FE Civil Engineering Economics Practice
Exam 2026–2027 | Comprehensive
Questions, Answers & Detailed
Rationales
1. An engineering project requires an initial investment of $120,000 and is
expected to generate annual net cash flows of $32,000 for 5 years. If the
interest rate is 8% per year, what is the approximate present worth of the
project?
A. $6,100
B. $7,800
C. $8,200
D. $9,400
Answer: $7,800
Rationale: The present worth of the annuity is
32,000(P/A,8%,5)32,000(P/A,8\%,5). Using P/A=3.9927P/A=3.9927, PW =
$127,766. Subtracting the $120,000 initial investment gives approximately
$7,766, or $7,800.
, 2. A machine costs $85,000 and has a salvage value of $10,000 after 6 years.
Using straight-line depreciation, what is the annual depreciation expense?
A. $10,500
B. $11,250
C. $12,500
D. $14,167
Answer: $12,500
Rationale: Straight-line depreciation is (85,000-10,000)/6=$12,500 per year.
3. An engineer deposits $5,000 at the end of each year for 10 years into an
account earning 6% annually. What is the approximate accumulated
amount immediately after the tenth deposit?
A. $55,000
B. $61,000
C. $65,905
D. $72,000
Answer: $65,905
Rationale: The future value of an ordinary annuity is F=A(F/A,i,n)F=A(F/A,i,n). At
6% for 10 years, F/A=13.181F/A=13.181, giving F=$65,905.
4. A project has a nominal annual interest rate of 9% compounded monthly.
What is its effective annual interest rate?
A. 9.00%
B. 9.27%
C. 9.38%
D. 9.75%
Answer: 9.38%
Rationale: ieff=(1+0.09/12)12−1=0.0938i_{eff}=(1+0.09/12)^{12}-1=0.0938, or
approximately 9.38%.
, 5. A project produces annual savings of $18,000 for 8 years. If the minimum
attractive rate of return is 10%, what is the approximate present worth of
these savings?
A. $82,000
B. $90,000
C. $96,000
D. $110,000
Answer: $96,000
Rationale: P=18,000(P/A,10%,8)P=18,000(P/A,10\%,8). With
P/A≈5.335P/A\approx5.335, the present worth is approximately $96,030.
6. An engineering firm is evaluating a $250,000 project with a useful life of 10
years and no salvage value. If the required rate of return is 8%, what annual
equivalent cost corresponds to the initial investment?
A. $25,000
B. $30,000
C. $37,255
D. $42,500
Answer: $37,255
Rationale: A=P(A/P,8%,10)A=P(A/P,8\%,10). Since
A/P≈0.14903A/P\approx0.14903, the annual equivalent is approximately
$37,258.
7. An investment of $40,000 grows to $60,000 in 5 years. What annual
effective rate of return was earned?
A. 6.5%
B. 7.8%
C. 8.45%
D. 10.0%
Answer: 8.45%
, Rationale: Solve 60,000=40,000(1+i)560,000=40,000(1+i)^5. Thus
i=(1.5)1/5−1≈8.45%i=(1.5)^{1/5}-1\approx8.45\%.
8. A construction project has cash flows of −$500,000 initially, followed by
$150,000 annually for 5 years. At 10%, what is the approximate net present
value?
A. −$32,000
B. −$65,000
C. $68,600
D. $125,000
Answer: $68,600
Rationale: The present value of the five annual receipts is
150,000(3.7908)=$568,620. NPV = $568,620 − $500,000 = $68,620.
9. Two mutually exclusive projects have different useful lives. Which
economic-analysis method is generally appropriate when replacement is
assumed indefinitely?
A. Simple payback period
B. Capitalized cost only
C. Equivalent annual worth
D. Accounting rate of return
Answer: Equivalent annual worth
Rationale: Equivalent annual worth allows mutually exclusive alternatives with
unequal lives to be compared on a common annual basis when repeatability is
assumed.
10.A $100,000 asset is depreciated using straight-line depreciation over 10
years with zero salvage value. What is its book value after 6 years?
A. $20,000
B. $40,000
Exam 2026–2027 | Comprehensive
Questions, Answers & Detailed
Rationales
1. An engineering project requires an initial investment of $120,000 and is
expected to generate annual net cash flows of $32,000 for 5 years. If the
interest rate is 8% per year, what is the approximate present worth of the
project?
A. $6,100
B. $7,800
C. $8,200
D. $9,400
Answer: $7,800
Rationale: The present worth of the annuity is
32,000(P/A,8%,5)32,000(P/A,8\%,5). Using P/A=3.9927P/A=3.9927, PW =
$127,766. Subtracting the $120,000 initial investment gives approximately
$7,766, or $7,800.
, 2. A machine costs $85,000 and has a salvage value of $10,000 after 6 years.
Using straight-line depreciation, what is the annual depreciation expense?
A. $10,500
B. $11,250
C. $12,500
D. $14,167
Answer: $12,500
Rationale: Straight-line depreciation is (85,000-10,000)/6=$12,500 per year.
3. An engineer deposits $5,000 at the end of each year for 10 years into an
account earning 6% annually. What is the approximate accumulated
amount immediately after the tenth deposit?
A. $55,000
B. $61,000
C. $65,905
D. $72,000
Answer: $65,905
Rationale: The future value of an ordinary annuity is F=A(F/A,i,n)F=A(F/A,i,n). At
6% for 10 years, F/A=13.181F/A=13.181, giving F=$65,905.
4. A project has a nominal annual interest rate of 9% compounded monthly.
What is its effective annual interest rate?
A. 9.00%
B. 9.27%
C. 9.38%
D. 9.75%
Answer: 9.38%
Rationale: ieff=(1+0.09/12)12−1=0.0938i_{eff}=(1+0.09/12)^{12}-1=0.0938, or
approximately 9.38%.
, 5. A project produces annual savings of $18,000 for 8 years. If the minimum
attractive rate of return is 10%, what is the approximate present worth of
these savings?
A. $82,000
B. $90,000
C. $96,000
D. $110,000
Answer: $96,000
Rationale: P=18,000(P/A,10%,8)P=18,000(P/A,10\%,8). With
P/A≈5.335P/A\approx5.335, the present worth is approximately $96,030.
6. An engineering firm is evaluating a $250,000 project with a useful life of 10
years and no salvage value. If the required rate of return is 8%, what annual
equivalent cost corresponds to the initial investment?
A. $25,000
B. $30,000
C. $37,255
D. $42,500
Answer: $37,255
Rationale: A=P(A/P,8%,10)A=P(A/P,8\%,10). Since
A/P≈0.14903A/P\approx0.14903, the annual equivalent is approximately
$37,258.
7. An investment of $40,000 grows to $60,000 in 5 years. What annual
effective rate of return was earned?
A. 6.5%
B. 7.8%
C. 8.45%
D. 10.0%
Answer: 8.45%
, Rationale: Solve 60,000=40,000(1+i)560,000=40,000(1+i)^5. Thus
i=(1.5)1/5−1≈8.45%i=(1.5)^{1/5}-1\approx8.45\%.
8. A construction project has cash flows of −$500,000 initially, followed by
$150,000 annually for 5 years. At 10%, what is the approximate net present
value?
A. −$32,000
B. −$65,000
C. $68,600
D. $125,000
Answer: $68,600
Rationale: The present value of the five annual receipts is
150,000(3.7908)=$568,620. NPV = $568,620 − $500,000 = $68,620.
9. Two mutually exclusive projects have different useful lives. Which
economic-analysis method is generally appropriate when replacement is
assumed indefinitely?
A. Simple payback period
B. Capitalized cost only
C. Equivalent annual worth
D. Accounting rate of return
Answer: Equivalent annual worth
Rationale: Equivalent annual worth allows mutually exclusive alternatives with
unequal lives to be compared on a common annual basis when repeatability is
assumed.
10.A $100,000 asset is depreciated using straight-line depreciation over 10
years with zero salvage value. What is its book value after 6 years?
A. $20,000
B. $40,000
