## Question 6:

### Part 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Odd degree nodes: $C$, $H$, $I$, $P$ | M1 | Sets up model by identifying the problem as a route inspection problem and noting that $C$, $H$, $I$ and $P$ are odd-degree nodes |
| Shortest distances: $C$-$H$: $575$, $I$-$P$: $850$, $C$-$I$: $1000$, $H$-$P$: $700$, $C$-$P$: $850$, $H$-$I$: $1075$ | M1 | Uses the model to find at least one correct total for a pair of shortest distances |
| Pairings: $(C\text{-}H)(I\text{-}P) = 1425^*$, $(C\text{-}I)(H\text{-}P) = 1700$, $(C\text{-}P)(H\text{-}I) = 1925$ | A1 | Finds all three correct totals for the pairs of shortest distances |
| Minimum distance the traffic warden can travel whilst monitoring is $9175 + 1425 = 10\,600$ m | M1 | Determines their correct minimum total distance that the gritter truck will cover during the journey |
| The shortest possible time to grit all roads is $\frac{10600}{5 \times 60} = 35.\dot{3}$ minutes | M1 | Determines their correct least possible time from their minimum total distance |
| Therefore the gritter truck has gritted all of the roads at least once by $7{:}36$ pm, to the nearest minute | A1 | Determines the correct time to the nearest minute. Allow $7{:}35$ pm or $7{:}36$ pm, must have 'pm' |

**Subtotal: 6 marks**

### Part 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| The gritter truck could start at one of the odd degree nodes and finish at another odd degree node. Doing this, the gritter truck would still grit all roads but cover less distance | E1 | Explains that the gritter truck could start at one odd degree node and finish at a different odd degree node, with explicit mention of two of the four odd nodes |
| Meaning it would take less time to grit all the roads | E1 | Explains that the gritter truck would then need to travel less distance and therefore take less time to grit all of the roads |

**Subtotal: 2 marks**

**Question 6 Total: 8 marks**