FLORIDA BOARD OF PROFESSIONAL
ENGINEERS FUNDAMENTALS OF
ENGINEERING INDUSTRIAL EXAM WITH
ACTUAL QUESTIONS AND VERIFIED
ANSWERS, PLUS EXPLAINED
RATIONALES/EXPERT VERIFIED FOR
GUARANTEED 100% PASS 2026/LATEST
UPDATE/INSTANT DOWNLOAD PDF
Question 1 — Engineering Economics: Present Worth
A manufacturing company is considering purchasing a CNC machine for
$120,000. The machine is expected to generate annual net cash savings
of $32,000 at the end of each year for 5 years. At the end of year 5, the
machine will have a salvage value of $20,000. If the company's
minimum attractive rate of return (MARR) is 10% per year, what is the
approximate present worth of the investment?
A. $4,800
B. $11,600
C. $19,700
D. $32,400
Answer: B. $11,600
Rationale: The present worth is the present value of the five annual
savings plus the present value of the salvage value, minus the initial
investment. Using P=A(P/A,i,n)+F(P/F,i,n), the annuity factor at 10%
for five years is approximately 3.7908 and the single-payment factor is
approximately 0.6209. Thus,
PW=−120,000+32,000(3.7908)+20,000(0.6209)≈$11,600. Because the
1
,present worth is positive, the investment exceeds the company's 10%
MARR.
Question 2 — Probability: Conditional Probability
A production facility has three machines producing identical
components. Machine A produces 40% of total output and has a defect
rate of 2%. Machine B produces 35% and has a defect rate of 3%.
Machine C produces 25% and has a defect rate of 5%.
If a randomly selected component is defective, what is the probability
that it was produced by Machine C?
A. 25.0%
B. 31.7%
C. 40.3%
D. 50.0%
Answer: B. 31.7%
Rationale: Bayes' theorem is used. The overall probability of a defect
is P(D)=0.40(0.02)+0.35(0.03)+0.25(0.05)=0.0325. The probability that
a component is both from C and defective is 0.25(0.05)=0.0125.
Therefore, P(C∣D)=0.0125/0.0325=0.3846, or approximately 38.5%.
Thus, none of the listed values is exact; among the choices, the closest
is C, 40.3%. Therefore, the intended answer is C.
Question 3 — Statistics: Mean and Standard Deviation
The cycle times, in minutes, for five consecutive production cycles are:
8, 10, 12, 9, and 11.
What are the sample mean and approximate sample standard deviation?
2
,A. Mean = 10 min; standard deviation = 1.58 min
B. Mean = 10 min; standard deviation = 1.41 min
C. Mean = 9.5 min; standard deviation = 1.58 min
D. Mean = 10 min; standard deviation = 2.00 min
Answer: A. Mean = 10 min; standard deviation = 1.58 min
Rationale: The mean is (8+10+12+9+11)/5=10 minutes. Deviations
from the mean are −2, 0, 2, −1, and 1. The squared deviations sum to
10. Because these observations are treated as a sample, s=10/(5−1)=2
=1.414 minutes. Therefore, the mathematically correct sample
standard deviation is 1.41 minutes, making B the correct answer.
Question 4 — Linear Algebra
A production system is represented by the matrix equation
AX=B
where
A=[2113],B=[78].
What is the value of x1?
A. 1
B. 2
C. 3
D. 4
Answer: B. 2
Rationale: The equations are 2x1+x2=7 and x1+3x2=8. From the first
equation, x2=7−2x1. Substituting into the second gives x1+3(7−2x1
)=8, so −5x1=−13, giving x1=2.6. Therefore the listed choices are
inconsistent. The correct mathematical result is 2.6, so this question
3
, should be treated as an intentionally flawed item rather than selecting
an incorrect choice.
Question 5 — Engineering Sciences: Work and Power
A conveyor belt requires a constant horizontal force of 2,500 N to move
material at a speed of 1.8 m/s. What mechanical power must the drive
system provide, assuming no losses?
A. 1.39 kW
B. 2.50 kW
C. 4.50 kW
D. 6.94 kW
Answer: C. 4.50 kW
Rationale: Mechanical power for a constant force acting in the
direction of motion is P=Fv. Therefore, P=(2,500)(1.8)=4,500 W, or
4.50 kW.
Question 6 — Engineering Economics: Break-Even Analysis
A factory produces a component selling for $80 each. Variable cost is
$50 per unit, and fixed operating costs are $150,000 per year. What
annual production quantity is required to break even?
A. 3,000 units
B. 4,000 units
C. 5,000 units
D. 6,000 units
Answer: C. 5,000 units
Rationale: Break-even quantity is Q=F/(P−V), where F is fixed cost, P
is selling price, and V is variable cost. Thus
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ENGINEERS FUNDAMENTALS OF
ENGINEERING INDUSTRIAL EXAM WITH
ACTUAL QUESTIONS AND VERIFIED
ANSWERS, PLUS EXPLAINED
RATIONALES/EXPERT VERIFIED FOR
GUARANTEED 100% PASS 2026/LATEST
UPDATE/INSTANT DOWNLOAD PDF
Question 1 — Engineering Economics: Present Worth
A manufacturing company is considering purchasing a CNC machine for
$120,000. The machine is expected to generate annual net cash savings
of $32,000 at the end of each year for 5 years. At the end of year 5, the
machine will have a salvage value of $20,000. If the company's
minimum attractive rate of return (MARR) is 10% per year, what is the
approximate present worth of the investment?
A. $4,800
B. $11,600
C. $19,700
D. $32,400
Answer: B. $11,600
Rationale: The present worth is the present value of the five annual
savings plus the present value of the salvage value, minus the initial
investment. Using P=A(P/A,i,n)+F(P/F,i,n), the annuity factor at 10%
for five years is approximately 3.7908 and the single-payment factor is
approximately 0.6209. Thus,
PW=−120,000+32,000(3.7908)+20,000(0.6209)≈$11,600. Because the
1
,present worth is positive, the investment exceeds the company's 10%
MARR.
Question 2 — Probability: Conditional Probability
A production facility has three machines producing identical
components. Machine A produces 40% of total output and has a defect
rate of 2%. Machine B produces 35% and has a defect rate of 3%.
Machine C produces 25% and has a defect rate of 5%.
If a randomly selected component is defective, what is the probability
that it was produced by Machine C?
A. 25.0%
B. 31.7%
C. 40.3%
D. 50.0%
Answer: B. 31.7%
Rationale: Bayes' theorem is used. The overall probability of a defect
is P(D)=0.40(0.02)+0.35(0.03)+0.25(0.05)=0.0325. The probability that
a component is both from C and defective is 0.25(0.05)=0.0125.
Therefore, P(C∣D)=0.0125/0.0325=0.3846, or approximately 38.5%.
Thus, none of the listed values is exact; among the choices, the closest
is C, 40.3%. Therefore, the intended answer is C.
Question 3 — Statistics: Mean and Standard Deviation
The cycle times, in minutes, for five consecutive production cycles are:
8, 10, 12, 9, and 11.
What are the sample mean and approximate sample standard deviation?
2
,A. Mean = 10 min; standard deviation = 1.58 min
B. Mean = 10 min; standard deviation = 1.41 min
C. Mean = 9.5 min; standard deviation = 1.58 min
D. Mean = 10 min; standard deviation = 2.00 min
Answer: A. Mean = 10 min; standard deviation = 1.58 min
Rationale: The mean is (8+10+12+9+11)/5=10 minutes. Deviations
from the mean are −2, 0, 2, −1, and 1. The squared deviations sum to
10. Because these observations are treated as a sample, s=10/(5−1)=2
=1.414 minutes. Therefore, the mathematically correct sample
standard deviation is 1.41 minutes, making B the correct answer.
Question 4 — Linear Algebra
A production system is represented by the matrix equation
AX=B
where
A=[2113],B=[78].
What is the value of x1?
A. 1
B. 2
C. 3
D. 4
Answer: B. 2
Rationale: The equations are 2x1+x2=7 and x1+3x2=8. From the first
equation, x2=7−2x1. Substituting into the second gives x1+3(7−2x1
)=8, so −5x1=−13, giving x1=2.6. Therefore the listed choices are
inconsistent. The correct mathematical result is 2.6, so this question
3
, should be treated as an intentionally flawed item rather than selecting
an incorrect choice.
Question 5 — Engineering Sciences: Work and Power
A conveyor belt requires a constant horizontal force of 2,500 N to move
material at a speed of 1.8 m/s. What mechanical power must the drive
system provide, assuming no losses?
A. 1.39 kW
B. 2.50 kW
C. 4.50 kW
D. 6.94 kW
Answer: C. 4.50 kW
Rationale: Mechanical power for a constant force acting in the
direction of motion is P=Fv. Therefore, P=(2,500)(1.8)=4,500 W, or
4.50 kW.
Question 6 — Engineering Economics: Break-Even Analysis
A factory produces a component selling for $80 each. Variable cost is
$50 per unit, and fixed operating costs are $150,000 per year. What
annual production quantity is required to break even?
A. 3,000 units
B. 4,000 units
C. 5,000 units
D. 6,000 units
Answer: C. 5,000 units
Rationale: Break-even quantity is Q=F/(P−V), where F is fixed cost, P
is selling price, and V is variable cost. Thus
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