ITPEC IP April 2025 Question 53
ITPEC IP April 2025 — Question 53 of 100
1 day — the three-day delay makes path A-D one day longer than the original critical path.
The original path durations are A + D = 2 + 4 = 6, B + E = 4 + 3 = 7, and C + F = 7 + 1 = 8 days, so C-F determines the eight-day project. Adding three days across A and D changes A-D to 9 days. The new project duration is therefore 9, an increase of 9 - 8 = 1 day.
Answer (a)
Why not others: (b), (c), and (d) treat more of the activity delay as project delay without accounting for the two days of float originally available on A-D.
Key rule: A delayed noncritical path consumes its total float first; only the portion exceeding float delays the project's finish.
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