 vertices, starting and ending at A, BE, CE, GI and HI appearing twice, A(2), B(2), C(2), D(1), E(3), F(2), G(2), H(2), I(3), J(2)). f2B1ft: For $227.2 +$ their smallest repeat out of a choice of at least two totals seen in (e) – this mark is dependent on M1 in (e) |
| Finishing vertex: C; Reduction in lengths: $51.2 – (10.3 + 15.2) = 25.7$ (m) | B1 B1 | g1B1: CAO (C). g2B1: CAO (25.7) – note that the correct answer can come from incorrect working e.g. $11.2 + 14.5 = 25.7$ is B0 (just adding BE and EC together) so this answer need to be checked carefully – correct method is $51.2 – (10.3 + 15.2)$ (subtracting BF and FG from 51.2) but give bod on a correct answer of 25.7 with no working |
| **Total: 18 marks** | | |

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