Quick Answer
The answer is 1 day. This is correct because total float, also known as slack, measures how long a task can be delayed without affecting the overall project finish date, and it is calculated by subtracting the duration of a non-critical path from the duration of the critical path. In this network diagram, the critical path (Path 1) totals 14 days, while Task D lies on Path 2, which totals 13 days, so the total float for Task D is 14 minus 13, equaling 1 day. On the CompTIA Project+ PK0-005 exam, this concept tests your ability to identify the critical path and compute float for tasks on parallel paths, often appearing in exhibit-based questions where you must first determine which path is longest. A common trap is confusing total float with free float or assuming all tasks have float, but remember that only tasks not on the critical path have positive total float. Memory tip: “Float is the gap between the path and the longest lap.”
This PK0-005 practice question tests your understanding of project life cycle phases. Read the scenario carefully and evaluate each option against the stated constraints before committing to an answer. After answering, compare your reasoning against the explanation and wrong-answer breakdown below. Once you have made your selection, read the full explanation to reinforce the concept and understand why each distractor is designed to mislead on exam day.
Project Network Diagram - Path Durations: Path 1: Task A (5d) -> Task C (3d) -> Task E (2d) -> Task G (4d) = 14 days Path 2: Task B (4d) -> Task D (4d) -> Task F (3d) -> Task H (2d) = 13 days No dependencies between paths.
Refer to the exhibit. What is the total float for Task D?
Project Network Diagram - Path Durations: Path 1: Task A (5d) -> Task C (3d) -> Task E (2d) -> Task G (4d) = 14 days Path 2: Task B (4d) -> Task D (4d) -> Task F (3d) -> Task H (2d) = 13 days No dependencies between paths.
0 days
Why wrong: Task D is not on the critical path, so float is not zero.
2 days
Why wrong: Incorrect calculation.
3 days
Why wrong: Incorrect calculation.
1 day
Float = 14 - 13 = 1 day.
Answer choices
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
1 day
The critical path is Path 1 (14 days). Task D is on Path 2 with a duration of 13 days. Total float = critical path duration - non-critical path duration = 14 - 13 = 1 day.
Key principle: NAT direction and interface roles matter as much as the IP address mapping. Inside/outside designation controls which traffic is translated.
Answer analysis
For each option: why learners choose it and why it is or isn't the right answer here.
0 days
Why it's wrong here
Task D is not on the critical path, so float is not zero.
2 days
Why it's wrong here
Incorrect calculation.
3 days
Why it's wrong here
Incorrect calculation.
1 day
Why this is correct
Float = 14 - 13 = 1 day.
Related concept
Static NAT maps one inside address to one outside address.
Common exam traps
NAT is not only about the public address. The inside/outside interface roles and the ACL or rule that matches traffic are just as important.
Detailed technical explanation
NAT questions usually test address translation, overload/PAT behaviour, static mappings and whether the right traffic is being translated. Read the interface direction and address terms carefully.
Key takeaway
NAT direction and interface roles matter as much as the IP address mapping. Inside/outside designation controls which traffic is translated.
Real-world example
A small business has 20 workstations on the 192.168.1.0/24 network and one public IP from its ISP. The router uses PAT (NAT overload) so all 20 devices share one public address using different source ports. NAT questions test whether you understand the four address terms and which direction each translation applies.
What to study next
Review the four NAT address types (inside local, inside global, outside local, outside global), PAT port overload, and static vs dynamic NAT use cases. Then practise related PK0-005 NAT questions on configuration and troubleshooting.
Project Life Cycle Phases — study guide chapter
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FAQ
Project Life Cycle Phases — This question tests Project Life Cycle Phases — Static NAT maps one inside address to one outside address..
The correct answer is: 1 day — The critical path is Path 1 (14 days). Task D is on Path 2 with a duration of 13 days. Total float = critical path duration - non-critical path duration = 14 - 13 = 1 day.
Review the four NAT address types (inside local, inside global, outside local, outside global), PAT port overload, and static vs dynamic NAT use cases. Then practise related PK0-005 NAT questions on configuration and troubleshooting.
Static NAT maps one inside address to one outside address.
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Same concept, more angles
These questions test the same concept from different angles. Work through them to make sure you can recognise it however the exam phrases it.
Variation 1. Which phase has the most total float?
mediumWhy D: Phase D is correct because in a typical project schedule, the phase with the most total float is the one that can be delayed the longest without affecting the project's overall completion date. Total float is calculated as the difference between the late start and early start (or late finish and early finish) of an activity. Phase C has the highest total float value, indicating it has the greatest scheduling flexibility.
Last reviewed: Jun 24, 2026
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