ICAR AIEEA-PG — Agricultural Engineering & Technology (Major Subject Group 10) · India · National Testing Agency for ICAR
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
ICAR AIEEA-PG — Agricultural Engineering (Unit V)
Definitions (field efficiency, timeliness, fixed versus variable cost), formulas and short calculations of capacity, cost and depreciation.
How our questions for ICAR AIEEA-PG spread across the units of Farm Machinery Management & Economics. Difficulty: 16 easy, 28 medium, 0 hard.
| Fixed costs: depreciation, interest, taxes, insurance and housing | 7 | |
| Depreciation methods: straight line, declining balance, sum-of-years digits | 7 | |
| Theoretical and effective field capacity | 4 | |
| Field efficiency and timeliness | 4 | |
| Variable costs: repair and maintenance, fuel and labour | 4 | |
| Break-even use and custom-hire comparison | 4 | |
| Replacement decisions | 4 | |
| Tractor-implement matching | 4 | |
| Mechanisation policy and subsidies | 3 | |
| Machinery selection and sizing | 2 | |
| Custom hiring centre economics | 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.
For the same machine and operator, why is field efficiency usually higher in a long rectangular field than in a short, irregular plot?
B. Headland turning time is a smaller share of total time in long fields
Turning time per turn is roughly fixed; a long field needs fewer turns per hectare, so productive time is a larger share. Draft, overlap and refilling needs do not depend on field length in this way.
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).
Which cost per hectare falls as the size (width) of a machine increases, other things being equal?
B. Operator labour cost
A wider machine finishes in fewer hours, so labour (and timeliness) cost per season falls. Depreciation, interest, insurance and housing all rise with the higher price of a bigger machine.
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).
What is the annual rate used in double-declining-balance depreciation for a machine with an 8-year life?
C. 25 %
Double-declining rate = 2/L = 2/8 = 25 % of the opening book value. 12.5 % is the straight-line rate 1/L; 20 % would be the rate for a 10-year life. 18.75 % is the 150 % declining-balance rate (1.5/8).
If the custom-hire rate for an operation rises while the owner's costs stay the same, the break-even use for ownership:
B. Falls to a smaller area
x* = FC/(R − VC). A larger R increases the denominator and lowers x*. It never reaches zero while FC > 0. So ownership pays at a smaller area than before.
The economic life of a machine is the age at which:
B. Its average annual cost of ownership and operation is lowest
Economic life minimises average annual cost. A machine can often still be repaired well after that age, and book value depends on the accounting method.
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).
While ploughing, a two-wheel-drive tractor shows 30 % wheel slip. What is the most appropriate first action?
A. Add ballast or reduce the draft load until slip falls into the recommended range
30 % slip wastes power and wears tyres; more weight on the driving wheels or a lighter load reduces it. Removing ballast would raise slip further. Too little slip, on the other hand, signals excess ballast.
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