GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)
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
GATE — Agricultural Engineering (AG)
Numerical answer items on theoretical and effective field capacity, field efficiency, cost of operation per hour, depreciation by straight line, declining balance and sum-of-years digits, and break-even area or hours for owning against custom hire.
How our questions for GATE spread across the units of Farm Machinery Management & Economics. Difficulty: 2 easy, 23 medium, 9 hard.
| Depreciation methods: straight line, declining balance, sum-of-years digits | 7 | |
| Theoretical and effective field capacity | 4 | |
| Tractor-implement matching | 4 | |
| 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 | |
| Machinery selection and sizing | 2 | |
| Custom hiring centre economics | 2 |
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).
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 rotavator-tractor set burns 14 L/h of diesel at 90 per litre and covers 0.7 ha/h effectively. What is the fuel cost per hectare?
C. 1,800/ha
Fuel per hectare = 14 / 0.7 = 20 L; cost = 20 × 90 = 1,800 per hectare. Multiplying by capacity gives 882; 1,260 is cost per hour, not per hectare.
A combine costs 500,000, has an 8-year life and 50,000 salvage. What is its book value at the end of year 4 by sum-of-years digits?
A. 175,000
Fractions for years 1–4 = (8 + 7 + 6 + 5)/36 = 26/36; accumulated depreciation = 26/36 × 450,000 = 325,000; book value = 500,000 − 325,000 = 175,000. Straight line gives 275,000; ascending digits (1 + 2 + 3 + 4 = 10) give 375,000.
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).
As a machine ages, how do its annual depreciation and annual repair cost usually change?
B. Depreciation falls and repair cost rises, so average cost is U-shaped
Market value drops most in the early years and repairs grow with accumulated wear. Their sum averaged over years owned is U-shaped, with the minimum at the economic life.
A 3-bottom plough with 30 cm bottoms works at 15 cm depth in soil with a specific resistance of 0.8 kgf/cm², at 4.5 km/h. What drawbar power is needed?
D. 13.24 kW
Cross-section = 3 × 30 × 15 = 1,350 cm²; draft = 0.8 × 1,350 = 1,080 kgf = 10.59 kN. Power = 10.59 × 4.5 / 3.6 = 13.24 kW. Forgetting the 3.6 conversion gives 47.7 kW; 10.59 is the draft in kN, not power; 1.35 treats kgf as newtons, leaving out g (1,080 N × 1.25 m/s = 1.35 kW).
A tractor can deliver 15 kW at the drawbar at 6 km/h. A harrow has a draft of 3.0 kN per metre of width. What is the widest harrow it can pull at that speed?
D. 3.0 m
Available pull = power × 3.6 / speed = 15 × 3.6 / 6 = 9.0 kN; width = 9.0 / 3.0 = 3.0 m. 5.0 m divides power by draft per metre, ignoring speed. 30 m is a factor-of-ten slip; 0.83 m forgets the 3.6 conversion (15/6/3.0).
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