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ICAR AICE-JRF/SRF Farm Machinery Management & Economics: practice questions

ICAR AICE-JRF/SRF (PhD) — Agricultural Engineering subjects 38–43 · India · National Testing Agency for ICAR

Original practice questions written to the published syllabus · not past papers
10worked here, free
25written to this exam in this subject
11syllabus units

About ICAR AICE-JRF/SRF

ICAR AICE-JRF/SRF (PhD) — Agricultural Engineering subjects 38–43 is a PhD, fellowship and lectureship exam run by National Testing Agency for ICAR, for candidates in India. LAN-based computer test, 2 hours, 120 MCQs: 20 Part A (general knowledge and reasoning), 50 Part B (core group), 50 Part C (specialisation); four options.

Source: icar.org.in.

Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.

Farm Machinery Management & Economics in ICAR AICE-JRF/SRF

Optimum machine size, timeliness cost modelling, replacement theory, system selection and mechanisation policy analysis.

Choosing, sizing, costing and replacing machines, and running a custom hiring centre that pays.

Our 25 ICAR AICE-JRF/SRF questions in this subject: 0 easy, 8 medium, 17 hard, across 11 units.

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

These 10 were written to ICAR AICE-JRF/SRF and are not shown on any other exam's page.

Question 1 · hard

A combine with a 4.2 m header is driven at 5 km/h. Because of overlap only 95 % of the header width is used, and turning and unloading take 20 % of total field time. What is its effective field capacity?

  1. 1.99 ha/h
  2. 1.60 ha/h
  3. 2.10 ha/h
  4. 1.68 ha/h
Answer and working

B. 1.60 ha/h

Effective width = 4.2 × 0.95 = 3.99 m; capacity while cutting = 5 × 3.99 / 10 = 1.995 ha/h; productive time is 80 % of field time, so effective capacity = 1.995 × 0.8 = 1.60 ha/h (field efficiency 76 %). 2.10 is theoretical; 1.68 ignores overlap; 1.99 ignores lost time.

Question 2 · hard

Wheat must be sown on 50 ha with a drill of 1.25 ha/h effective capacity, working 8 h per day; 75 % of days are workable. Yield falls by 0.8 % of 5 t/ha for each day of delay, and grain is worth 20,000 per tonne. If sowing starts on the optimum date, what is the timeliness cost?

  1. 133,333
  2. 266,667
  3. 100,000
  4. 66,667
Answer and working

A. 133,333

Days needed D = 50 / (1.25 × 8 × 0.75) = 6.667. Starting at the optimum, the average delay is D/2, so cost = K × A × Y × V × D / 2 = 0.008 × 50 × 5 × 20,000 × 6.667 / 2 = 133,333. Using z = 4 (operation centred on the optimum) gives 66,667; ignoring workable-day probability gives 100,000. 266,667 forgets to halve the delay (0.008 × 50 × 5 × 20,000 × 6.667).

Question 3 · medium

In the simple linear timeliness model, if an operation of fixed duration is centred on the optimum date instead of starting on it, the timeliness cost:

  1. Falls to zero, because the average date equals the optimum
  2. Halves, because the average deviation falls from D/2 to D/4
  3. Is unchanged, because the total duration is the same
  4. Doubles, because work is done both before and after the optimum
Answer and working

B. Halves, because the average deviation falls from D/2 to D/4

Starting at the optimum, work spans 0 to D days late, average D/2. Centred, it spans −D/2 to +D/2, average absolute deviation D/4, so the loss halves. It is not zero, because every day away from the optimum—early or late—costs yield.

Question 4 · hard

For the same $80,000 tractor (RF1 = 0.007, RF2 = 2.0, illustrative), what repair and maintenance cost is expected for the year in which it goes from 3,000 to 4,000 h?

  1. $3,920
  2. $5,040
  3. $560
  4. $8,960
Answer and working

A. $3,920

Annual cost = accumulated at 4,000 h − accumulated at 3,000 h = 80,000 × 0.007 × (16 − 9) = 3,920. 8,960 is the accumulated total since new, not the cost for that year. 5,040 is the accumulated cost at 3,000 h (80,000 × 0.007 × 9); 560 squares the difference in hours instead of differencing the squares (80,000 × 0.007 × 1²).

Question 5 · hard

A tractor costs 800,000 and is expected to be worth 80,000 after 10 years. What fixed declining-balance rate reduces the book value exactly to the salvage value at the end of the life?

  1. 20.6 %
  2. 9.0 %
  3. 10.0 %
  4. 20.0 %
Answer and working

A. 20.6 %

(1 − r)¹⁰ = S/P = 0.1, so r = 1 − 0.1^0.1 = 0.2057 = 20.6 %. 20 % is the double-declining rate (2/L), which does not end at S; 10 % is S/P; 9 % is the straight-line charge as a share of price.

Question 6 · hard

For the same reaper, hired machines usually arrive late, causing a timeliness loss worth 300 per hectare. Including this loss, what is the new break-even area?

  1. 50.0 ha
  2. 23.7 ha
  3. 75.0 ha
  4. 37.5 ha
Answer and working

D. 37.5 ha

The effective hire cost is 1,600 + 300 = 1,900 per hectare. Break-even = 45,000/(1,900 − 700) = 37.5 ha. Subtracting the loss from the rate (1,300) gives 75 ha, the wrong direction: late hire makes owning more attractive, so break-even falls. 23.7 ha divides by the effective hire cost alone (45,000/1,900).

Question 7 · hard

An owner's variable cost for an operation is 1,200 per hectare, and custom operators charge 1,100 per hectare including operator and fuel. What can be concluded, ignoring timeliness and quality?

  1. Break-even equals fixed cost divided by 1,100
  2. Ownership never pays
  3. Ownership pays only at very small areas
  4. Ownership pays above a break-even of FC/100 ha
Answer and working

B. Ownership never pays

Break-even x* = FC/(R − VC) is negative when R < VC: every hectare costs more to run than to hire, so owning loses more as use grows. Only non-cost factors (timeliness, availability, quality) could justify ownership.

Question 8 · hard

In the ASABE draft equation D = F_i [A + B S + C S²] W T, which factor adjusts the draft for soil texture?

  1. C
  2. A
  3. T
  4. F_i
Answer and working

D. F_i

F_i is the soil-texture adjustment factor (fine, medium, coarse soils). A, B and C are machine-specific constants; T is tillage depth and W width.

Question 9 · hard

A CHC invests 1,200,000 and expects net inflows of 300,000 a year for 6 years plus a salvage of 150,000 at the end of year 6. At 10 % discount rate, what is the net present value?

  1. 191,249
  2. 1,391,249
  3. 750,000
  4. 106,578
Answer and working

A. 191,249

PV of inflows = 300,000 × (P/A,10 %,6) = 300,000 × 4.3553 = 1,306,578; PV of salvage = 150,000 × 0.5645 = 84,671; NPV = 191,249. Undiscounted gives 750,000; omitting salvage gives 106,578; 1,391,249 is the present value of inflows before subtracting the investment.

Question 10 · medium

The FAO and African Union framework for Sustainable Agricultural Mechanization in Africa (2018) emphasises:

  1. Free distribution of tractors to all rural households by national governments
  2. Banning animal traction
  3. Private-sector-led hire services and appropriate, sustainable mechanisation
  4. State-owned tractor-hire fleets run directly by agriculture ministries
Answer and working

C. Private-sector-led hire services and appropriate, sustainable mechanisation

The framework draws on the poor record of earlier state tractor-hire schemes and favours private service providers, appropriate technology and environmental sustainability.

A study plan for Farm Machinery Management & Economics in ICAR AICE-JRF/SRF

This plan follows where our questions actually fall, so the time goes where the marks are likely to be. It is our reading of the syllabus, not the exam body's weighting.

  1. First, the heavy units (56% of the 25 questions): Replacement decisions (5, 20%); Custom hiring centre economics (3, 12%); Field efficiency and timeliness (3, 12%); Machinery selection and sizing (3, 12%).

    Read the manual chapters for these first and do every worked question you can find on them.

  2. Then the middle band: Tractor-implement matching (3, 12%); Break-even use and custom-hire comparison (2, 8%); Mechanisation policy and subsidies (2, 8%); Depreciation methods: straight line, declining balance, sum-of-years digits (1, 4%).

    Aim for one sitting per unit, then mixed questions across them.

  3. Then the lighter units: Fixed costs: depreciation, interest, taxes, insurance and housing (1, 4%); Theoretical and effective field capacity (1, 4%); Variable costs: repair and maintenance, fuel and labour (1, 4%).

    Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.

  4. Leave time for the hard ones: Break-even use and custom-hire comparison (2 hard of 2), Depreciation methods: straight line, declining balance, sum-of-years digits (1 hard of 1), Theoretical and effective field capacity (1 hard of 1).

    Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.

  5. Finish timed. Sit the subject as a timed set without notes, mark it, and send each wrong answer back to its unit above.

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