Agricultural Calculations, Research Methods and Field Statistics MCQs 2026
50 questions with detailed answers · 0 from past papers · 5 quiz batches available
Choose a Quiz Batch. Each batch has 10 questions from this topic, in order. Take them one by one to work through all 50 MCQs. Login to save your scores and see your best per batch.
Read each question, think about the answer, then click Show Answer to reveal the correct option and explanation. Load 10 at a time so it stays manageable — perfect for one-topic study sessions on the bus or during a break.
- Q1 medium
There are 45 missing hills in 1500 planted hills. Missing hill percentage is
💡 Explanation:Missing hill percentage = 45 / 1500 x 100 = 3 percent.
- Q2 medium
A dry soil core weighs 1300 g and has volume 1000 cm3. Bulk density is
💡 Explanation:Bulk density = oven dry mass / core volume = 1300 g / 1000 cm3 = 1.3 g cm-3.
- Q3 medium
If lime requirement is 2 t per hectare, the lime needed for 2.5 hectares is
💡 Explanation:Total lime = rate x area = 2 t per ha x 2.5 ha = 5.0 t.
- Q4 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
💡 Explanation:Pure live seed fraction = 0.90 x 0.80 = 0.72. Bulk seed rate = 100 / 0.72 = 138.9 kg per ha.
- Q5 medium
Urea contains 46 percent nitrogen. To apply 92 kg nitrogen per hectare, the required urea is
💡 Explanation:Required fertilizer = nutrient required / nutrient fraction = 92 / 0.46 = 200 kg urea per ha.
- Q6 medium
At 60 cm by 20 cm plant spacing, the approximate plant population per hectare is
💡 Explanation:Area per plant = 0.60 x 0.20 = 0.12 m2. Plants per ha = 10,000 / 0.12 = 83,333.
- Q7 medium
A herbicide dose is 1.5 kg active ingredient per hectare. If the formulation contains 50 percent active ingredient, product required is
💡 Explanation:Product required = active ingredient needed / fraction in product = 1.5 / 0.50 = 3.0 kg per ha.
- Q8 medium
A 12 m2 plot produced 4.8 kg grain. The equivalent yield per hectare is
💡 Explanation:Yield per m2 = 4.8 / 12 = 0.4 kg. Per hectare = 0.4 x 10,000 = 4000 kg per ha.
- Q9 medium
A grain lot weighs 1000 kg at 18 percent moisture. Its weight at 12 percent moisture is
💡 Explanation:Corrected weight = fresh weight x (100 - initial moisture) / (100 - final moisture) = 1000 x 82 / 88 = 931.8 kg.
- Q10 medium
If the mean yield is 40 q per ha and standard deviation is 4 q per ha, the coefficient of variation is
💡 Explanation:CV = standard deviation / mean x 100 = 4 / 40 x 100 = 10 percent.
- Q11 medium
For t = 2.10, error mean square = 16 and replications = 4, the LSD for comparing two means is
💡 Explanation:LSD = t x sqrt(2 x MSE / r) = 2.10 x sqrt(2 x 16 / 4) = 2.10 x sqrt(8) = 5.94.
- Q12 medium
There are 50 crop rows each 100 m long, and plants are 25 cm apart within rows. Total plants are
💡 Explanation:Total row length = 50 x 100 = 5000 m. Plant spacing = 0.25 m. Plants = 5000 / 0.25 = 20,000.
- Q13 medium
An irrigation depth of 75 mm is applied over 2 hectares. The water volume is
💡 Explanation:Depth = 75 mm = 0.075 m. Area = 2 ha = 20,000 m2. Volume = 0.075 x 20,000 = 1500 m3.
- Q14 medium
Six harvested rows are each 5 m long with 30 cm row spacing. Net plot area is
💡 Explanation:Net plot area = number of rows x row length x row spacing = 6 x 5 x 0.30 = 9.0 m2.
- Q15 medium
In a germination test, 172 seeds germinate out of 200 seeds. Germination percentage is
💡 Explanation:Germination percentage = 172 / 200 x 100 = 86 percent.
- Q16 medium
A crop gives 4 t grain and 6 t straw per hectare. Harvest index is
💡 Explanation:Harvest index = economic yield / biological yield x 100 = 4 / (4 + 6) x 100 = 40 percent.
- Q17 medium
If the difference between two treatment means is 8 and standard error of difference is 2, the calculated t value is
💡 Explanation:t = difference / standard error of difference = 8 / 2 = 4.0.
- Q18 medium
The mean of yields 3.2, 3.8, 4.0 and 3.0 t per ha is
💡 Explanation:Mean = (3.2 + 3.8 + 4.0 + 3.0) / 4 = 14.0 / 4 = 3.5 t per ha.
- Q19 medium
If standard deviation is 12 and sample size is 36, the standard error of mean is
💡 Explanation:Standard error = SD / sqrt(n) = 12 / sqrt(36) = 12 / 6 = 2.
- Q20 medium
In a randomized complete block design with 5 treatments and 4 replications, error degrees of freedom are
💡 Explanation:RCBD error df = (treatments - 1)(replications - 1) = (5 - 1)(4 - 1) = 4 x 3 = 12.
- Q21 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
💡 Explanation:Available water depth = (30 - 15)/100 x 1.3 x 30 cm = 5.85 cm = 58.5 mm.
- Q22 medium
DAP contains 46 percent P2O5. To supply 46 kg P2O5 per hectare, DAP required is
💡 Explanation:DAP required = P2O5 requirement / 0.46 = 46 / 0.46 = 100 kg per ha.
- Q23 medium
A pesticide recommendation is 2 ml per litre of water. For a 15 litre sprayer, pesticide needed is
💡 Explanation:Pesticide needed = 2 ml/L x 15 L = 30 ml.
- Q24 medium
If 18 plants are diseased among 120 observed plants, disease incidence is
💡 Explanation:Disease incidence = diseased plants / total plants x 100 = 18 / 120 x 100 = 15 percent.
- Q25 medium
For an experiment with 24 observations, total degrees of freedom are
💡 Explanation:Total degrees of freedom = N - 1 = 24 - 1 = 23.
- Q26 medium
If treatment mean square is 48 and error mean square is 12, the F value is
💡 Explanation:F value = treatment mean square / error mean square = 48 / 12 = 4.0.
- Q27 medium
Gross return is Rs 180,000 per ha and variable cost is Rs 115,000 per ha. Gross margin is
💡 Explanation:Gross margin = gross return - variable cost = 180,000 - 115,000 = Rs 65,000 per ha.
- Q28 medium
Gross return is Rs 240,000 and total cost is Rs 160,000. Benefit-cost ratio is
💡 Explanation:Benefit-cost ratio = gross return / total cost = 240,000 / 160,000 = 1.50.
- Q29 medium
A crop yield is 6000 kg and seasonal water use is 5000 m3. Water productivity is
💡 Explanation:Water productivity = crop yield / water used = 6000 / 5000 = 1.2 kg per m3.
- Q30 medium
In a tetrazolium test, 92 seeds stain viable out of 100 seeds tested. Seed viability is
💡 Explanation:Viability percentage = viable seeds / tested seeds x 100 = 92 / 100 x 100 = 92 percent.
- Q31 medium
A correlation coefficient of -0.80 between fertilizer rate and disease score indicates
💡 Explanation:The negative sign shows inverse direction and the magnitude 0.80 indicates a strong association.
- Q32 medium
A plot is 3.6 m wide and row spacing is 45 cm. Number of rows that fit across the width is
💡 Explanation:Row spacing = 45 cm = 0.45 m. Rows = plot width / spacing = 3.6 / 0.45 = 8 rows.
- Q33 medium
The sum of paired treatment differences from 10 plots is 30. The mean difference is
💡 Explanation:Mean difference = sum of differences / number of pairs = 30 / 10 = 3.0.
- Q34 medium
A weed survey records 40 weeds in a 2 m2 quadrat area. Weed density is
💡 Explanation:Weed density = number of weeds / area = 40 / 2 = 20 weeds per m2.
- Q35 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
💡 Explanation:Sprayer rate = 600 x output / (speed x width) = 600 x 0.5 / (5 x 0.5) = 300 / 2.5 = 120 L per ha.
- Q36 medium
Calcium ammonium nitrate contains 26 percent nitrogen. To supply 52 kg nitrogen per hectare, fertilizer required is
💡 Explanation:Fertilizer required = nitrogen required / nitrogen fraction = 52 / 0.26 = 200 kg per ha.
- Q37 medium
The main statistical purpose of replication in field experiments is to
💡 Explanation:Replication supplies independent observations, allowing estimation of experimental error and more precise treatment comparisons.
- Q38 medium
Yield increases from 2.5 t per ha to 3.0 t per ha. The percentage increase is
💡 Explanation:Increase = 3.0 - 2.5 = 0.5. Percentage increase = 0.5 / 2.5 x 100 = 20 percent.
- Q39 medium
The median of 12, 15, 18, 22 and 30 is
💡 Explanation:The values are already ordered. With five observations, the middle third value is 18.
- Q40 medium
The mode of 4, 5, 5, 6, 7, 7, 7 and 8 is
💡 Explanation:The mode is the most frequent value; 7 occurs three times, more than any other value.
- Q41 medium
If the maximum observation is 98 and the minimum is 62, the range is
💡 Explanation:Range = maximum - minimum = 98 - 62 = 36.
- Q42 medium
For observations 2, 4 and 6, the sample variance is
💡 Explanation:Mean = 4. Squared deviations are 4, 0 and 4, sum = 8. Sample variance = 8 / (3 - 1) = 4.
- Q43 medium
If 25 plants are diseased out of 200, the probability of selecting a diseased plant at random is
💡 Explanation:Probability = diseased plants / total plants = 25 / 200 = 0.125, equal to 12.5 percent.
- Q44 medium
If variance is 9, the standard deviation is
💡 Explanation:Standard deviation is the square root of variance, so sqrt(9) = 3.
- Q45 medium
In a split-plot design, compared with subplot treatments the main-plot treatments generally have
💡 Explanation:Main plots are larger and have fewer independent error degrees of freedom, so main-plot comparisons usually have lower precision.
- Q46 medium
A null hypothesis in an agricultural experiment usually states that treatments have
💡 Explanation:The null hypothesis is the default claim of no treatment effect or no difference, tested against an alternative hypothesis.
- Q47 medium
Rejecting a true null hypothesis is called
💡 Explanation:A Type I error occurs when a true null hypothesis is incorrectly rejected; its probability is alpha.
- Q48 medium
For mean 50, standard error 2 and t value 2.0, the 95 percent confidence interval is
💡 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.
- Q49 medium
If correlation coefficient r is 0.70, coefficient of determination is
💡 Explanation:Coefficient of determination = r squared = 0.70 x 0.70 = 0.49 = 49 percent.
- Q50 medium
Mean separation tests such as LSD should generally be applied after
💡 Explanation:ANOVA first tests whether treatment variation is significant; mean separation is then justified for comparing treatment means.