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 1Questions 1–10 of 50

Q1medium
A crop yield is 6000 kg and seasonal water use is 5000 m3. Water productivity is
1. A1.2 kg per m3
2. B0.83 kg per m3
3. C12.0 kg per m3
4. D0.12 kg per m3
💡 Explanation:
Water productivity = crop yield / water used = 6000 / 5000 = 1.2 kg per m3.
Q2medium
In a tetrazolium test, 92 seeds stain viable out of 100 seeds tested. Seed viability is
1. A8 percent
2. B84 percent
3. C92 percent
4. D100 percent
💡 Explanation:
Viability percentage = viable seeds / tested seeds x 100 = 92 / 100 x 100 = 92 percent.
Q3medium
A correlation coefficient of -0.80 between fertilizer rate and disease score indicates
1. AStrong positive association
2. BNo association
3. CA perfect positive association
4. DStrong negative association
💡 Explanation:
The negative sign shows inverse direction and the magnitude 0.80 indicates a strong association.
Q4medium
A plot is 3.6 m wide and row spacing is 45 cm. Number of rows that fit across the width is
1. A6 rows
2. B8 rows
3. C9 rows
4. D12 rows
💡 Explanation:
Row spacing = 45 cm = 0.45 m. Rows = plot width / spacing = 3.6 / 0.45 = 8 rows.
Q5medium
The sum of paired treatment differences from 10 plots is 30. The mean difference is
1. A3.0
2. B30.0
3. C10.0
4. D0.3
💡 Explanation:
Mean difference = sum of differences / number of pairs = 30 / 10 = 3.0.
Q6medium
A weed survey records 40 weeds in a 2 m2 quadrat area. Weed density is
1. A40 weeds per m2
2. B80 weeds per m2
3. C20 weeds per m2
4. D10 weeds per m2
💡 Explanation:
Weed density = number of weeds / area = 40 / 2 = 20 weeds per m2.
Q7medium
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. A60 L per ha
2. B120 L per ha
3. C300 L per ha
4. D12 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.
Q8medium
Calcium ammonium nitrate contains 26 percent nitrogen. To supply 52 kg nitrogen per hectare, fertilizer required is
1. A135 kg per ha
2. B100 kg per ha
3. C260 kg per ha
4. D200 kg per ha
💡 Explanation:
Fertilizer required = nitrogen required / nitrogen fraction = 52 / 0.26 = 200 kg per ha.
Q9medium
The main statistical purpose of replication in field experiments is to
1. AEstimate random error and improve precision
2. BRemove the need for randomization
3. CMake all plots identical
4. DIncrease soil heterogeneity
💡 Explanation:
Replication supplies independent observations, allowing estimation of experimental error and more precise treatment comparisons.
Q10medium
Yield increases from 2.5 t per ha to 3.0 t per ha. The percentage increase is
1. A16.7 percent
2. B20 percent
3. C25 percent
4. D5 percent
💡 Explanation:
Increase = 3.0 - 2.5 = 0.5. Percentage increase = 0.5 / 2.5 x 100 = 20 percent.
Q11medium
The median of 12, 15, 18, 22 and 30 is
1. A15
2. B19.4
3. C18
4. D22
💡 Explanation:
The values are already ordered. With five observations, the middle third value is 18.
Q12medium
The mode of 4, 5, 5, 6, 7, 7, 7 and 8 is
1. A5
2. B6
3. C8
4. D7
💡 Explanation:
The mode is the most frequent value; 7 occurs three times, more than any other value.
Q13medium
If the maximum observation is 98 and the minimum is 62, the range is
1. A36
2. B160
3. C98
4. D62
💡 Explanation:
Range = maximum - minimum = 98 - 62 = 36.
Q14medium
For observations 2, 4 and 6, the sample variance is
1. A2
2. B4
3. C8
4. D16
💡 Explanation:
Mean = 4. Squared deviations are 4, 0 and 4, sum = 8. Sample variance = 8 / (3 - 1) = 4.
Q15medium
If 25 plants are diseased out of 200, the probability of selecting a diseased plant at random is
1. A0.250
2. B0.800
3. C0.125
4. D0.075
💡 Explanation:
Probability = diseased plants / total plants = 25 / 200 = 0.125, equal to 12.5 percent.
Q16medium
If variance is 9, the standard deviation is
1. A3
2. B9
3. C18
4. D81
💡 Explanation:
Standard deviation is the square root of variance, so sqrt(9) = 3.
Q17medium
In a split-plot design, compared with subplot treatments the main-plot treatments generally have
1. ANo experimental error
2. BMore replications automatically
3. CNo randomization
4. DLower precision
💡 Explanation:
Main plots are larger and have fewer independent error degrees of freedom, so main-plot comparisons usually have lower precision.
Q18medium
A null hypothesis in an agricultural experiment usually states that treatments have
1. AMaximum possible yield
2. BNo real difference in effect
3. CUnequal replications
4. DGuaranteed superiority
💡 Explanation:
The null hypothesis is the default claim of no treatment effect or no difference, tested against an alternative hypothesis.
Q19medium
Rejecting a true null hypothesis is called
1. AType II error
2. BSampling frame
3. CType I error
4. DBlocking error
💡 Explanation:
A Type I error occurs when a true null hypothesis is incorrectly rejected; its probability is alpha.
Q20medium
For mean 50, standard error 2 and t value 2.0, the 95 percent confidence interval is
1. A46 to 54
2. B48 to 52
3. C50 to 54
4. D44 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.
Q21medium
If correlation coefficient r is 0.70, coefficient of determination is
1. A70 percent
2. B30 percent
3. C7 percent
4. D49 percent
💡 Explanation:
Coefficient of determination = r squared = 0.70 x 0.70 = 0.49 = 49 percent.
Q22medium
Mean separation tests such as LSD should generally be applied after
1. AIgnoring replication
2. BRemoving randomization
3. CA significant F-test
4. DCombining all blocks
💡 Explanation:
ANOVA first tests whether treatment variation is significant; mean separation is then justified for comparing treatment means.
Q23medium
A dry soil core weighs 1300 g and has volume 1000 cm3. Bulk density is
1. A0.77 g cm-3
2. B1.3 g cm-3
3. C2.3 g cm-3
4. D13.0 g cm-3
💡 Explanation:
Bulk density = oven dry mass / core volume = 1300 g / 1000 cm3 = 1.3 g cm-3.
Q24medium
If lime requirement is 2 t per hectare, the lime needed for 2.5 hectares is
1. A5.0 t
2. B2.5 t
3. C4.0 t
4. D0.8 t
💡 Explanation:
Total lime = rate x area = 2 t per ha x 2.5 ha = 5.0 t.
Q25medium
Six harvested rows are each 5 m long with 30 cm row spacing. Net plot area is
1. A30.0 m2
2. B1.5 m2
3. C6.0 m2
4. D9.0 m2
💡 Explanation:
Net plot area = number of rows x row length x row spacing = 6 x 5 x 0.30 = 9.0 m2.
Q26medium
There are 45 missing hills in 1500 planted hills. Missing hill percentage is
1. A30 percent
2. B4.5 percent
3. C3 percent
4. D1.5 percent
💡 Explanation:
Missing hill percentage = 45 / 1500 x 100 = 3 percent.
Q27medium
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. A111.1 kg per ha
2. B138.9 kg per ha
3. C125.0 kg per ha
4. D180.0 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.
Q28medium
Urea contains 46 percent nitrogen. To apply 92 kg nitrogen per hectare, the required urea is
1. A200 kg per ha
2. B184 kg per ha
3. C230 kg per ha
4. D92 kg per ha
💡 Explanation:
Required fertilizer = nutrient required / nutrient fraction = 92 / 0.46 = 200 kg urea per ha.
Q29medium
At 60 cm by 20 cm plant spacing, the approximate plant population per hectare is
1. A50,000 plants
2. B66,667 plants
3. C83,333 plants
4. D100,000 plants
💡 Explanation:
Area per plant = 0.60 x 0.20 = 0.12 m2. Plants per ha = 10,000 / 0.12 = 83,333.
Q30medium
A herbicide dose is 1.5 kg active ingredient per hectare. If the formulation contains 50 percent active ingredient, product required is
1. A0.75 kg per ha
2. B1.5 kg per ha
3. C2.0 kg per ha
4. D3.0 kg per ha
💡 Explanation:
Product required = active ingredient needed / fraction in product = 1.5 / 0.50 = 3.0 kg per ha.
Q31medium
A 12 m2 plot produced 4.8 kg grain. The equivalent yield per hectare is
1. A480 kg per ha
2. B4000 kg per ha
3. C2400 kg per ha
4. D5800 kg per ha
💡 Explanation:
Yield per m2 = 4.8 / 12 = 0.4 kg. Per hectare = 0.4 x 10,000 = 4000 kg per ha.
Q32medium
A grain lot weighs 1000 kg at 18 percent moisture. Its weight at 12 percent moisture is
1. A880.0 kg
2. B1068.2 kg
3. C931.8 kg
4. D820.0 kg
💡 Explanation:
Corrected weight = fresh weight x (100 - initial moisture) / (100 - final moisture) = 1000 x 82 / 88 = 931.8 kg.
Q33medium
If the mean yield is 40 q per ha and standard deviation is 4 q per ha, the coefficient of variation is
1. A10 percent
2. B4 percent
3. C20 percent
4. D16 percent
💡 Explanation:
CV = standard deviation / mean x 100 = 4 / 40 x 100 = 10 percent.
Q34medium
For t = 2.10, error mean square = 16 and replications = 4, the LSD for comparing two means is
1. A4.20
2. B8.40
3. C2.97
4. D5.94
💡 Explanation:
LSD = t x sqrt(2 x MSE / r) = 2.10 x sqrt(2 x 16 / 4) = 2.10 x sqrt(8) = 5.94.
Q35medium
There are 50 crop rows each 100 m long, and plants are 25 cm apart within rows. Total plants are
1. A20,000 plants
2. B12,500 plants
3. C25,000 plants
4. D10,000 plants
💡 Explanation:
Total row length = 50 x 100 = 5000 m. Plant spacing = 0.25 m. Plants = 5000 / 0.25 = 20,000.
Q36medium
An irrigation depth of 75 mm is applied over 2 hectares. The water volume is
1. A150 m3
2. B1500 m3
3. C750 m3
4. D15,000 m3
💡 Explanation:
Depth = 75 mm = 0.075 m. Area = 2 ha = 20,000 m2. Volume = 0.075 x 20,000 = 1500 m3.
Q37medium
In a germination test, 172 seeds germinate out of 200 seeds. Germination percentage is
1. A72 percent
2. B82 percent
3. C92 percent
4. D86 percent
💡 Explanation:
Germination percentage = 172 / 200 x 100 = 86 percent.
Q38medium
A crop gives 4 t grain and 6 t straw per hectare. Harvest index is
1. A60 percent
2. B66.7 percent
3. C40 percent
4. D25 percent
💡 Explanation:
Harvest index = economic yield / biological yield x 100 = 4 / (4 + 6) x 100 = 40 percent.
Q39medium
If the difference between two treatment means is 8 and standard error of difference is 2, the calculated t value is
1. A2.0
2. B4.0
3. C6.0
4. D0.25
💡 Explanation:
t = difference / standard error of difference = 8 / 2 = 4.0.
Q40medium
The mean of yields 3.2, 3.8, 4.0 and 3.0 t per ha is
1. A3.5 t per ha
2. B3.0 t per ha
3. C3.8 t per ha
4. D14.0 t per ha
💡 Explanation:
Mean = (3.2 + 3.8 + 4.0 + 3.0) / 4 = 14.0 / 4 = 3.5 t per ha.
Q41medium
If standard deviation is 12 and sample size is 36, the standard error of mean is
1. A12
2. B6
3. C3
4. D2
💡 Explanation:
Standard error = SD / sqrt(n) = 12 / sqrt(36) = 12 / 6 = 2.
Q42medium
In a randomized complete block design with 5 treatments and 4 replications, error degrees of freedom are
1. A20
2. B12
3. C15
4. D9
💡 Explanation:
RCBD error df = (treatments - 1)(replications - 1) = (5 - 1)(4 - 1) = 4 x 3 = 12.
Q43medium
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. A19.5 mm
2. B45.0 mm
3. C58.5 mm
4. D78.0 mm
💡 Explanation:
Available water depth = (30 - 15)/100 x 1.3 x 30 cm = 5.85 cm = 58.5 mm.
Q44medium
DAP contains 46 percent P2O5. To supply 46 kg P2O5 per hectare, DAP required is
1. A100 kg per ha
2. B46 kg per ha
3. C200 kg per ha
4. D92 kg per ha
💡 Explanation:
DAP required = P2O5 requirement / 0.46 = 46 / 0.46 = 100 kg per ha.
Q45medium
A pesticide recommendation is 2 ml per litre of water. For a 15 litre sprayer, pesticide needed is
1. A7.5 ml
2. B15 ml
3. C45 ml
4. D30 ml
💡 Explanation:
Pesticide needed = 2 ml/L x 15 L = 30 ml.
Q46medium
If 18 plants are diseased among 120 observed plants, disease incidence is
1. A18 percent
2. B15 percent
3. C12 percent
4. D20 percent
💡 Explanation:
Disease incidence = diseased plants / total plants x 100 = 18 / 120 x 100 = 15 percent.
Q47medium
For an experiment with 24 observations, total degrees of freedom are
1. A23
2. B24
3. C22
4. D25
💡 Explanation:
Total degrees of freedom = N - 1 = 24 - 1 = 23.
Q48medium
If treatment mean square is 48 and error mean square is 12, the F value is
1. A0.25
2. B36.0
3. C4.0
4. D60.0
💡 Explanation:
F value = treatment mean square / error mean square = 48 / 12 = 4.0.
Q49medium
Gross return is Rs 180,000 per ha and variable cost is Rs 115,000 per ha. Gross margin is
1. ARs 295,000 per ha
2. BRs 65,000 per ha
3. CRs 115,000 per ha
4. DRs 80,000 per ha
💡 Explanation:
Gross margin = gross return - variable cost = 180,000 - 115,000 = Rs 65,000 per ha.
Q50medium
Gross return is Rs 240,000 and total cost is Rs 160,000. Benefit-cost ratio is
1. A0.67
2. B2.50
3. C0.50
4. D1.50
💡 Explanation:
Benefit-cost ratio = gross return / total cost = 240,000 / 160,000 = 1.50.

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