## Question 1:

### Part (a)
**Middle right pivot(s):**

$$4 \quad 6.5 \quad 7 \quad 1.3 \quad 2 \quad \boxed{5} \quad 1.5 \quad 6 \quad 4.5 \quad 6 \quad 1$$
$$6.5 \quad 7 \quad \boxed{6} \quad 6 \quad \underline{5} \quad 4 \quad 1.3 \quad 2 \quad \boxed{1.5} \quad 4.5 \quad 1$$
$$6.5 \quad \boxed{7} \quad \underline{6} \quad 6 \quad \underline{5} \quad 4 \quad \boxed{2} \quad 4.5 \quad 1.5 \quad 1.3 \quad \boxed{1}$$
$$\underline{7} \quad 6.5 \quad \underline{6} \quad 6 \quad \underline{5} \quad 4 \quad \boxed{4.5} \quad \underline{2} \quad 1.5 \quad 1.3 \quad \underline{1}$$
$$\underline{7} \quad 6.5 \quad \underline{6} \quad 6 \quad \underline{5} \quad 4.5 \quad 4 \quad \underline{2} \quad 1.5 \quad 1.3 \quad \underline{1}$$

**Middle left pivot(s):**

$$4 \quad 6.5 \quad 7 \quad 1.3 \quad 2 \quad \boxed{5} \quad 1.5 \quad 6 \quad 4.5 \quad 6 \quad 1$$
$$6.5 \quad \boxed{7} \quad 6 \quad 6 \quad \underline{5} \quad 4 \quad 1.3 \quad \boxed{2} \quad 1.5 \quad 4.5 \quad 1$$
$$\underline{7} \quad 6.5 \quad \boxed{6} \quad 6 \quad \underline{5} \quad \boxed{4} \quad 4.5 \quad \underline{2} \quad 1.3 \quad 1.5 \quad 1$$
$$\underline{7} \quad 6.5 \quad \underline{6} \quad 6 \quad \underline{5} \quad 4.5 \quad \underline{4} \quad \underline{2} \quad 1.5 \quad \boxed{1.3} \quad 1$$
$$\underline{7} \quad 6.5 \quad \underline{6} \quad 6 \quad \underline{5} \quad 4.5 \quad \underline{4} \quad \underline{2} \quad 1.5 \quad \underline{1.3} \quad 1$$

| M1 for correct method shown; A1 A1 for correct passes | M1, A1, A1 | 1.1b |

**(3 marks)**

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### Part (b)

| Bin 1: $\boxed{7}$ $\quad 2 \quad 1$ | | |
|---|---|---|
| Bin 2: $\boxed{6.5}$ $\quad 1.5 \quad 1.3$ | | |
| Bin 3: $\boxed{6}$ $\quad 4$ | | |
| Bin 4: $\boxed{6}$ | | |
| Bin 5: $\boxed{5}$ $\quad \boxed{4.5}$ | | |

| Correct bin-packing shown | M1, A1(underlined), A1 | 1.1b |

**(3 marks)**

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### Part (c)

Lower bound for number of rolls required:

$$\frac{4 + 6.5 + 7 + 1.3 + 2 + 5 + 1.5 + 6 + 4.5 + 6 + 1}{10} = \frac{44.8}{10} = 4.48$$

Therefore the minimum number of bins is 5. Yes, the solution is optimal.

| $\frac{44.8}{10} = 4.48$ | M1 | 1.1b |
| Minimum number of bins is 5; Yes, solution is optimal | A1 | 2.2a |

**(2 marks)**

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### Part (d)

| e.g. The lowest two values should be in the two far right positions, and they are not. OR 1.6 is not in the correct position | B1 | 3.1a |

**(1 mark)**

**Total: (9 marks)**

# Question 1:

## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Quick sort: pivot $p$ selected, first pass gives $>p, p, <p$ | M1 | If choosing 1 pivot per iteration, award M1 only. Bubble sort is M0. If sorting into ascending order, pivots chosen for second pass M1 only. If inconsistent with middle left/middle right M1 only. |
| Second and third passes correct | A1 | |
| CSO including fourth pass | A1 | Miscopy/misread can score maximum of M1A0A0 |

## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| First six items placed correctly (values in boxes) with at least eight values placed | M1 | Allow cumulative totals for M1 only |
| First nine items placed correctly (values in boxes and underlined) | A1 | |
| CSO (therefore no repeated values) | A1 | |

## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Lower bound calculation attempted: $[37.5, 51.8]/10$ (accept sight of 4.48) **or** total wastage calculation attempted and compared with 10 **or** total space used and compared with 50 **or** clear statement that four values are over half bin capacity **and** one equals half, therefore minimum bins must be 5 | M1 | Either method acceptable |
| "Minimum 5 bins" with correct numerical argument (4.48) **or** correct numerical justification for why answer to (b) does use minimum number of bins | A1 | CSO. Dependent on correct bin packing in (b). Requires both a statement and a correct numerical argument |

## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Must mention either the two smallest values being in tenth and eleventh positions is not the case, or that second smallest value (1.6) is not in the correct position. Alternatively, may consider both ends: two largest numbers not in order at one end and two smallest numbers not in order at the other end | B1 | Allow candidates to only consider the right-hand end of the given list (condone omission of considering the left-hand end) |

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