QCAA General Mathematics Networks and decision mathematics 1
5 sample questions with marking guides and sample answers
Which statement is correct?
A minimum spanning tree must contain a loop.
A minimum spanning tree must contain a cycle.
Every network has only one minimum spanning tree.
A minimum spanning tree has one more vertex than the number of edges.
Reveal Answer
A minimum spanning tree must contain a loop.
By definition, a tree is an acyclic connected graph, meaning it cannot contain any loops or cycles.
A minimum spanning tree must contain a cycle.
A fundamental property of any spanning tree is that it connects all vertices without forming any cycles; if a cycle existed, an edge could be removed to reduce weight while maintaining connectivity.
Every network has only one minimum spanning tree.
While a network with distinct edge weights has a unique MST, networks with duplicate edge weights can have multiple different spanning trees that share the same minimum total weight.
A minimum spanning tree has one more vertex than the number of edges.
A tree with vertices always has exactly edges, so the number of vertices is always one greater than the number of edges.
The table shows information for a project with four activities.
| Activity | Duration (min) | Prerequisite | Earliest starting time | Latest starting time |
|---|---|---|---|---|
| W | 1 | — | 0 | 4 |
| X | 2 | — | 0 | 0 |
| Y | 3 | X | 2 | 2 |
| Z | 4 | W, Y | 5 | 5 |
What is the float time for activity W, in minutes?
0
1
4
5
Reveal Answer
0
A float of 0 indicates a critical activity where the earliest and latest start times are identical. For activity W, these times differ.
1
This value represents the duration of activity W (1 minute), not the available float time.
4
Float time is calculated as the difference between the Latest Starting Time (LST) and the Earliest Starting Time (EST). For activity W, minutes.
5
This value corresponds to the start times for activity Z, rather than the calculated float for activity W.
The precedence table below shows the 12 activities required to complete a project. The duration in days and immediate predecessors are shown.
| Activity | Duration | Immediate predecessors |
|---|---|---|
| 4 | – | |
| 6 | ||
| 8 | ||
| 3 | ||
| 9 | ||
| 6 | ||
| 7 | ||
| 12 | ||
| 6 | ||
| 4 | ||
| 3 | ||
| 9 |
The project is to be completed in minimum time.
The float time, in days, of Activity is
4
6
8
12
Reveal Answer
4
This is the earliest start time (EST) of Activity , not its float time.
6
This is the duration of Activity , not its float time.
8
The earliest start time for Activity is 4 and its latest finish time is 18. The float time is calculated as Latest Finish Time - Earliest Start Time - Duration, which is days.
12
This is the latest start time (LST) of Activity , not its float time.
Activities P and Q are the critical activities for a project.
| Activity | Duration | Prerequisite activity |
|---|---|---|
| P | 3 | – |
| Q | 6 | P |
What are the earliest starting time (EST) and latest starting time (LST) for Activity Q?
3 | 3
3 | 6
6 | 6
6 | 9
Reveal Answer
3 | 3
Activity Q depends on P (duration 3), so its Earliest Start Time (EST) is 3. Since Q is a critical activity, it has zero float, meaning its Latest Start Time (LST) must equal its EST ().
3 | 6
The EST is correct, but the LST is wrong. Because Q is a critical activity, there is no slack, so the Latest Start Time cannot be later than the Earliest Start Time.
6 | 6
This incorrectly identifies the EST as 6. Since P starts at 0 and takes 3 units of time, Q can begin at time 3, not 6.
6 | 9
Both values are incorrect. The EST is determined by the completion of P (time 3), and since Q is critical, the LST must also be 3.
The duration, in minutes, of all activities in a project are shown.
| Activity | P | Q | R | S | T | U | V |
|---|---|---|---|---|---|---|---|
| Duration | 38 | 42 | 32 | 34 | 16 | 14 | 26 |
The critical path for the project is PRSV.
What is the earliest completion time for the project if it starts at 11:00 am?
12:30 pm
1:10 pm
1:30 pm
2:10 pm
Reveal Answer
12:30 pm
This time implies a project duration of 90 minutes. However, the sum of the critical path activities (PRSV) is minutes.
1:10 pm
The earliest completion time is determined by the total duration of the critical path PRSV. Summing the durations gives minutes (2 hours and 10 minutes). Adding this to the 11:00 am start time results in 1:10 pm.
1:30 pm
This time implies a duration of 150 minutes. The correct duration based on the critical path is 130 minutes.
2:10 pm
This time implies a duration of 190 minutes (3 hours and 10 minutes), which is one hour longer than the actual critical path duration of 130 minutes.