Engineering Economics is 4-6 questions on the FE Mechanical exam, and it may be the most reliable scoring area on the paper. The problems follow a small number of repeating patterns, and every factor you need is tabulated in the handbook.
The reason candidates still lose points is a cash-flow diagram they never drew. Almost every error in this area is a timing error, not an arithmetic one.
Exam weight: NCEES lists Engineering Economics at 4-6 questions of the 110-question FE Mechanical exam. Work each problem below on paper first, then reveal the worked solution — reading a solution you have not attempted builds recognition, not recall.
What NCEES Tests in Engineering Economics
The specification covers time value of money, cost estimation, economic analyses including present worth and annual cost, comparison of alternatives, depreciation, and rate of return. Expect to convert between present, future and annual amounts, to choose between two machines with different lives, and to compute straight-line or MACRS depreciation.
Alternatives with unequal lives appear often in a mechanical setting, typically two pieces of equipment with different service intervals. Annual-cost comparison handles these without a least-common-multiple study period, which is why it is usually the intended route.
5 Free Engineering Economics Practice Problems
Each engineering economics problem below comes from the FEExamClick FE Mechanical question bank, with worked solutions, and matches the style, difficulty and format of the real exam. Attempt each one under a three-minute limit — roughly the pace the exam demands.
Problem 1 — A. Time value of money
A bridge requires 20,000 per year in perpetual maintenance. At an interest rate of 5% per year, the capitalized cost of this maintenance is most nearly:
Answer: B) 400,000
The capitalized cost of a perpetual uniform series is:
$$CC = \frac{A}{i} = \frac{20{,}000}{0.05} = 400{,}000$$
$$\boxed{CC = 400{,}000}$$
This is the present amount that, at 5%, generates 20,000 of interest each year forever.
Problem 2 — B. Cost types and breakdowns
A mechanical contractor estimates the following costs for an HVAC installation. The bid build-up is shown.
| Item | Cost |
| Materials | \(\$85{,}000\) |
| Direct labor (2,000 hrs at \(\$35\)/hr) | \(\$70{,}000\) |
| Equipment rental | \(\$15{,}000\) |
| Overhead (40% of direct labor) | ? |
| Profit (15% of the total bid price) | ? |
Answer: D) \(\$232{,}941\)
Handbook: this one is not in it. The Engineering Economics section (printed pp. 235–242) covers the interest factors, breakeven, depreciation, benefit-cost and decision trees — there is no bid build-up, no markup and no margin. The section headed Cost Estimation on printed p. 262 belongs to the Chemical Engineering chapter and deals with cost indexes and Lang factors, not contractor bids. Solve this one from definitions.
Step 1 – Build the cost subtotal. Overhead is quoted on direct labor only:
$$\text{Overhead} = 0.40 \times 70{,}000 = \$28{,}000$$
$$S = 85{,}000 + 70{,}000 + 15{,}000 + 28{,}000 = \$198{,}000$$
Step 2 – Read the profit basis carefully. This is the whole question. The profit is 15% of the total bid price \(B\), not 15% of the cost. So \(B\) appears on both sides:
$$B = S + 0.15B$$
Step 3 – Solve for B.
$$B - 0.15B = S \quad\Longrightarrow\quad B(1 - 0.15) = 198{,}000$$
$$B = \frac{198{,}000}{0.85} = \mathbf{\$232{,}941}$$
Step 4 – Answer: (D).
Margin is not markup. A 15% margin means profit is 15% of what you charge; a 15% markup means profit is 15% of what it costs you. They are different numbers and the gap widens fast: at 15% the margin bid is \(\$232{,}941\) and the markup bid is \(\$227{,}700\), but at a 40% rate the margin bid would be \(\$330{,}000\) against a markup bid of \(\$277{,}200\). Check the profit: \(0.15 \times 232{,}941 = \$34{,}941\), and \(198{,}000 + 34{,}941 = \$232{,}941\). It closes.
Step 5 – Eliminate the distractors. Each one drops or misplaces exactly one step:
(A) \(\$198{,}000\) is the cost subtotal with no profit added at all. It is a real intermediate value, which is what makes it tempting.
(B) \(\$200{,}000 = 170{,}000/0.85\) handles the margin correctly but omits the \(\$28{,}000\) of overhead.
(C) \(\$227{,}700 = 198{,}000 \times 1.15\) treats the 15% as a markup on cost. This is the misconception the question exists to test, so it is the option to watch: check that your profit really is 15% of your answer. Here \(0.15 \times 227{,}700 = \$34{,}155\), and \(198{,}000 + 34{,}155 = \$232{,}155 \neq \$227{,}700\) — the number fails its own definition.
Problem 3 — C. Economic analyses
A sensitivity analysis on a manufacturing plant project yields the following results, plotted in the figure. The base-case NPV is \(\$2{,}500{,}000\).
| Variable | Change | Resulting NPV |
| Equipment cost | +10% | \(\$1{,}300{,}000\) |
| Energy price | -25% | \(\$1{,}000{,}000\) |
| Maintenance cost | +20% | \(\$2{,}100{,}000\) |
Answer: A) Equipment cost, with a sensitivity ratio of 4.8
Handbook: not covered. The terms "sensitivity ratio", "spider" and "tornado" appear zero times in FE Reference Handbook 10.6, and so does "NPV" (p. 235 calls the quantity \(P\), present worth). The closest heading, Breakeven Analysis on p. 236, states the one-variable-at-a-time idea in words but gives no formula. The stem therefore defines the ratio for you.
Step 1 – Compute each percentage change in NPV from the \(\$2.5\) M base:
$$\text{Equipment cost:}\quad \frac{2.5 - 1.3}{2.5} = 48\%$$
$$\text{Energy price:}\quad \frac{2.5 - 1.0}{2.5} = 60\%$$
$$\text{Maintenance cost:}\quad \frac{2.5 - 2.1}{2.5} = 16\%$$
Step 2 – Divide each by the change in its own variable. This is the step the whole item turns on: the three variables were not probed by the same amount.
$$\text{Equipment cost:}\quad \frac{48\%}{10\%} = \mathbf{4.8}$$
$$\text{Energy price:}\quad \frac{60\%}{25\%} = 2.4$$
$$\text{Maintenance cost:}\quad \frac{16\%}{20\%} = 0.8$$
Step 3 – Answer: the highest ratio is the equipment cost at 4.8 — (A).
The trap, and what the figure is really showing. The energy price caused the biggest NPV drop in raw dollars: \(\$1.5\) M against the equipment cost's \(\$1.2\) M. Its line reaches the lowest point on the chart. But it took a 25% move to do that, while the equipment cost did nearly as much damage with a 10% move. Per point of change, the equipment cost is twice as dangerous.
On a spider diagram this is simply slope. The energy-price line goes further; the equipment-cost line goes steeper. Sensitivity is the steepness, not the reach — so read the gradient of each ray, never the depth of its endpoint.
A caution on the method. A high sensitivity ratio identifies the variable the NPV reacts to most, not necessarily the greatest risk. Risk is sensitivity combined with how much the variable is actually likely to move. A steep but stable input can matter less than a shallow but volatile one — which is why spider charts are a screening tool, not a decision.
Step 4 – Eliminate the distractors:
(B) 2.4 is the energy price's ratio, correctly computed. It is the right number for the wrong variable, and it is what you choose if you rank by the size of the NPV drop instead of by the ratio.
(C) 0.8 is the maintenance cost's ratio, also correct. A ratio below 1 means the NPV moves proportionally less than the input, so maintenance is the least threatening of the three, not the most.
(D) 0.21 inverts the equipment-cost ratio: \(10/48\) instead of \(48/10\). The right variable, the division upside down. A sensitivity ratio above 1 means amplification; getting 0.21 for the most sensitive input should look wrong immediately.
Problem 4 — A. Time value of money
Equipment costs \(\$40{,}000\), has a useful life of 10 years, a salvage value of \(\$5{,}000\), and annual operating costs of \(\$3{,}000\). An overhaul costing \(\$6{,}000\) is required at the end of year 5. Using an interest rate of 8%, the EUAC is most nearly:
Answer: A) \(\$9{,}200\)
Refer to the Nomenclature and Definitions section in the Engineering Economics chapter of the FE Reference Handbook.
Convert all costs to present worth then annualize:
$$P_{total} = 40{,}000 + 3{,}000(P/A, 8\%, 10) + 6{,}000(P/F, 8\%, 5) - 5{,}000(P/F, 8\%, 10)$$
\((P/A, 8\%, 10) = 6.7101\)
\((P/F, 8\%, 5) = 0.6806\)
\((P/F, 8\%, 10) = 0.4632\)
\(P_{total} = 40{,}000 + 3{,}000(6.7101) + 6{,}000(0.6806) - 5{,}000(0.4632)\)
\(= 40{,}000 + 20{,}130 + 4{,}084 - 2{,}316 = 61{,}898\)
EUAC = \(P_{total} \times (A/P, 8\%, 10)\)
$$(A/P, 8\%, 10) = 0.14903$$
EUAC \(= 61{,}898 \times 0.14903 = 9{,}225\), most nearly \(\$9{,}200\).
Omitting the salvage credit gives \(64{,}214 \times 0.14903 = 9{,}570\), which is option C.
Problem 5 — C. Economic analyses
The equivalent uniform annual cost (EUAC, in dollars) of keeping a machine for N years is: N = 3 gives 9,800; N = 4 gives 9,200; N = 5 gives 9,400; N = 6 gives 10,100. The economic life (the replacement period that minimizes EUAC) is most nearly:
Answer: B) 4 years
The economic life is the replacement period with the lowest EUAC.
N=3: 9,800 N=4: 9,200 (minimum) N=5: 9,400 N=6: 10,100
EUAC is minimized at N = 4.
$$\boxed{\text{Economic life} = 4\ \text{years}}$$
Using the FE Reference Handbook for Engineering Economics
The interest factor tables are the core of this section, and the notation is what trips people. Read (P/A, i, n) as: given A, find P. The letter before the slash is what you want, the letter after is what you have. Locate the tables once during practice so that on exam day you go straight to the right interest rate rather than scrolling through every page of factors.
Four Mistakes That Cost Points
- Not drawing the cash-flow diagram. Thirty seconds of sketching arrows on a timeline prevents the most common error in this area, which is applying a factor at the wrong point in time. Every distractor set contains the answer you get by being one period off.
- Confusing nominal and effective interest. A 12 percent nominal rate compounded monthly is not 12 percent effective. When compounding is stated as anything other than annual, convert before touching the factor tables.
- Comparing alternatives with unequal lives by present worth. Present worth is only valid over a common study period. With a 5-year and a 7-year machine, compare equivalent uniform annual cost instead, or extend to a common multiple.
- Forgetting salvage value in depreciation. Straight-line depreciation uses cost minus salvage over the life. MACRS ignores salvage entirely. Applying the wrong convention produces an answer that is present among the options.
Frequently Asked Questions
How many engineering economics questions are on the FE Mechanical exam?
NCEES specifies 4-6 questions out of 110. The same knowledge area appears on the FE Civil exam at 5-8 questions, so mechanical candidates see slightly fewer.
Do I need to memorise the interest factor formulas?
No. Both the closed-form expressions and the tabulated factors are in the FE Reference Handbook. What you must know is which factor the question calls for, and that comes from the cash-flow diagram, not from memory.
Is MACRS depreciation on the FE Mechanical exam?
It can appear. The handbook provides the MACRS recovery percentages, so the task is reading the correct year from the correct property class. Straight-line depreciation is more common and should be automatic.
What is the fastest way to prepare for this area?
Work twenty problems that all use different factors and force yourself to draw the timeline every time. The patterns repeat so tightly that after twenty problems you will recognise the structure of a new question before you finish reading it.
Keep Going
These topics feed into each other on the exam:
- FE Mechanical Ethics and Professional Practice practice problems — 4-6 questions on the exam
- FE Mechanical Probability and Statistics practice problems — 4-6 questions on the exam
- FE Mechanical Mechanical Design and Analysis practice problems — 10-15 questions on the exam
Done with engineering economics? Browse every knowledge area from the free FE Mechanical practice problem hub, see what the full bank covers on the FE Mechanical exam prep page, or plan your schedule with the FE study timeline. Or test yourself against the clock with the free FE Mechanical practice exam: 20 timed questions in the CBT format.