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FE Civil Construction Engineering
Practice Exam 2026–2027 | Full-Length
Questions, Answers & Detailed
Rationales
1. A contractor has an activity with an original duration of 12 days and a normal
direct cost of $18,000. The crash duration is 8 days with a crash direct cost of
$24,000. Assuming a linear cost-duration relationship, what is the cost slope?
A. $750/day
B. $1,000/day
C. $1,500/day
D. $2,000/day
Answer: $1,500/day
Rationale: The cost slope is (Crash Cost − Normal Cost)/(Normal Duration −
Crash Duration) = ($24,000 − $18,000)/(12 − 8) = $1,500 per day.
2. A construction project has an optimistic activity duration of 6 days, most likely
duration of 9 days, and pessimistic duration of 18 days. Using the PERT expected-
duration equation, what is the expected duration?
,A. 9.0 days
B. 9.5 days
C. 10.0 days
D. 11.0 days
Answer: 10.0 days
Rationale: PERT expected duration is te = (a + 4m + b)/6 = (6 + 4(9) + 18)/6 =
60/6 = 10 days.
3. An excavation requires 4,800 bank cubic yards of soil to be hauled. The soil has
a swell factor of 20%. How many loose cubic yards must be hauled?
A. 4,000 yd³
B. 4,800 yd³
C. 5,760 yd³
D. 6,000 yd³
Answer: 5,760 yd³
Rationale: A 20% swell means loose volume = bank volume × 1.20 = 4,800 × 1.20
= 5,760 yd³.
4. A contractor owns a piece of equipment with a purchase price of $180,000,
estimated salvage value of $30,000, and useful life of 6,000 operating hours.
Using straight-line depreciation based on operating hours, what is the
depreciation cost per operating hour?
A. $20/hr
B. $25/hr
C. $30/hr
D. $35/hr
Answer: $25/hr
,Rationale: Depreciable value = $180,000 − $30,000 = $150,000. Dividing by 6,000
hours gives $25/hr.
5. A crawler excavator has a bucket capacity of 2.0 bank yd³. The bucket fill factor
is 0.90, and the excavator completes 30 cycles per hour. What is the estimated
production?
A. 45 bank yd³/hr
B. 54 bank yd³/hr
C. 60 bank yd³/hr
D. 67 bank yd³/hr
Answer: 54 bank yd³/hr
Rationale: Production = bucket capacity × fill factor × cycles/hr = 2.0 × 0.90 × 30
= 54 bank yd³/hr.
6. A haul truck travels 2.5 miles loaded at 20 mph and returns empty at 30 mph. If
loading and dumping together require 6 minutes, what is the approximate cycle
time?
A. 12 min
B. 15 min
C. 18 min
D. 21 min
Answer: 18 minutes
Rationale: Loaded travel = 2.5/20 × 60 = 7.5 min. Empty travel = 2.5/30 × 60 =
5.0 min. Adding 6 minutes gives an 18.5-minute cycle, approximately 18–19
minutes.
, 7. In a CPM network, an activity has an early start of day 12, early finish of day 18,
late start of day 15, and late finish of day 21. What is the total float?
A. 0 days
B. 3 days
C. 6 days
D. 9 days
Answer: 3 days
Rationale: Total float = LS − ES = 15 − 12 = 3 days, equivalently LF − EF = 21 − 18 =
3 days.
8. Which activity is most directly associated with determining the project's critical
path?
A. Activity with the highest direct cost
B. Activity with the greatest resource requirement
C. Activity with zero total float
D. Activity with the longest individual duration
Answer: Activity with zero total float
Rationale: Critical-path activities generally have zero total float under the
standard CPM convention. A long individual duration does not necessarily make
an activity critical.
9. A concrete placement requires 120 yd³. The contractor's concrete pump can
place 30 yd³/hr, while concrete trucks can supply 45 yd³/hr. Ignoring setup and
interruptions, what controls the placement rate?
A. Truck supply
B. Pump capacity
FE Civil Construction Engineering
Practice Exam 2026–2027 | Full-Length
Questions, Answers & Detailed
Rationales
1. A contractor has an activity with an original duration of 12 days and a normal
direct cost of $18,000. The crash duration is 8 days with a crash direct cost of
$24,000. Assuming a linear cost-duration relationship, what is the cost slope?
A. $750/day
B. $1,000/day
C. $1,500/day
D. $2,000/day
Answer: $1,500/day
Rationale: The cost slope is (Crash Cost − Normal Cost)/(Normal Duration −
Crash Duration) = ($24,000 − $18,000)/(12 − 8) = $1,500 per day.
2. A construction project has an optimistic activity duration of 6 days, most likely
duration of 9 days, and pessimistic duration of 18 days. Using the PERT expected-
duration equation, what is the expected duration?
,A. 9.0 days
B. 9.5 days
C. 10.0 days
D. 11.0 days
Answer: 10.0 days
Rationale: PERT expected duration is te = (a + 4m + b)/6 = (6 + 4(9) + 18)/6 =
60/6 = 10 days.
3. An excavation requires 4,800 bank cubic yards of soil to be hauled. The soil has
a swell factor of 20%. How many loose cubic yards must be hauled?
A. 4,000 yd³
B. 4,800 yd³
C. 5,760 yd³
D. 6,000 yd³
Answer: 5,760 yd³
Rationale: A 20% swell means loose volume = bank volume × 1.20 = 4,800 × 1.20
= 5,760 yd³.
4. A contractor owns a piece of equipment with a purchase price of $180,000,
estimated salvage value of $30,000, and useful life of 6,000 operating hours.
Using straight-line depreciation based on operating hours, what is the
depreciation cost per operating hour?
A. $20/hr
B. $25/hr
C. $30/hr
D. $35/hr
Answer: $25/hr
,Rationale: Depreciable value = $180,000 − $30,000 = $150,000. Dividing by 6,000
hours gives $25/hr.
5. A crawler excavator has a bucket capacity of 2.0 bank yd³. The bucket fill factor
is 0.90, and the excavator completes 30 cycles per hour. What is the estimated
production?
A. 45 bank yd³/hr
B. 54 bank yd³/hr
C. 60 bank yd³/hr
D. 67 bank yd³/hr
Answer: 54 bank yd³/hr
Rationale: Production = bucket capacity × fill factor × cycles/hr = 2.0 × 0.90 × 30
= 54 bank yd³/hr.
6. A haul truck travels 2.5 miles loaded at 20 mph and returns empty at 30 mph. If
loading and dumping together require 6 minutes, what is the approximate cycle
time?
A. 12 min
B. 15 min
C. 18 min
D. 21 min
Answer: 18 minutes
Rationale: Loaded travel = 2.5/20 × 60 = 7.5 min. Empty travel = 2.5/30 × 60 =
5.0 min. Adding 6 minutes gives an 18.5-minute cycle, approximately 18–19
minutes.
, 7. In a CPM network, an activity has an early start of day 12, early finish of day 18,
late start of day 15, and late finish of day 21. What is the total float?
A. 0 days
B. 3 days
C. 6 days
D. 9 days
Answer: 3 days
Rationale: Total float = LS − ES = 15 − 12 = 3 days, equivalently LF − EF = 21 − 18 =
3 days.
8. Which activity is most directly associated with determining the project's critical
path?
A. Activity with the highest direct cost
B. Activity with the greatest resource requirement
C. Activity with zero total float
D. Activity with the longest individual duration
Answer: Activity with zero total float
Rationale: Critical-path activities generally have zero total float under the
standard CPM convention. A long individual duration does not necessarily make
an activity critical.
9. A concrete placement requires 120 yd³. The contractor's concrete pump can
place 30 yd³/hr, while concrete trucks can supply 45 yd³/hr. Ignoring setup and
interruptions, what controls the placement rate?
A. Truck supply
B. Pump capacity
