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10 FE practice problems: comparing alternatives, with solutions

By Dr. Mohamed Elansary, PhD Environmental EngineeringUpdated

These ten original problems practice comparing alternatives, a topic from the FE exam specification, using the Engineering Economics, Benefit-Cost Analysis part of the FE Reference Handbook 10.6. Each problem gives the situation and the values with units. Work it with the handbook PDF open and commit to an answer before you open the solution. Every solution shows the handbook page, the equation, the substitution with units, a size check, and the mistake behind each wrong option, so a wrong pick tells you exactly what to fix.

How to use these problems

Give each problem an honest attempt before you open the solution: write the given values with units, find the equation in the FE Reference Handbook PDF, and commit to an answer. Then compare line by line. If you picked a wrong option, read the note for that option; each one names the mistake that produces it.

The problems

Problem 1 · FE, Comparing alternatives

A county flood-control project costs $690,000 now and $26,000 per year to operate. It prevents damages worth $138,000 per year for 23 years. At a MARR of 6% per year, the conventional benefit-cost ratio is most nearly:

  • A 2.00
  • B 1.68
  • C 2.46
  • D 0.595

Handbook: Engineering Economics, Benefit-Cost Analysis, FE Reference Handbook 10.6

Show the worked solution
Answer
B (1.68)
Given
P = $690,000, C = $26,000, B = $138,000, n = 23 years, i = 6% per year
Find
conventional benefit-cost ratio
Handbook
Engineering Economics, Benefit-Cost Analysis and factor formulas, pages 237 and 240
Equation
B/C = PW(benefits)/PW(costs), with (P/A, i, n) = [(1 + i)ⁿ - 1]/[i(1 + i)ⁿ]
Substitute
  1. (P/A, 6%, 23) = 12.3034 (factor table, i = 6%)
  2. PW of benefits = $138,000 × 12.3034 = $1,698,000
  3. PW of costs = $690,000 + $26,000 × 12.3034 = $1,010,000
  4. B/C = $1,698,000/$1,010,000 = 1.68
Result
1.68, 3 significant figures
Check
B/C ≥ 1, so the project is justified at 6%.
Why the others are wrong
  • A: is the modified ratio (benefits minus O&M over first cost), not the conventional one asked for
  • C: left the yearly operating cost out of the cost side
  • D: is the cost-to-benefit ratio (inverted)

Problem 2 · FE, Comparing alternatives

A treatment plant buys a centrifuge for $60,000. It has a 14-year life and a salvage value of $2,000. Using straight-line depreciation, the book value at the end of year 4 is most nearly:

  • A $16,600
  • B $4,140
  • C $43,400
  • D $39,300

Handbook: Engineering Economics, Depreciation and Book Value, FE Reference Handbook 10.6

Show the worked solution
Answer
C ($43,400)
Given
P = $60,000, S = $2,000, n = 14 years, k = 4 years
Find
book value after year 4 ($)
Handbook
Engineering Economics, Depreciation (Straight Line) and Book Value, page 236
Equation
Dj = (P - S)/n; BV = P - Σ Dj
Substitute
  1. Dj = (P - S)/n = ($60,000 - $2,000)/14 = $4,143 per year
  2. BV after 4 years = P - 4 × Dj = $60,000 - 4 × $4,143 = $43,400
Result
$43,400, 3 significant figures
Check
the book value falls in equal steps from $60,000 to the salvage value $2,000 at year 14.
Why the others are wrong
  • A: is the accumulated depreciation, not the book value
  • B: is the yearly depreciation charge, not the book value
  • D: is the book value after 5 years

Problem 3 · FE, Comparing alternatives

A treatment plant buys a centrifuge for $92,000. It has a 6-year life and a salvage value of $10,000. Using straight-line depreciation, the book value at the end of year 5 is most nearly:

  • A $10,000
  • B $78,300
  • C $23,700
  • D $13,700

Handbook: Engineering Economics, Depreciation and Book Value, FE Reference Handbook 10.6

Show the worked solution
Answer
C ($23,700)
Given
P = $92,000, S = $10,000, n = 6 years, k = 5 years
Find
book value after year 5 ($)
Handbook
Engineering Economics, Depreciation (Straight Line) and Book Value, page 236
Equation
Dj = (P - S)/n; BV = P - Σ Dj
Substitute
  1. Dj = (P - S)/n = ($92,000 - $10,000)/6 = $13,670 per year
  2. BV after 5 years = P - 5 × Dj = $92,000 - 5 × $13,670 = $23,700
Result
$23,700, 3 significant figures
Check
the book value falls in equal steps from $92,000 to the salvage value $10,000 at year 6.
Why the others are wrong
  • A: is the book value after 6 years
  • B: subtracted only one year of depreciation
  • D: is the yearly depreciation charge, not the book value

Problem 4 · FE, Comparing alternatives

A treatment plant buys a centrifuge for $62,000. It has a 11-year life and a salvage value of $6,000. Using straight-line depreciation, the book value at the end of year 9 is most nearly:

  • A $45,800
  • B $56,900
  • C $16,200
  • D $5,090

Handbook: Engineering Economics, Depreciation and Book Value, FE Reference Handbook 10.6

Show the worked solution
Answer
C ($16,200)
Given
P = $62,000, S = $6,000, n = 11 years, k = 9 years
Find
book value after year 9 ($)
Handbook
Engineering Economics, Depreciation (Straight Line) and Book Value, page 236
Equation
Dj = (P - S)/n; BV = P - Σ Dj
Substitute
  1. Dj = (P - S)/n = ($62,000 - $6,000)/11 = $5,091 per year
  2. BV after 9 years = P - 9 × Dj = $62,000 - 9 × $5,091 = $16,200
Result
$16,200, 3 significant figures
Check
the book value falls in equal steps from $62,000 to the salvage value $6,000 at year 11.
Why the others are wrong
  • A: is the accumulated depreciation, not the book value
  • B: subtracted only one year of depreciation
  • D: is the yearly depreciation charge, not the book value

Problem 5 · FE, Comparing alternatives

A precast plant has fixed costs of $311,000 per year. Each manhole section sells for $144 and costs $11 to make. The breakeven production is most nearly:

  • A 2,160 units/yr
  • B 2,010 units/yr
  • C 2,340 units/yr
  • D 28,300 units/yr

Handbook: Engineering Economics, Breakeven Analysis, FE Reference Handbook 10.6

Show the worked solution
Answer
C (2,340 units/yr)
Given
F = $311,000, price = $144, v = $11
Find
breakeven quantity (units per year)
Handbook
Engineering Economics, Breakeven Analysis, page 236
Equation
Q = F/(price - variable cost per unit)
Substitute
  1. At breakeven, revenue = cost: price·Q = F + v·Q
  2. Q = F/(price - v) = $311,000/($144 - $11) = 2,340 units per year
Result
2,340 units/yr, 3 significant figures
Check
at Q = 2,340, revenue $337,000 equals cost $337,000.
Why the others are wrong
  • A: divided the fixed cost by the price, ignoring the variable cost
  • B: added the variable cost to the price
  • D: divided the fixed cost by the variable cost

Problem 6 · FE, Comparing alternatives

A county flood-control project costs $670,000 now and $4,500 per year to operate. It prevents damages worth $36,000 per year for 21 years. At a MARR of 6% per year, the conventional benefit-cost ratio is most nearly:

  • A 0.0534
  • B 0.989
  • C 1.71
  • D 0.586

Handbook: Engineering Economics, Benefit-Cost Analysis, FE Reference Handbook 10.6

Show the worked solution
Answer
D (0.586)
Given
P = $670,000, C = $4,500, B = $36,000, n = 21 years, i = 6% per year
Find
conventional benefit-cost ratio
Handbook
Engineering Economics, Benefit-Cost Analysis and factor formulas, pages 237 and 240
Equation
B/C = PW(benefits)/PW(costs), with (P/A, i, n) = [(1 + i)ⁿ - 1]/[i(1 + i)ⁿ]
Substitute
  1. (P/A, 6%, 21) = 11.7641 (factor table, i = 6%)
  2. PW of benefits = $36,000 × 11.7641 = $423,500
  3. PW of costs = $670,000 + $4,500 × 11.7641 = $722,900
  4. B/C = $423,500/$722,900 = 0.586
Result
0.586, 3 significant figures
Check
B/C < 1, so the project is not justified at 6%.
Why the others are wrong
  • A: divided one year of benefit by the first cost plus one year of cost
  • B: used undiscounted totals
  • C: is the cost-to-benefit ratio (inverted)

Problem 7 · FE, Comparing alternatives

A precast plant has fixed costs of $559,000 per year. Each manhole section sells for $138 and costs $91 to make. The breakeven production is most nearly:

  • A 11,900 units/yr
  • B 4,050 units/yr
  • C 6,140 units/yr
  • D 2,440 units/yr

Handbook: Engineering Economics, Breakeven Analysis, FE Reference Handbook 10.6

Show the worked solution
Answer
A (11,900 units/yr)
Given
F = $559,000, price = $138, v = $91
Find
breakeven quantity (units per year)
Handbook
Engineering Economics, Breakeven Analysis, page 236
Equation
Q = F/(price - variable cost per unit)
Substitute
  1. At breakeven, revenue = cost: price·Q = F + v·Q
  2. Q = F/(price - v) = $559,000/($138 - $91) = 11,900 units per year
Result
11,900 units/yr, 3 significant figures
Check
at Q = 11,900, revenue $1,640,000 equals cost $1,640,000.
Why the others are wrong
  • B: added one unit's variable cost to the fixed cost
  • C: divided the fixed cost by the variable cost
  • D: added the variable cost to the price

Problem 8 · FE, Comparing alternatives

A precast plant has fixed costs of $906,000 per year. Each manhole section sells for $96 and costs $54 to make. The breakeven production is most nearly:

  • A 16,800 units/yr
  • B 6,040 units/yr
  • C 21,600 units/yr
  • D 9,440 units/yr

Handbook: Engineering Economics, Breakeven Analysis, FE Reference Handbook 10.6

Show the worked solution
Answer
C (21,600 units/yr)
Given
F = $906,000, price = $96, v = $54
Find
breakeven quantity (units per year)
Handbook
Engineering Economics, Breakeven Analysis, page 236
Equation
Q = F/(price - variable cost per unit)
Substitute
  1. At breakeven, revenue = cost: price·Q = F + v·Q
  2. Q = F/(price - v) = $906,000/($96 - $54) = 21,600 units per year
Result
21,600 units/yr, 3 significant figures
Check
at Q = 21,600, revenue $2,070,000 equals cost $2,070,000.
Why the others are wrong
  • A: divided the fixed cost by the variable cost
  • B: added the variable cost to the price
  • D: divided the fixed cost by the price, ignoring the variable cost

Problem 9 · FE, Comparing alternatives

A county flood-control project costs $490,000 now and $7,000 per year to operate. It prevents damages worth $64,000 per year for 13 years. At a MARR of 4% per year, the conventional benefit-cost ratio is most nearly:

  • A 0.129
  • B 1.14
  • C 1.30
  • D 1.43

Handbook: Engineering Economics, Benefit-Cost Analysis, FE Reference Handbook 10.6

Show the worked solution
Answer
B (1.14)
Given
P = $490,000, C = $7,000, B = $64,000, n = 13 years, i = 4% per year
Find
conventional benefit-cost ratio
Handbook
Engineering Economics, Benefit-Cost Analysis and factor formulas, pages 237 and 240
Equation
B/C = PW(benefits)/PW(costs), with (P/A, i, n) = [(1 + i)ⁿ - 1]/[i(1 + i)ⁿ]
Substitute
  1. (P/A, 4%, 13) = 9.9856 (factor table, i = 4%)
  2. PW of benefits = $64,000 × 9.9856 = $639,100
  3. PW of costs = $490,000 + $7,000 × 9.9856 = $559,900
  4. B/C = $639,100/$559,900 = 1.14
Result
1.14, 3 significant figures
Check
B/C ≥ 1, so the project is justified at 4%.
Why the others are wrong
  • A: divided one year of benefit by the first cost plus one year of cost
  • C: left the yearly operating cost out of the cost side
  • D: used undiscounted totals

Problem 10 · FE, Comparing alternatives

A precast plant has fixed costs of $508,000 per year. Each manhole section sells for $119 and costs $58 to make. The breakeven production is most nearly:

  • A 8,760 units/yr
  • B 8,330 units/yr
  • C 4,270 units/yr
  • D 2,870 units/yr

Handbook: Engineering Economics, Breakeven Analysis, FE Reference Handbook 10.6

Show the worked solution
Answer
B (8,330 units/yr)
Given
F = $508,000, price = $119, v = $58
Find
breakeven quantity (units per year)
Handbook
Engineering Economics, Breakeven Analysis, page 236
Equation
Q = F/(price - variable cost per unit)
Substitute
  1. At breakeven, revenue = cost: price·Q = F + v·Q
  2. Q = F/(price - v) = $508,000/($119 - $58) = 8,330 units per year
Result
8,330 units/yr, 3 significant figures
Check
at Q = 8,330, revenue $991,000 equals cost $991,000.
Why the others are wrong
  • A: divided the fixed cost by the variable cost
  • C: divided the fixed cost by the price, ignoring the variable cost
  • D: added the variable cost to the price

A new problem posts every day inside the lab, with the full worked solution the same evening.

Frequently asked questions

Where is comparing alternatives in the FE Reference Handbook?

Look in the Engineering Economics, Benefit-Cost Analysis part of FE Reference Handbook 10.6. Each solution gives the exact page, so you can practice finding it in the PDF the way you will on exam day.

Are these real FE exam questions?

No. They are original problems generated from the handbook formulas and checked by code. Real exam questions are confidential, and sharing them breaks the NCEES agreement every examinee accepts.

How do I check my answer before opening the solution?

Check the units of your result and whether its size makes sense for the situation. Each solution ends with the same kind of check, so you can compare your habit with ours.

Sources

  1. NCEES FE Civil CBT exam specifications (PDF). Retrieved October 3, 2026.
  2. NCEES FE Environmental CBT exam specifications (PDF). Retrieved October 3, 2026.
  3. NCEES FE Mechanical CBT exam specifications (PDF). Retrieved October 3, 2026.
  4. NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.