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Agricultural Calculations, Research Methods and Field Statistics MCQs 2026

50 questions with detailed answers · 0 from past papers · 5 quiz batches available

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Page 1 of 1 Questions 1–10 of 50
  1. Q1 medium

    A crop yield is 6000 kg and seasonal water use is 5000 m3. Water productivity is

    1. A 1.2 kg per m3
    2. B 0.83 kg per m3
    3. C 12.0 kg per m3
    4. D 0.12 kg per m3
    💡 Explanation:

    Water productivity = crop yield / water used = 6000 / 5000 = 1.2 kg per m3.

  2. Q2 medium

    In a tetrazolium test, 92 seeds stain viable out of 100 seeds tested. Seed viability is

    1. A 8 percent
    2. B 84 percent
    3. C 92 percent
    4. D 100 percent
    💡 Explanation:

    Viability percentage = viable seeds / tested seeds x 100 = 92 / 100 x 100 = 92 percent.

  3. Q3 medium

    A correlation coefficient of -0.80 between fertilizer rate and disease score indicates

    1. A Strong positive association
    2. B No association
    3. C A perfect positive association
    4. D Strong negative association
    💡 Explanation:

    The negative sign shows inverse direction and the magnitude 0.80 indicates a strong association.

  4. Q4 medium

    A plot is 3.6 m wide and row spacing is 45 cm. Number of rows that fit across the width is

    1. A 6 rows
    2. B 8 rows
    3. C 9 rows
    4. D 12 rows
    💡 Explanation:

    Row spacing = 45 cm = 0.45 m. Rows = plot width / spacing = 3.6 / 0.45 = 8 rows.

  5. Q5 medium

    The sum of paired treatment differences from 10 plots is 30. The mean difference is

    1. A 3.0
    2. B 30.0
    3. C 10.0
    4. D 0.3
    💡 Explanation:

    Mean difference = sum of differences / number of pairs = 30 / 10 = 3.0.

  6. Q6 medium

    A weed survey records 40 weeds in a 2 m2 quadrat area. Weed density is

    1. A 40 weeds per m2
    2. B 80 weeds per m2
    3. C 20 weeds per m2
    4. D 10 weeds per m2
    💡 Explanation:

    Weed density = number of weeds / area = 40 / 2 = 20 weeds per m2.

  7. Q7 medium

    A sprayer has nozzle output 0.5 L per minute, speed 5 km per hour and spray width 0.5 m. Application rate is

    1. A 60 L per ha
    2. B 120 L per ha
    3. C 300 L per ha
    4. D 12 L per ha
    💡 Explanation:

    Sprayer rate = 600 x output / (speed x width) = 600 x 0.5 / (5 x 0.5) = 300 / 2.5 = 120 L per ha.

  8. Q8 medium

    Calcium ammonium nitrate contains 26 percent nitrogen. To supply 52 kg nitrogen per hectare, fertilizer required is

    1. A 135 kg per ha
    2. B 100 kg per ha
    3. C 260 kg per ha
    4. D 200 kg per ha
    💡 Explanation:

    Fertilizer required = nitrogen required / nitrogen fraction = 52 / 0.26 = 200 kg per ha.

  9. Q9 medium

    The main statistical purpose of replication in field experiments is to

    1. A Estimate random error and improve precision
    2. B Remove the need for randomization
    3. C Make all plots identical
    4. D Increase soil heterogeneity
    💡 Explanation:

    Replication supplies independent observations, allowing estimation of experimental error and more precise treatment comparisons.

  10. Q10 medium

    Yield increases from 2.5 t per ha to 3.0 t per ha. The percentage increase is

    1. A 16.7 percent
    2. B 20 percent
    3. C 25 percent
    4. D 5 percent
    💡 Explanation:

    Increase = 3.0 - 2.5 = 0.5. Percentage increase = 0.5 / 2.5 x 100 = 20 percent.

  11. Q11 medium

    The median of 12, 15, 18, 22 and 30 is

    1. A 15
    2. B 19.4
    3. C 18
    4. D 22
    💡 Explanation:

    The values are already ordered. With five observations, the middle third value is 18.

  12. Q12 medium

    The mode of 4, 5, 5, 6, 7, 7, 7 and 8 is

    1. A 5
    2. B 6
    3. C 8
    4. D 7
    💡 Explanation:

    The mode is the most frequent value; 7 occurs three times, more than any other value.

  13. Q13 medium

    If the maximum observation is 98 and the minimum is 62, the range is

    1. A 36
    2. B 160
    3. C 98
    4. D 62
    💡 Explanation:

    Range = maximum - minimum = 98 - 62 = 36.

  14. Q14 medium

    For observations 2, 4 and 6, the sample variance is

    1. A 2
    2. B 4
    3. C 8
    4. D 16
    💡 Explanation:

    Mean = 4. Squared deviations are 4, 0 and 4, sum = 8. Sample variance = 8 / (3 - 1) = 4.

  15. Q15 medium

    If 25 plants are diseased out of 200, the probability of selecting a diseased plant at random is

    1. A 0.250
    2. B 0.800
    3. C 0.125
    4. D 0.075
    💡 Explanation:

    Probability = diseased plants / total plants = 25 / 200 = 0.125, equal to 12.5 percent.

  16. Q16 medium

    If variance is 9, the standard deviation is

    1. A 3
    2. B 9
    3. C 18
    4. D 81
    💡 Explanation:

    Standard deviation is the square root of variance, so sqrt(9) = 3.

  17. Q17 medium

    In a split-plot design, compared with subplot treatments the main-plot treatments generally have

    1. A No experimental error
    2. B More replications automatically
    3. C No randomization
    4. D Lower precision
    💡 Explanation:

    Main plots are larger and have fewer independent error degrees of freedom, so main-plot comparisons usually have lower precision.

  18. Q18 medium

    A null hypothesis in an agricultural experiment usually states that treatments have

    1. A Maximum possible yield
    2. B No real difference in effect
    3. C Unequal replications
    4. D Guaranteed superiority
    💡 Explanation:

    The null hypothesis is the default claim of no treatment effect or no difference, tested against an alternative hypothesis.

  19. Q19 medium

    Rejecting a true null hypothesis is called

    1. A Type II error
    2. B Sampling frame
    3. C Type I error
    4. D Blocking error
    💡 Explanation:

    A Type I error occurs when a true null hypothesis is incorrectly rejected; its probability is alpha.

  20. Q20 medium

    For mean 50, standard error 2 and t value 2.0, the 95 percent confidence interval is

    1. A 46 to 54
    2. B 48 to 52
    3. C 50 to 54
    4. D 44 to 56
    💡 Explanation:

    Confidence interval = mean plus or minus t x SE = 50 plus or minus 2.0 x 2 = 50 plus or minus 4, giving 46 to 54.

  21. Q21 medium

    If correlation coefficient r is 0.70, coefficient of determination is

    1. A 70 percent
    2. B 30 percent
    3. C 7 percent
    4. D 49 percent
    💡 Explanation:

    Coefficient of determination = r squared = 0.70 x 0.70 = 0.49 = 49 percent.

  22. Q22 medium

    Mean separation tests such as LSD should generally be applied after

    1. A Ignoring replication
    2. B Removing randomization
    3. C A significant F-test
    4. D Combining all blocks
    💡 Explanation:

    ANOVA first tests whether treatment variation is significant; mean separation is then justified for comparing treatment means.

  23. Q23 medium

    A dry soil core weighs 1300 g and has volume 1000 cm3. Bulk density is

    1. A 0.77 g cm-3
    2. B 1.3 g cm-3
    3. C 2.3 g cm-3
    4. D 13.0 g cm-3
    💡 Explanation:

    Bulk density = oven dry mass / core volume = 1300 g / 1000 cm3 = 1.3 g cm-3.

  24. Q24 medium

    If lime requirement is 2 t per hectare, the lime needed for 2.5 hectares is

    1. A 5.0 t
    2. B 2.5 t
    3. C 4.0 t
    4. D 0.8 t
    💡 Explanation:

    Total lime = rate x area = 2 t per ha x 2.5 ha = 5.0 t.

  25. Q25 medium

    Six harvested rows are each 5 m long with 30 cm row spacing. Net plot area is

    1. A 30.0 m2
    2. B 1.5 m2
    3. C 6.0 m2
    4. D 9.0 m2
    💡 Explanation:

    Net plot area = number of rows x row length x row spacing = 6 x 5 x 0.30 = 9.0 m2.

  26. Q26 medium

    There are 45 missing hills in 1500 planted hills. Missing hill percentage is

    1. A 30 percent
    2. B 4.5 percent
    3. C 3 percent
    4. D 1.5 percent
    💡 Explanation:

    Missing hill percentage = 45 / 1500 x 100 = 3 percent.

  27. Q27 medium

    A crop needs 100 kg pure viable seed per hectare. If seed purity is 90 percent and germination is 80 percent, the bulk seed rate is

    1. A 111.1 kg per ha
    2. B 138.9 kg per ha
    3. C 125.0 kg per ha
    4. D 180.0 kg per ha
    💡 Explanation:

    Pure live seed fraction = 0.90 x 0.80 = 0.72. Bulk seed rate = 100 / 0.72 = 138.9 kg per ha.

  28. Q28 medium

    Urea contains 46 percent nitrogen. To apply 92 kg nitrogen per hectare, the required urea is

    1. A 200 kg per ha
    2. B 184 kg per ha
    3. C 230 kg per ha
    4. D 92 kg per ha
    💡 Explanation:

    Required fertilizer = nutrient required / nutrient fraction = 92 / 0.46 = 200 kg urea per ha.

  29. Q29 medium

    At 60 cm by 20 cm plant spacing, the approximate plant population per hectare is

    1. A 50,000 plants
    2. B 66,667 plants
    3. C 83,333 plants
    4. D 100,000 plants
    💡 Explanation:

    Area per plant = 0.60 x 0.20 = 0.12 m2. Plants per ha = 10,000 / 0.12 = 83,333.

  30. Q30 medium

    A herbicide dose is 1.5 kg active ingredient per hectare. If the formulation contains 50 percent active ingredient, product required is

    1. A 0.75 kg per ha
    2. B 1.5 kg per ha
    3. C 2.0 kg per ha
    4. D 3.0 kg per ha
    💡 Explanation:

    Product required = active ingredient needed / fraction in product = 1.5 / 0.50 = 3.0 kg per ha.

  31. Q31 medium

    A 12 m2 plot produced 4.8 kg grain. The equivalent yield per hectare is

    1. A 480 kg per ha
    2. B 4000 kg per ha
    3. C 2400 kg per ha
    4. D 5800 kg per ha
    💡 Explanation:

    Yield per m2 = 4.8 / 12 = 0.4 kg. Per hectare = 0.4 x 10,000 = 4000 kg per ha.

  32. Q32 medium

    A grain lot weighs 1000 kg at 18 percent moisture. Its weight at 12 percent moisture is

    1. A 880.0 kg
    2. B 1068.2 kg
    3. C 931.8 kg
    4. D 820.0 kg
    💡 Explanation:

    Corrected weight = fresh weight x (100 - initial moisture) / (100 - final moisture) = 1000 x 82 / 88 = 931.8 kg.

  33. Q33 medium

    If the mean yield is 40 q per ha and standard deviation is 4 q per ha, the coefficient of variation is

    1. A 10 percent
    2. B 4 percent
    3. C 20 percent
    4. D 16 percent
    💡 Explanation:

    CV = standard deviation / mean x 100 = 4 / 40 x 100 = 10 percent.

  34. Q34 medium

    For t = 2.10, error mean square = 16 and replications = 4, the LSD for comparing two means is

    1. A 4.20
    2. B 8.40
    3. C 2.97
    4. D 5.94
    💡 Explanation:

    LSD = t x sqrt(2 x MSE / r) = 2.10 x sqrt(2 x 16 / 4) = 2.10 x sqrt(8) = 5.94.

  35. Q35 medium

    There are 50 crop rows each 100 m long, and plants are 25 cm apart within rows. Total plants are

    1. A 20,000 plants
    2. B 12,500 plants
    3. C 25,000 plants
    4. D 10,000 plants
    💡 Explanation:

    Total row length = 50 x 100 = 5000 m. Plant spacing = 0.25 m. Plants = 5000 / 0.25 = 20,000.

  36. Q36 medium

    An irrigation depth of 75 mm is applied over 2 hectares. The water volume is

    1. A 150 m3
    2. B 1500 m3
    3. C 750 m3
    4. D 15,000 m3
    💡 Explanation:

    Depth = 75 mm = 0.075 m. Area = 2 ha = 20,000 m2. Volume = 0.075 x 20,000 = 1500 m3.

  37. Q37 medium

    In a germination test, 172 seeds germinate out of 200 seeds. Germination percentage is

    1. A 72 percent
    2. B 82 percent
    3. C 92 percent
    4. D 86 percent
    💡 Explanation:

    Germination percentage = 172 / 200 x 100 = 86 percent.

  38. Q38 medium

    A crop gives 4 t grain and 6 t straw per hectare. Harvest index is

    1. A 60 percent
    2. B 66.7 percent
    3. C 40 percent
    4. D 25 percent
    💡 Explanation:

    Harvest index = economic yield / biological yield x 100 = 4 / (4 + 6) x 100 = 40 percent.

  39. Q39 medium

    If the difference between two treatment means is 8 and standard error of difference is 2, the calculated t value is

    1. A 2.0
    2. B 4.0
    3. C 6.0
    4. D 0.25
    💡 Explanation:

    t = difference / standard error of difference = 8 / 2 = 4.0.

  40. Q40 medium

    The mean of yields 3.2, 3.8, 4.0 and 3.0 t per ha is

    1. A 3.5 t per ha
    2. B 3.0 t per ha
    3. C 3.8 t per ha
    4. D 14.0 t per ha
    💡 Explanation:

    Mean = (3.2 + 3.8 + 4.0 + 3.0) / 4 = 14.0 / 4 = 3.5 t per ha.

  41. Q41 medium

    If standard deviation is 12 and sample size is 36, the standard error of mean is

    1. A 12
    2. B 6
    3. C 3
    4. D 2
    💡 Explanation:

    Standard error = SD / sqrt(n) = 12 / sqrt(36) = 12 / 6 = 2.

  42. Q42 medium

    In a randomized complete block design with 5 treatments and 4 replications, error degrees of freedom are

    1. A 20
    2. B 12
    3. C 15
    4. D 9
    💡 Explanation:

    RCBD error df = (treatments - 1)(replications - 1) = (5 - 1)(4 - 1) = 4 x 3 = 12.

  43. Q43 medium

    A soil has field capacity 30 percent, permanent wilting point 15 percent, bulk density 1.3 g cm-3 and root depth 30 cm. Available water is

    1. A 19.5 mm
    2. B 45.0 mm
    3. C 58.5 mm
    4. D 78.0 mm
    💡 Explanation:

    Available water depth = (30 - 15)/100 x 1.3 x 30 cm = 5.85 cm = 58.5 mm.

  44. Q44 medium

    DAP contains 46 percent P2O5. To supply 46 kg P2O5 per hectare, DAP required is

    1. A 100 kg per ha
    2. B 46 kg per ha
    3. C 200 kg per ha
    4. D 92 kg per ha
    💡 Explanation:

    DAP required = P2O5 requirement / 0.46 = 46 / 0.46 = 100 kg per ha.

  45. Q45 medium

    A pesticide recommendation is 2 ml per litre of water. For a 15 litre sprayer, pesticide needed is

    1. A 7.5 ml
    2. B 15 ml
    3. C 45 ml
    4. D 30 ml
    💡 Explanation:

    Pesticide needed = 2 ml/L x 15 L = 30 ml.

  46. Q46 medium

    If 18 plants are diseased among 120 observed plants, disease incidence is

    1. A 18 percent
    2. B 15 percent
    3. C 12 percent
    4. D 20 percent
    💡 Explanation:

    Disease incidence = diseased plants / total plants x 100 = 18 / 120 x 100 = 15 percent.

  47. Q47 medium

    For an experiment with 24 observations, total degrees of freedom are

    1. A 23
    2. B 24
    3. C 22
    4. D 25
    💡 Explanation:

    Total degrees of freedom = N - 1 = 24 - 1 = 23.

  48. Q48 medium

    If treatment mean square is 48 and error mean square is 12, the F value is

    1. A 0.25
    2. B 36.0
    3. C 4.0
    4. D 60.0
    💡 Explanation:

    F value = treatment mean square / error mean square = 48 / 12 = 4.0.

  49. Q49 medium

    Gross return is Rs 180,000 per ha and variable cost is Rs 115,000 per ha. Gross margin is

    1. A Rs 295,000 per ha
    2. B Rs 65,000 per ha
    3. C Rs 115,000 per ha
    4. D Rs 80,000 per ha
    💡 Explanation:

    Gross margin = gross return - variable cost = 180,000 - 115,000 = Rs 65,000 per ha.

  50. Q50 medium

    Gross return is Rs 240,000 and total cost is Rs 160,000. Benefit-cost ratio is

    1. A 0.67
    2. B 2.50
    3. C 0.50
    4. D 1.50
    💡 Explanation:

    Benefit-cost ratio = gross return / total cost = 240,000 / 160,000 = 1.50.

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