hardMultiple ChoiceObjective-mapped
PK0-005 Total Float Practice Question
In a network diagram, Activity A has early start (ES) of 0 and early finish (EF) of 5. Activity B depends on A and has ES=5, EF=10. Activity C also depends on A and has ES=5, EF=12. Activity D depends on B and C and has ES=12, EF=15. What is the total float for Activity C?
⚠ Common exam trap
Candidates may mistakenly think that because Activity B has slack (float of 2 on path A-B-D), Activity C also has float. However, C is on the critical path and has no float.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
0
The critical path is A-C-D with duration 5+7+3=15. Activity C is on this longest path, so its total float is 0. Total float is calculated as late start minus early start (5-5=0) or late finish minus early finish (12-12=0). Since C is critical, it has zero float. Option B (0) is correct.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
2
Why it's wrong here
2 is incorrect. This might be the free float of Activity B, not C.
- ✓
0
Why this is correct
0 is correct. Activity C is on the critical path, so its total float is zero.
- ✗
5
Why it's wrong here
5 is incorrect. That is the duration of Activity A, not the float of C.
- ✗
3
Why it's wrong here
3 is incorrect. That is the duration of Activity D, not the float of C.
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