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 (4-5%) 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 problem below comes from the PECivilClick FE Mechanical question bank 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
What is the future value of 10,000 invested for 5 years at 8% annual interest, compounded annually?
A) 15,000
B) 14,000
C) 14,693
D) 13,500
Answer: C) 14,693
Using the Single Payment Compound Amount factor \((F/P, i\%, n)\):
\(F = P(F/P, i\%, n) = P(1 + i)^n\)
From the factor table at \(i = 8\%\), \(n = 5\):
\((F/P, 8\%, 5) = 1.4693\)
\(F = 10,000 \times 1.4693\)
\(\boxed{F = 14,693}\)
Problem 2 — A. Time value of money
A machine costs 15,000 and has a salvage value of 1,000 after 8 years. If the interest rate is 12%, what is the equivalent uniform annual cost (capital recovery)?
A) 3,500
B) 3,200
C) 2,500
D) 2,938
Answer: D) 2,938
The equivalent uniform annual cost (EUAC) is calculated using:
\(EUAC = P(A/P, i\%, n) - S(A/F, i\%, n)\)
From the factor table at \(i = 12\%\), \(n = 8\):
\((A/P, 12\%, 8) = 0.2013\), \((A/F, 12\%, 8) = 0.0813\)
\(EUAC = 15,000 \times 0.2013 - 1,000 \times 0.0813\)
\(EUAC = 3,019.50 - 81.30\)
\(\boxed{EUAC = 2,938}\)
Problem 3 — A. Time value of money
A loan of 10,000 is to be repaid in 5 equal annual payments at 12% interest. What is the annual payment amount?
A) 2,200
B) 2,774
C) 3,000
D) 2,500
Answer: B) 2,774
Using the Capital Recovery factor \((A/P, i\%, n)\):
\(A = P(A/P, i\%, n)\)
From the factor table at \(i = 12\%\), \(n = 5\):
\((A/P, 12\%, 5) = 0.2774\)
\(A = 10,000 \times 0.2774\)
\(\boxed{A = 2,774}\)
Problem 4 — A. Time value of money
If you deposit 1,000 per year for 15 years at 8% annual interest, what will be the accumulated amount?
A) 27,152
B) 25,000
C) 20,000
D) 30,000
Answer: A) 27,152
Using the Uniform Series Compound Amount factor \((F/A, i\%, n)\):
\(F = A(F/A, i\%, n)\)
From the factor table at \(i = 8\%\), \(n = 15\):
\((F/A, 8\%, 15) = 27.1521\)
\(F = 1,000 \times 27.1521\)
\(\boxed{F = 27,152}\)
Problem 5 — A. Time value of money
A company needs 10,000 in 10 years. If the interest rate is 6% per year, how much must be deposited today?
A) 7,500
B) 5,584
C) 6,000
D) 8,000
Answer: B) 5,584
Using the Single Payment Present Worth factor \((P/F, i\%, n)\):
\(P = F(P/F, i\%, n) = F(1 + i)^{-n}\)
From the factor table at \(i = 6\%\), \(n = 10\):
\((P/F, 6\%, 10) = 0.5584\)
\(P = 10,000 \times 0.5584\)
\(\boxed{P = 5,584}\)
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, roughly 4-5 percent. 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
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.