ASRB NET and ARS / SMS / STO (Agricultural Scientists Recruitment Board) · India · Agricultural Scientists Recruitment Board
Original practice questions written to the published syllabus · not past papersASRB NET and ARS / SMS / STO (Agricultural Scientists Recruitment Board) is a government recruitment exam run by Agricultural Scientists Recruitment Board, for candidates in India. NET: 150 MCQs, 150 marks, 2 hours, -1/3 per wrong answer. ARS: preliminary 150, then descriptive mains 300, then interview 45.
Official site: pw.live.
We have not been able to confirm these details against the exam body's own specification, so fees, dates and pass marks are left out on purpose. Check the current format, eligibility, fees and dates on the official site.
Breeds, housing, feeding, breeding and record keeping for cattle, buffalo, sheep and goats: the husbandry a field veterinarian advises on every day.
Our 23 ASRB NET and ARS / SMS / STO questions in this subject: 0 easy, 2 medium, 21 hard, across 14 units.
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 ASRB NET and ARS / SMS / STO and are not shown on any other exam's page.
A state planner wants annual estimates of milk production by district. Which data source is designed for this purpose in India?
D. The Integrated Sample Survey (Basic Animal Husbandry Statistics)
Production of milk, eggs, meat and wool is estimated every year from the Integrated Sample Survey and published by DAHD as Basic Animal Husbandry Statistics. The livestock census counts animals, not production, and only about every five years; FAOSTAT compiles national totals and does not give district data.
In designing heat abatement for a high-yielding Holstein-Friesian herd, which sequence follows the usual order of priority?
B. Shade, then ventilation, then evaporative cooling with fans
Shade removes radiant heat load cheaply and is the first step; ventilation then removes heat and moisture; evaporative cooling (sprinklers or foggers combined with fans) adds the greatest cooling. Wetting cows without air movement in humid conditions can raise humidity around the animal without effective cooling.
Why is heat detection harder in buffalo than in cattle?
B. Buffalo show weak signs, often at night, and frequent silent heat
Buffalo heat is weakly expressed and frequently nocturnal, and silent heat is common, especially in summer. They cycle through much of the year though breeding peaks from late monsoon to winter; oestrus lasts about a day, and ovulation follows its end.
Why should dairy heifers not be over-fattened before puberty?
A. Excess pre-pubertal fat in the udder can reduce future milk-secreting tissue
Rapid fattening before puberty has been associated with more fat and less secretory parenchyma in the mammary gland. Better-fed heifers actually reach puberty earlier, and fatness does not cause permanent anoestrus.
A cow calves on day 0. Test-day yields are 14, 16, 15, 13, 12, 11, 10, 9, 8 and 7 kg on days 15, 45, 75, 105, 135, 165, 195, 225, 255 and 285. Using the test interval method to day 305, her lactation yield is:
A. 3,485 kg
First period: 15 x 14 = 210. Nine 30-day intervals, each credited with the mean of its two tests: 30 x (15 + 15.5 + 14 + 12.5 + 11.5 + 10.5 + 9.5 + 8.5 + 7.5) = 30 x 104.5 = 3,135. Last period: 20 x 7 = 140. Total = 210 + 3,135 + 140 = 3,485 kg. 3,507.5 kg multiplies the simple mean of the tests (11.5) by 305 and ignores interval lengths; 3,135 kg leaves out the first and last periods; 3,345 kg leaves out the last period.
Why is dentition-based age estimation less reliable in zebu cattle and buffalo than in well-fed taurine cattle?
B. Eruption is later and more variable, influenced by breed and nutrition
Late-maturing zebu, poorly fed cattle and buffalo erupt permanent incisors later, with buffalo reaching full mouth at around five years; standard taurine tables overestimate their youthfulness. They do have deciduous incisors and incisiform canines.
A 100-cow herd replaces 25% of its cows each year and loses 10% of heifers between birth and first calving. How many heifer calves must be reared each year?
B. About 28
Heifers to rear = 25 / (1 - 0.10) = 25 / 0.9 = 27.8, about 28. Multiplying 25 by 0.9 gives about 23, the wrong direction; 35 adds 10 heifers instead of 10%.
A cow bought for 80,000 is expected to be sold for 20,000 after 6 years. Annual straight-line depreciation is:
C. 10,000
(80,000 - 20,000) / 6 = 60,000 / 6 = 10,000. 13,333 is 80,000/6; 16,667 is (80,000 + 20,000)/6; 3,333 is 20,000/6.
A dairy unit has annual fixed costs of 120,000, sells milk at 40 per litre and has variable costs of 25 per litre. How many litres must it sell to break even?
D. 8,000
Break-even quantity = fixed cost / (price - variable cost) = 120,000 / (40 - 25) = 120,000 / 15 = 8,000 litres. 3,000 divides by price only; 4,800 divides by variable cost only; 1,846 divides by price plus variable cost (65) instead of the margin.
A dairy project has gross returns of 540,000 and total costs of 450,000 in a year. Its benefit-cost ratio and interpretation are:
B. 1.2; the enterprise pays
BCR = returns / costs = 540,000 / 450,000 = 1.2, and a ratio above 1 means returns exceed costs. 0.83 inverts the ratio; 90,000 is net return, not a ratio.
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.
Read the manual chapters for these first and do every worked question you can find on them.
Aim for one sitting per unit, then mixed questions across them.
Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.
Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.
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