EDEXCEL DECISION MATHEMATICS D1 (R) (6689) – JUNE 2014
Question
Number
1. (a)
(b)
(c)
(d)
2. (a)
(b)
(c)
FINAL MARK SCHEME
Scheme
Bin 1: 31 10 19
Bin 2: 38 12
Bin 3: 45
Bin 4: 47
Bin 5: 35
Bin 6: 28
e.g. middle right
31 10 38 45
31 38 45 47
47 31 38 45
47 45 31 38
47 45 38 35
47 45 38 35
Marks
M1 A1 A1
(3)
19
35
35
35
31
31
47 35
28
28 19 10
28 19 12
28 19 12
28 19 12
28 19 12
12
12
10
10
10
10
Pivot 19
Pivots 47, 12
Pivots 45 (10)
Pivot 35
Pivot 28 (38)
(sort complete)
M1
A1
A1ft
A1
(4)
M1 A1
(2)
Bin 1: 47 12
Bin 2: 45 10
Bin 3: 38 19
Bin 4: 35
Bin 5: 31 28
265
 4.417 so yes 5 bins is optimal
60
M1 A1
(2)
11 marks
e. g. Activities 1 and 3 both can only be done by Hugo
B2, 1, 0
J to 1 should be chosen
e. g. J to 1 would release H to do 3.
e. g. if H is retrained then tasks 1 and 3 can still only be done by H.
M1
A–2=P–4=C–5=J–1=H–3
Change status A = 2 – P = 4 – C = 5 – J = 1 – H = 3
Complete matching: A = 2, C = 5, H = 3, J = 1 and P = 4
M1
A1
A1
7 marks
1
A1
(2)
(2)
(3)
EDEXCEL DECISION MATHEMATICS D1 (R) (6689) – JUNE 2014
Question
Number
FINAL MARK SCHEME
Scheme
M1
A1 (SACFB)
A1 (DE)
A1ft (GT)
3. (a)
(b)
4. (a)
Marks
Shortest path S to T: SACFDGT
Time of shortest path S to T is 72 (minutes)
A1
A1ft
Shortest paths S to T including E: SACFDET
Time is 75 (minutes), so 3 (minutes) longer.
DB1
B1, B1ft
9 marks
B(E)D + FI
= 32 + 38 = 70
B(C)F + D(E)I = 25 + 36 = 61*
B(E)I + D(E)F = 20 + 52 = 72
Length = 359 + 61 = 420
M1 A1
A1
A1
A1ft
(5)
M1 A1
(2)
420
 120 = 3360 (seconds)
15
(b)
Time taken =
(c)
e.g. If we start at an odd vertex we will finish at another odd vertex. This
removes the need to repeat the route between them. So we just have to
consider one repeated route rather than two.
B2,1,0
Choose to repeat the shortest route BI (20)
Therefore start at D (and finish at F)
New length = 359 + 20 = 379
379
 120  2 119 = 3270 (seconds)
Time taken =
15
B1
B1
B1
(d)
(3)
(2)
B1
13 marks
2
(6)
(4)
EDEXCEL DECISION MATHEMATICS D1 (R) (6689) – JUNE 2014
Question
Number
FINAL MARK SCHEME
Scheme
B1
B1
B1
B1 R (4)
5. (a)
(b)
(c)
(d)
Marks
Drawing an objective line accept reciprocal gradient
correct objective line minimum length equivalent to (0, 10) to (15,0)
V labelled correctly
M1
A1
A1
7
13 

V  49 , 61 
17 
 17
M1 A1(2)
Testing the correct inequalities for points with integer coordinates
(50, 61)
3
(3)
M1
A1
(2)
11 marks
EDEXCEL DECISION MATHEMATICS D1 (R) (6689) – JUNE 2014
Question
Number
FINAL MARK SCHEME
Scheme
Marks
M1 (7
activities +
1 dummy)
A1 (start +
ABCE)
6. (i)
A1 (DFG +
1st dummy)
A1 (HIJ +
2nd dummy)
A1cso
(ii)
1st dummy – G depends on A only, but D depends on A and C.
2nd dummy – This is so that H and I will not share the same start and end events
or so that H and I can be uniquely described in terms of their end events.
B1
B1
7 marks
4
EDEXCEL DECISION MATHEMATICS D1 (R) (6689) – JUNE 2014
Question
Number
Scheme
FINAL MARK SCHEME
Marks
M1 A1
7. (a)
M1 A1 (4)
(b)
(c)
(d)
M1 A1
(2)
Total float for D = 12 – 4 – 4 = 4
52
 2.36 so 3 workers
22
M1 A1 (2)
e.g.
M1
A1
8.
A1
(3)
11 marks
B1
Minimise C  3x  2 y
Subject to:
x  y  1000
1
 x  y   x, simplifies to y  3x
4
2x  y
( x, y  0 )
B1
M1 A1
M1 A1
6 marks
5