FE Industrial & Systems practice questions with answers

Answer: C) 6

• AIncorrect. Include the +x term.
• BIncorrect. That's the integrand at x = 2.
• CCorrect. \([x^2 + x]_0^2 = 4 + 2\).
• DIncorrect. Recheck the antiderivative.

Answer: B) 250 kJ

Sensible heat is Q = mcΔT. Multiplying 5 kg by 0.5 kJ/(kg·K) by 100 K gives 250 kJ.

• AIncorrect. This leaves out the specific heat (5 × 100 = 500). The heat required is Q = mcΔT, so c must be included.
• BCorrect. Q = mcΔT = 5 kg × 0.5 kJ/(kg·K) × 100 K = 250 kJ.
• CIncorrect. This divides by the specific heat instead of multiplying (5 × 100 / 0.5 = 1,000).
• DIncorrect. This leaves out the mass (0.5 × 100 = 50), which gives the heat per kilogram rather than for the full 5 kg.

Answer: A) Permitting misrepresentation of an associate's professional qualifications

The NCEES Model Rules bar licensees from falsifying or permitting misrepresentation of their own or their associates' professional qualifications. A PE who approves listing an unlicensed engineer as a 'P.E.' violates that rule, even when the engineer works under the PE's supervision.

• ACorrect. The Model Rules state that licensees shall not falsify or permit misrepresentation of their, or their associates', academic or professional qualifications. Listing an unlicensed engineer as a P.E. misstates that person's credentials, and the PE allowed it.
• BIncorrect. A conflict of interest involves a business or financial interest that could influence the licensee's judgment. Nothing in the scenario creates such an interest. The problem is a false statement of credentials.
• CIncorrect. Nothing was sealed, and the engineer works under the PE's supervision. Rules on sealing and responsible charge do not apply here. The violation is the false P.E. title in the proposal.
• DIncorrect. Working under a PE's supervision lets an unlicensed engineer do engineering work, but it does not give that person the P.E. title. Only a licensed professional engineer may use it, so presenting the engineer as a P.E. misrepresents their qualifications.

Answer: C) $2,312

• AIncorrect. That divides.
• BIncorrect. Discounting lowers it.
• CCorrect. 500 × 4.623.
• DIncorrect. That's a future worth.

Answer: B) $2,874

EAC = P(A/P,i,n) − S(A/F,i,n) = 3,292 − 418 = $2,874. The annualized salvage value is a credit against the annualized first cost.

• AIncorrect. This annualizes the first cost but leaves out the credit for the $2,000 salvage value.
• BCorrect. EAC = 10,000(0.3292) − 2,000(0.2092) = 3,292 − 418 = $2,874. Equivalently, (10,000 − 2,000)(0.3292) + 2,000(0.12) = $2,874.
• CIncorrect. This applies the A/P factor to the net cost of $8,000 and leaves out the interest on the money tied up in the salvage value (2,000 × 0.12 = $240).
• DIncorrect. This adds the annualized salvage value instead of subtracting it. Salvage is a cash inflow, so it lowers the annual cost.

Answer: A) 2

• ACorrect. (130 − 100)/15.
• BIncorrect. Divide by σ.
• CIncorrect. That's inverted.
• DIncorrect. Subtract the mean first.

Answer: B) 9

• AIncorrect. Subtract 1.
• BCorrect. df = n − 1 = 10 − 1.
• CIncorrect. Subtract 1.
• DIncorrect. Only one parameter is estimated.

Answer: A) 0.5 hour

• ACorrect. W = 1/(μ − λ) = 1/2.
• BIncorrect. That's utilization, not time.
• CIncorrect. That's the mean service time.
• DIncorrect. That's L, the number in the system.

Answer: A) Smooths resource use, possibly extending the schedule

• ACorrect. It reduces peaks and valleys.
• BIncorrect. It may lengthen it.
• CIncorrect. It uses float.
• DIncorrect. It addresses them.

Answer: B) Keeps output constant

• AIncorrect. That's chase.
• BCorrect. Inventory absorbs demand swings; a chase strategy matches output to demand.
• CIncorrect. That's one tactic.
• DIncorrect. That isn't level.

Answer: B) Lowest total cost at that volume

• AIncorrect. Variable cost matters at high volume.
• BCorrect. Total cost = fixed + variable × volume.
• CIncorrect. That raises total cost.
• DIncorrect. That isn't the criterion.

Answer: A) 1 to 2 weeks

• ACorrect. Gradual exposure builds tolerance.
• BIncorrect. That's too short.
• CIncorrect. It's much faster.
• DIncorrect. People do acclimatize.

Answer: B) 2.53 minutes

Normal time = observed time × rating = 2.0 × 1.10 = 2.20 min. Standard time = normal time × (1 + allowance) = 2.20 × 1.15 ≈ 2.53 min.

• AIncorrect. 2.0 × 1.10 = 2.20 min is the normal time. The allowances have not been added yet.
• BCorrect. Normal time = 2.0 × 1.10 = 2.20 min. Standard time = 2.20 × (1 + 0.15) ≈ 2.53 min.
• CIncorrect. 2.0 × 1.15 = 2.30 min applies the allowances but leaves out the 110% performance rating.
• DIncorrect. 2.20 / 1.15 ≈ 1.91 min divides the normal time by the allowance factor. Allowances increase the time, so you multiply by 1.15.

Answer: B) Rank complaint types to show which few account for most

A Pareto chart ranks categories by frequency in descending order to show the vital few that cause most problems. Capability studies, control charts and fishbone diagrams do other jobs.

• AIncorrect. Comparing process spread with specification limits is process capability analysis (for example, Cp and Cpk), not a Pareto chart.
• BCorrect. A Pareto chart puts the categories in descending order of frequency, usually with a cumulative line. This shows the vital few categories that cause most of the problem, so the team can focus there first.
• CIncorrect. Plotting a statistic over time against control limits is what a control chart does. A Pareto chart has no time axis and no control limits.
• DIncorrect. Grouping possible causes into branches (for example, man, machine, method, material) is a cause-and-effect (fishbone) diagram.

Answer: B) 150 hours

• AIncorrect. Redundancy adds life.
• BCorrect. For two active parallel units, MTBF = 1.5/λ = 1.5/0.01.
• CIncorrect. That's for perfect standby.
• DIncorrect. That's series.