Rajasthan PSC — Assistant Agriculture Engineer (Agriculture Engineering syllabus) · India · Rajasthan Public Service Commission
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 Rajasthan PSC spread across the units of Farm Machinery Management & Economics. Difficulty: 25 easy, 14 medium, 3 hard.
| Custom hiring centre economics | 7 | |
| Mechanisation policy and subsidies | 6 | |
| Break-even use and custom-hire comparison | 5 | |
| Theoretical and effective field capacity | 4 | |
| Variable costs: repair and maintenance, fuel and labour | 4 | |
| Field efficiency and timeliness | 3 | |
| Machinery selection and sizing | 3 | |
| Fixed costs: depreciation, interest, taxes, insurance and housing | 3 | |
| Tractor-implement matching | 3 | |
| Depreciation methods: straight line, declining balance, sum-of-years digits | 2 | |
| Replacement decisions | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
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.
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.
A power tiller costs 650,000, has a life of 10 years and a salvage value of 10 % of cost. What is its annual straight-line depreciation?
B. 58,500
Depreciation = (P − S)/L = (650,000 − 65,000)/10 = 58,500. Ignoring salvage gives 65,000; 71,500 adds salvage instead of subtracting it. 52,000 subtracts the salvage twice ((650,000 − 130,000)/10).
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.
After a 40 % capital subsidy, a CHC's net investment is 2,000,000 and its expected net cash inflow is 450,000 per year. What is the simple payback period?
B. 4.44 years
Payback = investment / annual net inflow = 2,000,000 / 450,000 = 4.44 years. Inverting gives 0.225, the simple return rate, not the payback; 7.41 years uses the gross cost before the 40 % subsidy, which the owner did not pay. 2.22 years uses half the investment, as for an average-investment return (1,000,000/450,000).
Under India's mechanisation subsidy schemes, a machine is generally eligible for subsidy only if it:
A. Has been tested and reported on by a recognised testing institution
Scheme guidelines generally require test reports from recognised institutes, such as the Farm Machinery Training and Testing Institutes, to protect buyers and public money. Exact rules are in the current operational guidelines.
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