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 papersChoosing, sizing, costing and replacing machines, and running a custom hiring centre that pays.
In the exams: GATE AG Sections 2 and 3 (machinery cost in 5 of 8 papers 2019–26); ICAR AIEEA-PG Unit V; NCEES PE (ASAE D497/EP496 supplied); PRC ABE machinery management
How our questions for Agricultural and Biosystems Engineers Licensure Examination spread across the units of Farm Machinery Management & Economics. Difficulty: 4 easy, 17 medium, 5 hard.
| Field efficiency and timeliness | 3 | |
| Fixed costs: depreciation, interest, taxes, insurance and housing | 3 | |
| Variable costs: repair and maintenance, fuel and labour | 3 | |
| Break-even use and custom-hire comparison | 3 | |
| Replacement decisions | 3 | |
| Custom hiring centre economics | 3 | |
| Theoretical and effective field capacity | 2 | |
| Depreciation methods: straight line, declining balance, sum-of-years digits | 2 | |
| Tractor-implement matching | 2 | |
| Machinery selection and sizing | 1 | |
| Mechanisation policy and subsidies | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A 3-bottom mouldboard plough with 35 cm bottoms works at 4.5 km/h with 80 % field efficiency. How many hours of field time are needed to plough 12 ha?
C. 31.7 h
Width = 3 × 0.35 = 1.05 m; theoretical capacity = 4.5 × 1.05 / 10 = 0.4725 ha/h; effective = 0.378 ha/h; time = 12 / 0.378 = 31.7 h. 25.4 h ignores efficiency; 20.3 h multiplies area by efficiency instead of dividing, which makes the job look shorter when losses should make it longer. 3.17 h is a factor-of-ten slip.
A farm must harvest 60 ha with a machine of 1.5 ha/h effective capacity, working 10 h per day. Only 60 % of days are fit for harvest. How many calendar days should be planned?
B. 7 days
Days = 60 / (1.5 × 10 × 0.6) = 6.67, rounded up to 7. Ignoring the workable-day probability gives 60 / 15 = 4 days, which assumes every day is fit to work. 10 days divides the 4 fit-day answer by the 40 % unfit fraction instead of the 60 % fit fraction (4/0.4); 3 days multiplies by 0.6 instead of dividing (2.4, rounded up).
For the same power tiller (price 650,000, salvage 65,000) with interest at 12 %, what is the annual interest charge on average investment?
C. 42,900
Average investment = (P + S)/2 = (650,000 + 65,000)/2 = 357,500; interest = 357,500 × 0.12 = 42,900. Charging on full price gives 78,000; using (P − S)/2 gives 35,100. 70,200 charges interest on P − S (585,000 × 0.12).
Why is housing cost charged in a machinery budget even for a farm that keeps its machines in the open?
D. Unhoused machines deteriorate faster and lose resale value
The housing charge stands for either the cost of a shed or the extra depreciation and repair from weather exposure. It is a fixed cost, not a tax and not linked to hours.
An operator is paid 75 per hour. The machine's effective field capacity is 0.5 ha/h, and paid labour hours are 10 % more than machine field hours. What is the labour cost per hectare?
D. 165/ha
Machine hours per hectare = 1/0.5 = 2 h; labour hours = 2 × 1.1 = 2.2 h; cost = 2.2 × 75 = 165 per hectare. 150 omits the allowance; 41.25 multiplies by capacity instead of dividing. 135 deducts the 10 % instead of adding it (2 × 0.9 × 75).
A machine's resale value is 580,000 at the end of year 3 and 500,000 at the end of year 4, and its repairs in year 4 cost 110,000. Its minimum average annual cost (over 3 years) is 183,333. What is the marginal cost of keeping it for year 4 (loss in resale value during year 4 plus repairs in year 4)?
D. 190,000
Marginal cost = (580,000 − 500,000) + 110,000 = 190,000. It exceeds the minimum average of 183,333, confirming replacement after year 3. 110,000 is repairs only; 185,000 is the 4-year average, not the marginal cost. 610,000 adds the resale value itself to the repairs (500,000 + 110,000).
An old combine is cheap to own but breaks down for several days in most harvests, causing shattering losses. In a replacement analysis, these losses should be:
A. Added to the old machine's cost of keeping it another year
Downtime in the critical window causes a timeliness loss that is a direct consequence of keeping the old machine, so it belongs in its marginal cost. Leaving it out biases the decision toward keeping unreliable machines.
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