D1 June 2009 Q2
2.
| 32 | 45 | 17 | 23 | 38 | 28 | 16 | 9 | 12 | 10 |
The numbers in the list above represent the lengths, in metres, of ten lengths of fabric. They are to be cut from rolls of fabric of length 60m.
| Scheme | Marks |
|---|---|
| \(\dfrac{230}{60} = 3.8\dot{3}\) so 4 needed | M1 A1 |
| (2) |
Notes
(a) 1M1: Their 230 divided by 60, some evidence of correct method 3.8 enough.
1A1: cso 4.
| Scheme | Marks |
|---|---|
| Bin 1: 32 17 9 Bin 2: 45 12 Bin 3: 23 28 Bin 4: 38 16 Bin 5: 10 | M1 A1 A1 A1 |
| (4) |
Notes
(b) 1M1: Use of first fit. Probably 32, 45 and 17 correctly placed.
1A1: 32, 45, 17, 23, 38 and 28 placed correctly
2A1: 32, 45, 17, 23, 38, 28, 16, 9 placed correctly.
3A1: cao
Special case for (b) misread using first fit decreasing.
Give M1A1 (max)
Bin 1: 45 12
Bin 2: 38 17
Bin 3: 32 28
Bin 4: 23 16 10 9
M1 for placing 45, 38, 32, 28 and 23 correctly
A1 for cao.
| Scheme | Marks |
|---|---|
| e.g. Bin 1: 32 28 Bin 2: 38 12 10 Bin 3: 45 9 Bin 4: 23 17 16 | M1 A1 A1 |
| (3) | |
| (9 marks) |
Notes
(c) 1M1: Use of full bin – at least one full bin found and 5 numbers placed.
1A1: 2 full bins found
Eg [32+28 and 38+12+10] [23+28+9 and 16+12+32]
[32+28 and 23+16+12+9] [38+12+10 and 23+28+9]
2A1: A 4 bin solution found.