Decision 1 Worked Solutions June 2005

1
The Further Mathematics Network
Worked Solutions to MEI Decision Maths 1
January 2005
2
Head Tail
(i)
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8
3
Head Tail
4 7 7 2 6 9 2 4
3
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8
2
6
1
5
(ii)
To finish in 3 tosses, you must end up at 9 Trace
back through tree
4
1
9
4
Head Tail
4 7 7 2 6 9 2 4
3
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8
2
6
1
5
(ii)
To finish in 3 tosses, you must end up at 9 Trace
back through tree
4
1
5 (T) 9
5
Head Tail
4 7 7 2 6 9 2 4
3
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8
2
6
1
5
(ii)
To finish in 3 tosses, you must end up at 9 Trace
back through tree
4
1
4 (H) 5 (T) 9
6
Head Tail
4 7 7 2 6 9 2 4
3
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8
2
6
1
5
(ii)
To finish in 3 tosses, you must end up at 9 Trace
back through tree
4
1
1 (T) 4 (H) 5 (T) 9
7
Head Tail
(i)
4 7 7 2 6 9 2 4
3
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8
2
6
1
5
(iii)
Possible positions after 3 tosses 2
4
1
8
Head Tail
(i)
4 7 7 2 6 9 2 4
3
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8
2
6
1
5
(iii)
Possible positions after 3 tosses 2, 4
4
1
9
Head Tail
(i)
4 7 7 2 6 9 2 4
3
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8
2
6
1
5
(iii)
Possible positions after 3 tosses 2, 4, 6
4
1
10
Head Tail
(i)
4 7 7 2 6 9 2 4
3
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8
2
6
1
5
(iii)
Possible positions after 3 tosses 2, 4, 6, 7
4
1
11
Head Tail
(i)
4 7 7 2 6 9 2 4
3
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8
2
6
1
5
(iii)
Possible positions after 3 tosses 2, 4, 6, 7,
9
4
1
12
Head Tail
(i)
4 7 7 2 6 9 2 4
3
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8
2
6
1
5
(iv)
Game can go on for ever with infinite loops e.g.
1 ? 4 ? 1 ? 4 ? 1 ? 1 ? 2 ? 6 ? 2 ? 6 ?
4
1
13
DIJKSTRAS ALGORITHM
A
B
5
2
C
10
5
6
2
E
D
1
2
4
1
F
G
14
1
0
Perm label
Order
A
B
5
Temp labels
2
C
10
5
6
2
E
D
1
2
4
1
F
G
15
1
0
Perm label
Order
A
B
5
5
Temp labels
2
C
10
5
6
2
E
D
10
1
2
4
1
F
G
16
2
5
1
0
Perm label
Order
A
B
5
5
Temp labels
2
C
10
5
6
2
E
D
10
1
2
4
1
F
G
17
2
5
1
0
Perm label
Order
A
B
5
5
Temp labels
2
C
10
5
6
2
7
E
D
10
10
1
2
4
1
F
G
18
2
5
1
0
Perm label
Order
A
B
5
5
Temp labels
2
C
10
5
6
7
3
2
7
E
D
10
10
1
2
4
1
F
G
19
2
5
1
0
Perm label
Order
A
B
5
5
Temp labels
2
C
10
5
6
3
7
2
7
E
D
10
10 9
1
2
4
1
F
G
13
20
2
5
1
0
Perm label
Order
A
B
5
5
Temp labels
2
C
10
5
6
3
7
2
7
9
4
E
D
10
10 9
1
2
4
1
F
G
13
21
2
5
1
0
Perm label
Order
A
B
5
5
Temp labels
2
C
10
5
6
3
7
2
7
9
4
E
D
10
10 9
1
2
4
1
F
G
11
13
22
2
5
1
0
Perm label
Order
A
B
5
5
Temp labels
2
C
10
5
6
3
7
2
7
9
10
4
5
E
D
10
10 9
1
2
4
1
F
G
11
13
23
2
5
1
0
Perm label
Order
A
B
5
5
Temp labels
2
C
10
5
6
3
7
2
7
9
10
4
5
E
D
10
10 9
1
2
4
1
F
G
11
13
24
2
5
1
0
Perm label
Order
A
B
5
5
Temp labels
2
C
10
5
6
3
7
2
7
9
10
4
5
E
D
10
10 9
1
2
4
11
6
1
F
G
11
13
25
2
5
1
0
Perm label
Order
A
B
5
5
Temp labels
2
C
10
5
6
3
7
2
7
9
5
10
4
E
D
10
10 9
1
2
4
6
11
1
F
G
11
13 12
26
2
5
1
0
Perm label
Order
A
B
5
5
Temp labels
2
C
10
5
6
3
7
2
7
9
5
10
4
E
D
10
10 9
1
2
4
6
11
7
12
1
F
G
11
13 12
27
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28
MINIMUM CONNECTOR Kruskals Algorithm
A
B
5
2
C
10
5
6
2
E
D
1
2
4
1
F
G
Total length 1 1 2
29
MINIMUM CONNECTOR Kruskals Algorithm
A
B
5
2
C
10
5
6
2
E
D
1
2
4
1
F
G
Total length 1 1 2 2
30
MINIMUM CONNECTOR Kruskals Algorithm
A
B
5
2
C
10
5
6
2
E
D
1
2
4
1
F
G
Total length 1 1 2 2 2
31
MINIMUM CONNECTOR Kruskals Algorithm
A
B
5
2
C
10
5
6
2
E
D
1
2
4
1
F
G
Total length 1 1 2 2 2 5 13
32
MINIMUM CONNECTOR Kruskals Algorithm
A
B
5
2
C
10
5
6
2
E
D
1
2
4
1
F
G
Minimum connector length for network 13
33
ALGORITHMS Highest Common Factor
(i)
34
ALGORITHMS Highest Common Factor
(i)
35
ALGORITHMS Highest Common Factor
(i)
The HCF of 84 and 660 is 12
36
ALGORITHMS Highest Common Factor
(ii)
37
ALGORITHMS Highest Common Factor
(ii)
38
ALGORITHMS Highest Common Factor
(ii)
The HCF of 84 and 660 is 12
39
ALGORITHMS Highest Common Factor
(iii)
12 42 130
40
ALGORITHMS Highest Common Factor
(iii)
12 42 130
6 30 212
41
ALGORITHMS Highest Common Factor
(iii)
12 42 130
6 30 212
6 30 212 30 2(42 130)
42
ALGORITHMS Highest Common Factor
(iii)
12 42 130
6 30 212
6 30 212 30 2(42 130)
30 242 230
43
ALGORITHMS Highest Common Factor
(iii)
12 42 130
6 30 212
6 30 212 30 2(42 130)
30 242 230
330 242
44
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45
CRITICAL PATH ANALYSIS Perms and Combs
G 3
E 3
J 3
B 2
A 5
D 5
F 15
H 5
K 10
L 10
N 4
5
10
I 20
C 3
M 15
46
CRITICAL PATH ANALYSIS Perms and Combs
0
5
10
25
30
50
60
70
75
30
Forward pass finding earliest event times
Minimum completion time 75 minutes
47
CRITICAL PATH ANALYSIS Perms and Combs
71
60
50
30
25
10
5
0
75
30
Backward pass finding latest event times
Minimum completion time 75 minutes
48
CRITICAL PATH ANALYSIS Perms and Combs
Critical activities
A 5 0 5
D 10 5 5
F 25 10 15
H 30 25 5
I 50 30 20
K 60 50 10
M 75 60 15
49
CRITICAL PATH ANALYSIS Perms and Combs
10
71
Total float for activity G 71 10 3 58
minutes
50
CRITICAL PATH ANALYSIS Perms and Combs
51
TWO DIGIT SIMULATION RULE
(i)
Since each fraction has a denominator of 7, and
14 7 98, use the first 98 pairs of two-digit
random numbers (00 to 97) and ignore the last two
pairs (98 and 99)
00 - 41
42 - 69
70 - 83
84 - 97
98 - 99
52
TWO DIGIT SIMULATION RULE
00 - 41
42 - 69
70 - 83
84 - 97
98 - 99
(ii)
Simulation of arrival times of five cars at the
first lights Arrival times are cumulative
inter-arrival times
2
2
25
15
2
2
4
29
44
46
53
Simulation of the passage of 10 cars through both
sets of traffic lights using the arrival times at
the 1st set of lights given in the table and the
following table showing when the lights are red
(iii)
54
29 23 52
52
29 0 29
54
29 2 31
31 23 54
70 0 70
29 4 33
33 23 56
44 23 67
70 2 72
44
70 4 74
46
46 23 69
48 23 71
70 6 76
48
65 0 65
88
65 23 88
78
78 23 101
101
103
80
80 23 103
85 23 108
120 0 120
85
55
(iii)
Simulation of the passage of 10 cars through both
sets of traffic lights
Mean delay at 1st lights (4 4 4 0 0
0 12 0 0 0)/10 2.4 seconds
Mean delay at 2nd lights (0 0 14 5 5
5 0 0 0 12)/10 4.1 seconds
(iv)
Output from simulation could be made more
reliable by increasing the number of repetitions.
56
LINEAR PROGRAMMING Making Jam
A recipe for jam states that the weight of sugar
used must be between the weight of fruit used and
four thirds of the weight of fruit used. Georgia
has 10 kg of fruit available and 11 kg of sugar.
Defining the variables
Let s represent the weight of sugar in kg
Let f represent the weight of fruit in kg
57
LINEAR PROGRAMMING Making Jam
A recipe for jam states that the weight of sugar
used must be between the weight of fruit used and
four thirds of the weight of fruit used. Georgia
has 10 kg of fruit available and 11 kg of sugar.
Formulating the inequalities (1)
Weight of sugar must be greater than or equal to
the weight of fruit
? s f
58
LINEAR PROGRAMMING Making Jam
A recipe for jam states that the weight of sugar
used must be between the weight of fruit used and
four thirds of the weight of fruit used. Georgia
has 10 kg of fruit available and 11 kg of sugar.
Formulating the inequalities (2)
Weight of sugar must be less than or equal to
four-thirds of the weight of fruit
59
LINEAR PROGRAMMING Making Jam
A recipe for jam states that the weight of sugar
used must be between the weight of fruit used and
four thirds of the weight of fruit used. Georgia
has 10 kg of fruit available and 11 kg of sugar.
Formulating the inequalities (3)
Weight of sugar must be less than or equal to 11
kg
? s 11
60
LINEAR PROGRAMMING Making Jam
A recipe for jam states that the weight of sugar
used must be between the weight of fruit used and
four thirds of the weight of fruit used. Georgia
has 10 kg of fruit available and 11 kg of sugar.
Formulating the inequalities (4)
Weight of fruit must be less than or equal to 10
kg
? f 10
61
LINEAR PROGRAMMING Making Jam
62
LINEAR PROGRAMMING Making Jam
s f
s f
63
LINEAR PROGRAMMING Making Jam
s f
64
LINEAR PROGRAMMING Making Jam
s f
s 11
s 11
65
LINEAR PROGRAMMING Making Jam
f 10
s f
s 11
f 10
66
LINEAR PROGRAMMING Making Jam
(10, 11)
(8.25, 11)
(10, 10)
(0, 0)
67
LINEAR PROGRAMMING Making Jam

• To produce as much jam as possible,
• maximise f s
• when
• f 10, s 11,
• f s 21
• Hence 21 kg of jam

(10, 11)
68
LINEAR PROGRAMMING Making Jam
(B) To produce as much jam as possible, with
lowest proportion of sugar maximise f
s subject to s f s f 10, f s
20 Hence 20 kg of jam
(10, 10)
69
LINEAR PROGRAMMING Making Jam
(8.25, 11)
70
LINEAR PROGRAMMING Making Jam
(9.5, 11)
(D) To produce as much jam as possible, within a
budget of 15 maximise f s subject to f
0.5s 15 f 9.5, s 11, f s
20.5 Hence 20.5 kg of jam
f 0.5s 15