NABARD Assistant Manager Grade A (RDBS) · India · National Bank for Agriculture and Rural Development
Original practice questions written to the published syllabus · not past papersShort business courses on individual input and livestock industries: seed, fertiliser, agro-chemicals, dairy, feed, poultry, farm power and renewable energy.
In the exams: agricultural development bank officer recruitment, animal husbandry, fisheries and farm engineering (heavy); agricultural field officer recruitment in banking, crops, soils, horticulture and animal husbandry (heavy)
NABARD Assistant Manager Grade A (RDBS) — Agriculture & Rural Development
Animal husbandry, poultry, dairy and fisheries economics; farm machinery and farm power costing; renewable energy in agriculture; model bankable projects with unit cost, BCR, IRR and repayment.
How our questions for NABARD Assistant Manager Grade A spread across the units of Sector-specific Agribusiness Electives. Difficulty: 4 easy, 20 medium, 8 hard.
| Renewable energy institutions (ABM 529) | 5 | |
| Poultry and hatchery business (ABM 522) | 4 | |
| Dairy plant management and dairy marketing (ABM 527) | 4 | |
| Farm machinery business | 4 | |
| Livestock business | 4 | |
| Farm power and its cost analysis (ABM 517) | 3 | |
| Floriculture business | 3 | |
| Fertiliser manufacture and demand (ABM 512) | 2 | |
| Agro-chemicals and the Insecticides Act (ABM 513) | 1 | |
| Seed programmes and certification (ABM 514) | 1 | |
| Feed plant management and regulation (ABM 520) | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A district grows 150,000 ha of a cereal. The recommended dose is 100 kg N/ha and 70% of the area is fertilised at that dose. If all nitrogen is supplied as urea (46% N), what is the district's urea demand?
D. 22,826 t
N demand = 150,000 x 100 x 0.70 = 10,500,000 kg = 10,500 t N. Urea = 10,500 / 0.46 = 22,826 t. 10,500 t is nutrient, not product; 32,609 t ignores the 70% adoption; 4,830 t multiplies by 0.46 instead of dividing.
A state sows 200,000 ha of a crop at a seed rate of 25 kg/ha and targets a seed replacement rate of 40%. How much quality seed must be supplied?
D. 2,000 t
Seed need = area x seed rate x SRR = 200,000 x 25 x 0.40 = 2,000,000 kg = 2,000 t. 5,000 t ignores the SRR (it is the whole seed requirement); 800 t applies the 40% twice; 20,000 t is a kg-to-tonne slip.
A hatchery sets 12,000 eggs, 88% are fertile, and 9,500 saleable chicks hatch. What is hatchability on fertile eggs?
C. 90.0%
Fertile eggs = 12,000 x 0.88 = 10,560; hatchability on fertile = 9,500 / 10,560 = 90.0%. 79.2% is hatchability on total eggs set; 88.0% is fertility; 10.0% is the share of fertile eggs that failed to hatch.
A broiler flock of 10,000 chicks placed sells 9,600 birds at an average 2.4 kg and consumed 36,864 kg of feed. What is the flock FCR based on live weight sold?
D. 1.60
Live weight sold = 9,600 x 2.4 = 23,040 kg; FCR = 36,864 / 23,040 = 1.60. 1.54 uses chicks placed instead of birds sold; 0.625 inverts the ratio; 3.84 is feed per bird, not per kg.
Under two-axis pricing a dairy pays 750 per kg of fat and 320 per kg of SNF. What is the price per kg of milk testing 4.0% fat and 8.6% SNF?
A. 57.52
Price per kg = (4.0 x 750 + 8.6 x 320)/100 = (3,000 + 2,752)/100 = 57.52. 30.00 is the fat payment only; 27.52 is the SNF payment only; 5,752 is the price per 100 kg read as per kg.
During the flush season, milk procurement exceeds liquid-milk sales. What do dairy plants usually do with the surplus?
B. Convert it into storable products such as skim milk powder and butter
Plants dry surplus skim milk into SMP and churn fat into butter or ghee, then reconstitute or sell these in the lean season. Chilled raw milk cannot be held for months, and cutting the price to zero would break supply relationships.
A 2,000 m2 greenhouse yields 150 stems/m2 a year; 60% are top grade selling at 10 per stem and 40% lower grade at 4 per stem. What is annual revenue?
C. 2,280,000
Stems = 2,000 x 150 = 300,000; average price = 0.6 x 10 + 0.4 x 4 = 7.6; revenue = 300,000 x 7.6 = 2,280,000. 3,000,000 prices every stem at top grade; 2,100,000 uses the simple average of 10 and 4. 1,920,000 swaps the grade shares (0.6 x 4 + 0.4 x 10 = 6.4).
Which feature best describes loose-flower markets in South Asia compared with export cut-flower markets?
A. Open-field growing, lower value per unit, sharp festival price peaks
Loose flowers for garlands and rituals are mostly open-field, high volume and low unit value, traded in open wholesale markets with festival peaks. Greenhouse production, clock auctions and supermarket contracts characterise the export cut-flower trade.
A solar pump costs 375,000; a 60% subsidy leaves the farmer paying 150,000. It replaces 800 L of diesel a year at 90 per litre and saves 6,000 a year in engine maintenance. What is the farmer's simple payback?
D. 1.92 years
Annual saving = 800 x 90 + 6,000 = 78,000; payback = 150,000 / 78,000 = 1.92 years. 2.08 omits the maintenance saving; 4.81 uses the cost before subsidy; 0.52 inverts the ratio.
A 1 MW solar plant costs 30 million, has a 25-year life, a 10% discount rate (capital recovery factor 0.1102), O&M of 0.3 million a year and a 20% capacity utilisation factor. What is its levelised cost of electricity?
B. 2.06 per kWh
Generation = 1,000 kW x 8,760 h x 0.20 = 1,752,000 kWh. Annual cost = 30,000,000 x 0.1102 + 300,000 = 3,606,000. LCOE = 3,606,000 / 1,752,000 = 2.06. 0.86 spreads capital straight-line with no interest; 1.89 leaves out O&M; 0.41 assumes the plant runs at 100% all year.
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