Agricultural and Biosystems Engineers Licensure Examination (ABELE) · Philippines · Professional Regulatory Board of Agricultural and Biosystems Engineering, PRC
Original practice questions written to the published syllabus · not past papersMoney and time in engineering decisions: discounting, costs, break-even, project appraisal, scheduling and starting a business.
In the exams: NCEES FE Engineering Economics 6–9 questions; NCEES PE engineering economics and project management; ICAR AICE subject 41 Unit 10; PRC ABE project management
PRC Agricultural and Biosystems Engineers Licensure Examination
Project management, machinery cost analysis, depreciation and break-even problems set within the allied-subjects papers, solved with a non-programmable calculator.
How our questions for Agricultural and Biosystems Engineers Licensure Examination spread across the units of Engineering Economics, Project & Business Management. Difficulty: 19 easy, 25 medium, 9 hard.
| Fixed and variable costs | 8 | |
| Break-even analysis | 8 | |
| Entrepreneurship and marketing | 8 | |
| NPV, IRR, benefit-cost ratio and payback | 6 | |
| Depreciation methods | 6 | |
| CPM and PERT | 6 | |
| Time value of money | 5 | |
| Replacement analysis | 4 | |
| Project appraisal and detailed project reports | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A grading line costs 240,000 and earns net receipts of 60,000 a year. What is its discounted payback period at 10 % a year, interpolating within the year?
B. 5.4 years
Cumulative discounted receipts: 60,000(P/A, 10 %, n) reaches 240,000 between year 5 (227,447) and year 6 (261,316); interpolation gives 5.4 years. 4.0 years is simple payback, 240,000/60,000, which ignores interest; discounted payback is always longer. 3.6 years compounds the receipts instead of discounting them (240,000/(60,000 × 1.1)).
A tractor costs 800,000, has a salvage value of 80,000 and a recovery period of 8 years. Using double declining balance, what is the depreciation in year 2?
A. 150,000
d = 2/n = 0.25. D1 = 0.25 × 800,000 = 200,000; BV1 = 600,000; D2 = 0.25 × 600,000 = 150,000. 200,000 is year 1; 135,000 wrongly applies the rate to B − S, but declining balance uses book value and ignores salvage until the floor is reached; 90,000 is straight line.
A spice-grinding unit has fixed costs of 600,000 a year, sells at 50 a kg and has variable costs of 30 a kg. What is its break-even output?
A. 30,000 kg
Q* = F/(p − v) = 600,000/(50 − 30) = 30,000 kg. 12,000 divides by price and 20,000 by variable cost; 7,500 adds them instead of subtracting. The contribution margin, p − v, is what covers fixed cost.
A spice-grinding unit has fixed costs of 600,000 a year, a price of 50 a kg and variable costs of 30 a kg. If it plans to sell 40,000 kg, what is its margin of safety as a percentage of planned sales?
A. 25.0 %
Break-even = 600,000/20 = 30,000 kg. Margin of safety = 40,000 − 30,000 = 10,000 kg = 10,000/40,000 = 25.0 % of planned sales. 33.3 % divides by break-even output instead; 75 % is break-even as a share of plan; 20 % is the contribution per kg read as a percentage.
An electric pump set has annual fixed costs of 60,000 and running cost 15 per hour; a diesel set has fixed costs of 20,000 and running cost 55 per hour. Which statement is correct?
B. The electric set is cheaper for any use above 1,000 hours of pumping a year
Equal annual cost: 60,000 + 15h = 20,000 + 55h, so h = 40,000/40 = 1,000 h. Above that, the electric set's lower running cost outweighs its higher fixed cost; below it, diesel is cheaper. Any first costs should be converted to annual amounts with the capital recovery factor before this comparison.
A pump still works reliably, but a new model uses 30 % less energy for the same duty. The loss in value of the old pump for this reason is called:
C. Obsolescence, or functional depreciation
Obsolescence is the loss of value caused by better technology or changed requirements, not by wear. Physical depreciation comes from use, corrosion and accidents. It is one of the main reasons assets are replaced before they wear out.
Activities and durations in days, with predecessors: A 4 (none), B 6 (none), C 5 (after A), D 3 (after A), E 4 (after B and D), F 2 (after C and E). What is the project duration?
D. 13 days
Forward pass: A 0–4, B 0–6, C 4–9, D 4–7, E starts at max(6, 7) = 7 and ends 11, F starts at max(9, 11) = 11 and ends 13. 24 days adds all durations as if in series; 11 days is path A–C–F and 12 days path B–E–F, both shorter than the critical path A–D–E–F.
An activity has a normal time of 10 days at a cost of 50,000 and a crash time of 6 days at a cost of 70,000. What is its cost slope?
A. 5,000 per day
Cost slope = (crash cost − normal cost)/(normal time − crash time) = (70,000 − 50,000)/(10 − 6) = 5,000 per day. 2,000 divides by the normal time; 7,000 divides the whole crash cost by 10; 11,667 divides crash cost by 6 days.
What does resource levelling do in project scheduling?
D. It uses the float of non-critical activities to smooth daily resource demand
Levelling uses the float of non-critical activities to reduce peaks and valleys in resource use without extending the project. Shortening critical activities with extra resources is crashing. If resources are too few even after levelling, resource-constrained scheduling extends the project.
The plan-do-check-act cycle of continuous improvement is associated with:
B. W. Edwards Deming
Deming popularised the PDCA (Shewhart) cycle as the core of quality improvement. Taylor founded scientific management, Fayol set out principles of administration, and Kotler is a marketing author.
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