Fundamentals of Engineering (FE)
Examination Practice Test - NCEES Style
Questions And Correct Answers (Verified
Answers) Plus Rationales 2026 Q&A |
Instant Download Pdf
1. Evaluate the limit as x approaches 0 of (sin 3x) / (tan 2x).
A. 0
B. 1
C. 3/2
D. ∞
Answer: C
Rationale: Using L’Hôpital’s rule or small-angle approximations, sin 3x ≈ 3x and
tan 2x ≈ 2x, so the ratio simplifies to 3x/2x = 3/2. The limit exists and is finite, not 0
or ∞.
2. If A and B are independent events with P(A)=0.4 and P(B)=0.3, what is P(A ∪
B)?
A. 0.12
B. 0.70
C. 0.58
D. 0.40
Answer: C
Rationale: For independent events, P(A∩B)=P(A)P(B)=0.12. By inclusion-exclusion,
P(A∪B)=P(A)+P(B)-P(A∩B)=0.4+0.3-0.12=0.58.
3. A project has an initial cost of $50,000 and annual benefits of $12,000 for
10 years. If the MARR is 8%, what is the net present worth (NPW) using
, single-payment present worth factors?
A. $30,500
B. **$30,520**
C. $25,000
D. $35,000
Answer: B
Rationale: NPW = -50,000 + 12,000(P/A, 8%, 10). (P/A, 8%, 10) = (1 - (1.08)^-
10)/0.08 ≈ 6.7101. 12,000 × 6.7101 = 80,521.2. Minus 50,000 = 30,521.2 ≈
$30,520.
4. A 10-m long simply supported beam carries a uniformly distributed load of
5 kN/m over its entire length. What is the maximum bending moment?
A. 62.5 kN·m
B. 125 kN·m
C. 50 kN·m
D. 250 kN·m
Answer: B
Rationale: For a simply supported beam with UDL w over length L, max moment
occurs at midspan: Mmax = wL²/8 = 5 × 10² / 8 = 500/8 = 62.5 kN·m. Wait,
calculation: 5100/8 = 62.5. Let me recalc: 5*100=500/8=62.5. So A is correct. Let
me correct: I will adjust question. Actually let me change the numbers to make B
correct. Let's say L=20m, w=5 kN/m. Then M=5*400/8=250. Let's fix. I'll revise
item 4. Let's rewrite properly. I'll keep track. I'll generate all items carefully.*
Self-correction: I will ensure calculations are correct. Let me write item 4 again:
4. A 12-m long simply supported beam carries a uniformly distributed load of
4 kN/m. What is the maximum shear force?
A. 48 kN
B. 24 kN
C. 24 kN (Wait, that's same as B. Let's do max bending moment. Let's use
L=10, w=10. M=10*100/8=125. So: A 62.5, B 125, C 250, D 50. Correct B.
Let's just write it correctly.)
, Let me rewrite item 4 properly:
4. A 10-m long simply supported beam carries a uniformly distributed load of
10 kN/m. What is the maximum bending moment?
A. 62.5 kN·m
B. 125 kN·m
C. 250 kN·m
D. 50 kN·m
Answer: B
Rationale: Maximum bending moment for simply supported beam with UDL
is wL²/8 = 10 × 10² / 8 = 1000/8 = 125 kN·m.
5. The derivative of f(x) = e^(2x) cos(3x) is:
A. e^(2x)[2cos(3x) - 3sin(3x)]
B. e^(2x)[2cos(3x) - 3sin(3x)]
C. e^(2x)[2cos(3x) + 3sin(3x)]
D. e^(2x)[-2sin(3x) - 3cos(3x)]
Answer: B
Rationale: Using product rule: f' = 2e^(2x)cos(3x) + e^(2x)(-3sin(3x)) =
e^(2x)[2cos(3x) - 3sin(3x)].
6. For a standard normal distribution, what is the approximate probability that
Z is between -1 and 1?
A. 0.50
B. 0.68
C. 0.68
D. 0.95
Answer: C
Rationale: The empirical rule states that approximately 68% of values lie within
one standard deviation of the mean for a normal distribution, so P(-1 < Z < 1) ≈
0.68.
7. In engineering ethics, a conflict of interest arises when:
A. An engineer works overtime.
Examination Practice Test - NCEES Style
Questions And Correct Answers (Verified
Answers) Plus Rationales 2026 Q&A |
Instant Download Pdf
1. Evaluate the limit as x approaches 0 of (sin 3x) / (tan 2x).
A. 0
B. 1
C. 3/2
D. ∞
Answer: C
Rationale: Using L’Hôpital’s rule or small-angle approximations, sin 3x ≈ 3x and
tan 2x ≈ 2x, so the ratio simplifies to 3x/2x = 3/2. The limit exists and is finite, not 0
or ∞.
2. If A and B are independent events with P(A)=0.4 and P(B)=0.3, what is P(A ∪
B)?
A. 0.12
B. 0.70
C. 0.58
D. 0.40
Answer: C
Rationale: For independent events, P(A∩B)=P(A)P(B)=0.12. By inclusion-exclusion,
P(A∪B)=P(A)+P(B)-P(A∩B)=0.4+0.3-0.12=0.58.
3. A project has an initial cost of $50,000 and annual benefits of $12,000 for
10 years. If the MARR is 8%, what is the net present worth (NPW) using
, single-payment present worth factors?
A. $30,500
B. **$30,520**
C. $25,000
D. $35,000
Answer: B
Rationale: NPW = -50,000 + 12,000(P/A, 8%, 10). (P/A, 8%, 10) = (1 - (1.08)^-
10)/0.08 ≈ 6.7101. 12,000 × 6.7101 = 80,521.2. Minus 50,000 = 30,521.2 ≈
$30,520.
4. A 10-m long simply supported beam carries a uniformly distributed load of
5 kN/m over its entire length. What is the maximum bending moment?
A. 62.5 kN·m
B. 125 kN·m
C. 50 kN·m
D. 250 kN·m
Answer: B
Rationale: For a simply supported beam with UDL w over length L, max moment
occurs at midspan: Mmax = wL²/8 = 5 × 10² / 8 = 500/8 = 62.5 kN·m. Wait,
calculation: 5100/8 = 62.5. Let me recalc: 5*100=500/8=62.5. So A is correct. Let
me correct: I will adjust question. Actually let me change the numbers to make B
correct. Let's say L=20m, w=5 kN/m. Then M=5*400/8=250. Let's fix. I'll revise
item 4. Let's rewrite properly. I'll keep track. I'll generate all items carefully.*
Self-correction: I will ensure calculations are correct. Let me write item 4 again:
4. A 12-m long simply supported beam carries a uniformly distributed load of
4 kN/m. What is the maximum shear force?
A. 48 kN
B. 24 kN
C. 24 kN (Wait, that's same as B. Let's do max bending moment. Let's use
L=10, w=10. M=10*100/8=125. So: A 62.5, B 125, C 250, D 50. Correct B.
Let's just write it correctly.)
, Let me rewrite item 4 properly:
4. A 10-m long simply supported beam carries a uniformly distributed load of
10 kN/m. What is the maximum bending moment?
A. 62.5 kN·m
B. 125 kN·m
C. 250 kN·m
D. 50 kN·m
Answer: B
Rationale: Maximum bending moment for simply supported beam with UDL
is wL²/8 = 10 × 10² / 8 = 1000/8 = 125 kN·m.
5. The derivative of f(x) = e^(2x) cos(3x) is:
A. e^(2x)[2cos(3x) - 3sin(3x)]
B. e^(2x)[2cos(3x) - 3sin(3x)]
C. e^(2x)[2cos(3x) + 3sin(3x)]
D. e^(2x)[-2sin(3x) - 3cos(3x)]
Answer: B
Rationale: Using product rule: f' = 2e^(2x)cos(3x) + e^(2x)(-3sin(3x)) =
e^(2x)[2cos(3x) - 3sin(3x)].
6. For a standard normal distribution, what is the approximate probability that
Z is between -1 and 1?
A. 0.50
B. 0.68
C. 0.68
D. 0.95
Answer: C
Rationale: The empirical rule states that approximately 68% of values lie within
one standard deviation of the mean for a normal distribution, so P(-1 < Z < 1) ≈
0.68.
7. In engineering ethics, a conflict of interest arises when:
A. An engineer works overtime.
