Datasets:
id stringlengths 13 20 | question stringlengths 28 257 | worked_solution stringlengths 80 666 | answer stringlengths 1 82 | family stringclasses 10
values | subtask stringclasses 18
values | answer_type stringclasses 2
values | answer_unit stringclasses 15
values | citation stringclasses 14
values | difficulty stringclasses 3
values | verified_by stringclasses 9
values | verification_note stringclasses 16
values | numeric_tolerance float64 0.01 0.02 ⌀ | exactly_gradable bool 2
classes | verifiable bool 1
class |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
f_units_0000602 | How many acres is 336.4 hectares? | 1 ha = 2.47105 ac.
336.4 x 2.47105 = 831.26 ac.
The final answer is $\boxed{831.26 ac}$. | 831.26 ac | units | ha_ac | numeric | ac | SI / US customary conversion factors | easy | closed-form | exact conversion constant | 0.01 | true | true |
f_seeding_0000400 | You want 280,000 established barley plants per hectare. Thousand-kernel weight is 18.5 g, germination 96% and expected field emergence 92%. What seeding rate in kg/ha? | Establishment = 96% x 92% = 88.32% of seeds sown.
Seeds to sow = 280,000 / 0.8832 = 317,029 seeds/ha.
A thousand-kernel weight of 18.5 g means one seed weighs 18.5/1000 g.
Rate = 317,029 x 18.5 / 1000 / 1000 = 5.87 kg/ha.
Check: 5.87 kg/ha at that TKW and establishment gives back 280,000 plants/ha.
The final answer is... | 5.87 kg/ha | seeding | rate_from_population | numeric | kg/ha | standard extension seeding-rate arithmetic | medium | inverse-recompute | population re-derived from the computed seed rate | 0.02 | true | true |
f_spray_0000912 | Your sprayer runs 1.95 L/min per nozzle at 16 km/h with nozzles 0.375 m apart. The label rate is 2.56 L/ha and the tank holds 600 L. What is the output in L/ha, and how much product goes in one tank? | Output L/ha = (L/min x 600) / (km/h x nozzle spacing in m).
= (1.95 x 600) / (16 x 0.375) = 195 L/ha.
One tank covers 600 / 195 = 3.08 ha.
Product per tank = 2.56 x 3.08 = 7.88 L.
The final answer is $\boxed{195 L/ha; 7.88 L per tank}$. | 195 L/ha; 7.88 L per tank | spraying | calibration | numeric | L/ha | sprayer calibration 600-rule | hard | area-roundtrip | product per tank cross-checked against hectares per tank | 0.02 | true | true |
f_grain_0000938 | Corn tests 52.4 lb/bu at 18.4% moisture. What is the expected test weight once it is dried to 15.5%? | Adjusted test weight = ((100 - MC_dry) / (100 - MC_wet)) x TW_wet.
= ((100 - 15.5) / (100 - 18.4)) x 52.4
= (84.5 / 81.6) x 52.4 = 54.26 lb/bu.
The final answer is $\boxed{54.26 lb/bu}$. | 54.26 lb/bu | grain | test_weight | numeric | lb/bu | Purdue — grain test weight considerations for corn | medium | closed-form | published test-weight identity | 0.02 | true | true |
f_grain_0002528 | A weigh wagon shows 8,900 lb of corn from one acre at 19.5% moisture. What is the yield in bushels per acre at the 15.5% standard? | Dry matter is conserved, so weight at 15.5% = 8,900 x (100 - 19.5) / (100 - 15.5)
= 8,900 x 80.5 / 84.5 = 8,478.7 lb.
Corn is 56 lb per bushel, so yield = 8,478.7 / 56 = 151.41 bu/ac.
The final answer is $\boxed{151.41 bu/ac}$. | 151.41 bu/ac | grain | yield_at_standard | numeric | bu/ac | OSU — bushels, test weights and calculations | medium | dry-matter-conservation | dry weight re-derived and re-divided | 0.02 | true | true |
f_npk_blend_0000929 | Your recommendation is 110-40-85 kg/ha of N-P2O5-K2O. Using DAP (18-46-0), potash (0-0-60) and urea (46-0-0), what rate of each do you apply? Meet the phosphate with DAP first and credit the nitrogen it carries. | Phosphate first. DAP is 46% P2O5, so DAP = 40 / 0.46 = 86.96 kg/ha.
That DAP also carries nitrogen: 86.96 x 18% = 15.65 kg N/ha.
Potash is 60% K2O, so potash = 85 / 0.60 = 141.67 kg/ha.
Remaining N = 110 - 15.65 = 94.35 kg/ha.
Urea is 46% N, so urea = 94.35 / 0.46 = 205.1 kg/ha.
Check: N 110, P2O5 40, K2O 85 — matches ... | DAP 86.96 kg/ha, potash 141.67 kg/ha, urea 205.1 kg/ha | fertilizer | npk_blend | numeric | kg/ha | WSU extension fertilizer calculations | hard | nutrient-balance | all three nutrient totals re-summed from the product rates | 0.02 | true | true |
f_livestock_0000238 | 120 head averaging 600 kg graze a 65 ha dryland field of 8 ha carrying 2,600 kg DM/ha. At 2.5% of bodyweight intake and 50% utilisation, how many grazing days does the paddock provide? | Intake per head = 600 x 2.5% = 15 kg DM/day.
Herd demand = 120 x 15 = 1,800 kg DM/day.
Usable forage = 2,600 x 8 x 50% = 10,400 kg DM.
Grazing days = 10,400 / 1,800 = 5.78 days.
The final answer is $\boxed{5.78 days}$. | 5.78 days | livestock | grazing_days | numeric | days | standard forage budgeting | medium | demand-supply-roundtrip | usable forage and herd demand recomputed separately | 0.02 | true | true |
f_grain_0000476 | A weigh wagon shows 8,925 lb of corn from one acre at 18.5% moisture. What is the yield in bushels per acre at the 15.5% standard? | Dry matter is conserved, so weight at 15.5% = 8,925 x (100 - 18.5) / (100 - 15.5)
= 8,925 x 81.5 / 84.5 = 8,608.14 lb.
Corn is 56 lb per bushel, so yield = 8,608.14 / 56 = 153.72 bu/ac.
The final answer is $\boxed{153.72 bu/ac}$. | 153.72 bu/ac | grain | yield_at_standard | numeric | bu/ac | OSU — bushels, test weights and calculations | medium | dry-matter-conservation | dry weight re-derived and re-divided | 0.02 | true | true |
f_seeding_0000430 | You want 400,000 established soybean plants per hectare. Thousand-kernel weight is 43.3 g, germination 96% and expected field emergence 85%. What seeding rate in kg/ha? | Establishment = 96% x 85% = 81.6% of seeds sown.
Seeds to sow = 400,000 / 0.816 = 490,196 seeds/ha.
A thousand-kernel weight of 43.3 g means one seed weighs 43.3/1000 g.
Rate = 490,196 x 43.3 / 1000 / 1000 = 21.23 kg/ha.
Check: 21.23 kg/ha at that TKW and establishment gives back 400,000 plants/ha.
The final answer is... | 21.23 kg/ha | seeding | rate_from_population | numeric | kg/ha | standard extension seeding-rate arithmetic | medium | inverse-recompute | population re-derived from the computed seed rate | 0.02 | true | true |
f_units_0000645 | Convert 249.8 kg/ha to lb/acre. | 1 kg = 2.20462 lb and 1 ha = 2.47105 ac.
249.8 x 2.20462 / 2.47105 = 222.87 lb/ac.
The final answer is $\boxed{222.87 lb/ac}$. | 222.87 lb/ac | units | kgha_lbac | numeric | lb/ac | SI / US customary conversion factors | easy | closed-form | exact conversion constant | 0.01 | true | true |
f_fert_rate_0001067 | A soil test for winter wheat on the north field calls for 110 kg N/ha. You are using UAN 32 (32-0-0). How much product do you need per hectare, and how much in total for 10 ha? | The label analysis UAN 32 (32-0-0) means the product is 32% N by weight.
Product rate = N required / (%N / 100) = 110 / (32/100) = 343.75 kg/ha.
Total for 10 ha = 343.75 x 10 = 3,437.5 kg.
Check: 343.75 kg/ha x 32% = 110 kg N/ha, which is the rate asked for.
The final answer is $\boxed{343.75 kg/ha (3,437.5 kg total)}... | 343.75 kg/ha (3,437.5 kg total) | fertilizer | single_nutrient_rate | numeric | kg/ha | WSU extension fertilizer calculations | easy | inverse-recompute | product rate pushed back through the label analysis | 0.02 | true | true |
f_gdd_0000352 | Accumulate growing degree days for corn over these 7 days (max/min air temperature in F): day 1 83/61F; day 2 68/41F; day 3 73/52F; day 4 84/65F; day 5 60/38F; day 6 68/60F; day 7 47/38F. What is the total GDD? | GDD per day = ((Tmax + Tmin)/2) - 50F, the base temperature for corn.
Before averaging, highs above 86F are counted as 86F, and lows below 50F are counted as 50F. A negative daily value is recorded as 0.
Day 1: 83/61 -> 22 GDD (running 22).
Day 2: 68/41 -> 9 GDD (running 31).
Day 3: 73/52 -> 12.5 GDD (running 43.... | 87 | degree_days | accumulate | numeric | GDD (F) | NDSU NDAWN | medium | per-day-resum | daily contributions re-summed independently of the accumulator | 0.01 | true | true |
f_economics_0001332 | A crop yields 10.4 t/ha and sells for 215 per tonne. Variable costs are 1380 per hectare. What is the gross margin per hectare? | Revenue = 10.4 x 215 = 2,236 per ha.
Gross margin = revenue - variable costs = 2,236 - 1380 = 856 per ha.
The final answer is $\boxed{856}$. | 856 | economics | gross_margin | numeric | currency/ha | gross margin budgeting | easy | closed-form | revenue re-multiplied | 0.02 | true | true |
f_grain_0001772 | Corn tests 50.9 lb/bu at 22.4% moisture. What is the expected test weight once it is dried to 15.5%? | Adjusted test weight = ((100 - MC_dry) / (100 - MC_wet)) x TW_wet.
= ((100 - 15.5) / (100 - 22.4)) x 50.9
= (84.5 / 77.6) x 50.9 = 55.43 lb/bu.
The final answer is $\boxed{55.43 lb/bu}$. | 55.43 lb/bu | grain | test_weight | numeric | lb/bu | Purdue — grain test weight considerations for corn | medium | closed-form | published test-weight identity | 0.02 | true | true |
f_npk_blend_0000105 | Your recommendation is 160-20-95 kg/ha of N-P2O5-K2O. Using DAP (18-46-0), potash (0-0-60) and urea (46-0-0), what rate of each do you apply? Meet the phosphate with DAP first and credit the nitrogen it carries. | Phosphate first. DAP is 46% P2O5, so DAP = 20 / 0.46 = 43.48 kg/ha.
That DAP also carries nitrogen: 43.48 x 18% = 7.83 kg N/ha.
Potash is 60% K2O, so potash = 95 / 0.60 = 158.33 kg/ha.
Remaining N = 160 - 7.83 = 152.17 kg/ha.
Urea is 46% N, so urea = 152.17 / 0.46 = 330.81 kg/ha.
Check: N 160, P2O5 20, K2O 95 — matches... | DAP 43.48 kg/ha, potash 158.33 kg/ha, urea 330.81 kg/ha | fertilizer | npk_blend | numeric | kg/ha | WSU extension fertilizer calculations | hard | nutrient-balance | all three nutrient totals re-summed from the product rates | 0.02 | true | true |
f_refusal_0001149 | The block is 18 ha. How many days can I graze the home paddock? | Grazing days need the forage dry matter available per hectare and a utilisation assumption. Neither is provided, and standing forage cannot be estimated from field size alone.
The details given do not substitute for it — none of them determine the value asked for.
The missing input is the current forage dry matter on o... | cannot be determined without the current forage dry matter on offer | refusal | insufficient_input | text | null | n/a — refusal by construction | medium | by-construction | the question deliberately omits an input the calculation requires | null | false | true |
f_refusal_0000921 | Soil is a peaty loam. Planting went in on mid-October. The block is 80 ha. What will my sunflower weigh once it is dried down? | Dried weight follows from the starting moisture and the target moisture. The starting moisture is not given, and dry-matter conservation cannot be applied without it.
The details given do not substitute for it — none of them determine the value asked for.
The missing input is the grain moisture at harvest.
The final a... | cannot be determined without the grain moisture at harvest | refusal | insufficient_input | text | null | n/a — refusal by construction | medium | by-construction | the question deliberately omits an input the calculation requires | null | false | true |
f_seeding_0000620 | You want 250,000 established canola plants per hectare. Thousand-kernel weight is 30.4 g, germination 95% and expected field emergence 92%. What seeding rate in kg/ha? | Establishment = 95% x 92% = 87.4% of seeds sown.
Seeds to sow = 250,000 / 0.874 = 286,041 seeds/ha.
A thousand-kernel weight of 30.4 g means one seed weighs 30.4/1000 g.
Rate = 286,041 x 30.4 / 1000 / 1000 = 8.7 kg/ha.
Check: 8.7 kg/ha at that TKW and establishment gives back 250,000 plants/ha.
The final answer is $\b... | 8.7 kg/ha | seeding | rate_from_population | numeric | kg/ha | standard extension seeding-rate arithmetic | medium | inverse-recompute | population re-derived from the computed seed rate | 0.02 | true | true |
f_grain_0002743 | A weigh wagon shows 7,800 lb of corn from one acre at 23.9% moisture. What is the yield in bushels per acre at the 15.5% standard? | Dry matter is conserved, so weight at 15.5% = 7,800 x (100 - 23.9) / (100 - 15.5)
= 7,800 x 76.1 / 84.5 = 7,024.62 lb.
Corn is 56 lb per bushel, so yield = 7,024.62 / 56 = 125.44 bu/ac.
The final answer is $\boxed{125.44 bu/ac}$. | 125.44 bu/ac | grain | yield_at_standard | numeric | bu/ac | OSU — bushels, test weights and calculations | medium | dry-matter-conservation | dry weight re-derived and re-divided | 0.02 | true | true |
f_seeding_0000500 | You want 180,000 established canola plants per hectare. Thousand-kernel weight is 27.6 g, germination 98% and expected field emergence 85%. What seeding rate in kg/ha? | Establishment = 98% x 85% = 83.3% of seeds sown.
Seeds to sow = 180,000 / 0.833 = 216,086 seeds/ha.
A thousand-kernel weight of 27.6 g means one seed weighs 27.6/1000 g.
Rate = 216,086 x 27.6 / 1000 / 1000 = 5.96 kg/ha.
Check: 5.96 kg/ha at that TKW and establishment gives back 180,000 plants/ha.
The final answer is $... | 5.96 kg/ha | seeding | rate_from_population | numeric | kg/ha | standard extension seeding-rate arithmetic | medium | inverse-recompute | population re-derived from the computed seed rate | 0.02 | true | true |
f_irrigation_0001804 | Sugarbeet is at mid-season on 20 ha. Reference evapotranspiration is 5.3 mm/day and the FAO-56 mid-season crop coefficient is 1.2. If you replace 5 days of crop water use with a system delivering 80 m3/h, how long must it run? | FAO-56: ETc = ETo x Kc = 5.3 x 1.2 = 6.36 mm/day.
Over 5 days the crop uses 6.36 x 5 = 31.8 mm.
1 mm applied over 1 ha is 10 m3, so 31.8 mm over 20 ha = 31.8 x 10 x 20 = 6,360 m3.
Run time = 6,360 / 80 = 79.5 hours.
The final answer is $\boxed{79.5 hours}$. | 79.5 hours | irrigation | etc_runtime | numeric | hours | FAO Irrigation & Drainage Paper 56 | hard | dimensional-roundtrip | depth reconstructed from the computed volume | 0.02 | true | true |
f_fert_rate_0001331 | A soil test for cotton on a 65 ha dryland field calls for 50 kg N/ha. You are using UAN 32 (32-0-0). How much product do you need per hectare, and how much in total for 15 ha? | The label analysis UAN 32 (32-0-0) means the product is 32% N by weight.
Product rate = N required / (%N / 100) = 50 / (32/100) = 156.25 kg/ha.
Total for 15 ha = 156.25 x 15 = 2,343.75 kg.
Check: 156.25 kg/ha x 32% = 50 kg N/ha, which is the rate asked for.
The final answer is $\boxed{156.25 kg/ha (2,343.75 kg total)}... | 156.25 kg/ha (2,343.75 kg total) | fertilizer | single_nutrient_rate | numeric | kg/ha | WSU extension fertilizer calculations | easy | inverse-recompute | product rate pushed back through the label analysis | 0.02 | true | true |
f_irrigation_0000423 | Tomato is at mid-season on 25 ha. Reference evapotranspiration is 7.1 mm/day and the FAO-56 mid-season crop coefficient is 1.15. If you replace 3 days of crop water use with a system delivering 100 m3/h, how long must it run? | FAO-56: ETc = ETo x Kc = 7.1 x 1.15 = 8.16 mm/day.
Over 3 days the crop uses 8.16 x 3 = 24.49 mm.
1 mm applied over 1 ha is 10 m3, so 24.49 mm over 25 ha = 24.49 x 10 x 25 = 6,123.75 m3.
Run time = 6,123.75 / 100 = 61.24 hours.
The final answer is $\boxed{61.24 hours}$. | 61.24 hours | irrigation | etc_runtime | numeric | hours | FAO Irrigation & Drainage Paper 56 | hard | dimensional-roundtrip | depth reconstructed from the computed volume | 0.02 | true | true |
f_grain_0002432 | A weigh wagon shows 8,800 lb of corn from one acre at 21.3% moisture. What is the yield in bushels per acre at the 15.5% standard? | Dry matter is conserved, so weight at 15.5% = 8,800 x (100 - 21.3) / (100 - 15.5)
= 8,800 x 78.7 / 84.5 = 8,195.98 lb.
Corn is 56 lb per bushel, so yield = 8,195.98 / 56 = 146.36 bu/ac.
The final answer is $\boxed{146.36 bu/ac}$. | 146.36 bu/ac | grain | yield_at_standard | numeric | bu/ac | OSU — bushels, test weights and calculations | medium | dry-matter-conservation | dry weight re-derived and re-divided | 0.02 | true | true |
f_irrigation_0000885 | Onion (dry) is at mid-season on 25 ha. Reference evapotranspiration is 8.4 mm/day and the FAO-56 mid-season crop coefficient is 1.05. If you replace 10 days of crop water use with a system delivering 250 m3/h, how long must it run? | FAO-56: ETc = ETo x Kc = 8.4 x 1.05 = 8.82 mm/day.
Over 10 days the crop uses 8.82 x 10 = 88.2 mm.
1 mm applied over 1 ha is 10 m3, so 88.2 mm over 25 ha = 88.2 x 10 x 25 = 22,050 m3.
Run time = 22,050 / 250 = 88.2 hours.
The final answer is $\boxed{88.2 hours}$. | 88.2 hours | irrigation | etc_runtime | numeric | hours | FAO Irrigation & Drainage Paper 56 | hard | dimensional-roundtrip | depth reconstructed from the computed volume | 0.02 | true | true |
f_fert_rate_0001655 | A soil test for potato on a 65 ha dryland field calls for 160 kg N/ha. You are using UAN 32 (32-0-0). How much product do you need per hectare, and how much in total for 10 ha? | The label analysis UAN 32 (32-0-0) means the product is 32% N by weight.
Product rate = N required / (%N / 100) = 160 / (32/100) = 500 kg/ha.
Total for 10 ha = 500 x 10 = 5,000 kg.
Check: 500 kg/ha x 32% = 160 kg N/ha, which is the rate asked for.
The final answer is $\boxed{500 kg/ha (5,000 kg total)}$. | 500 kg/ha (5,000 kg total) | fertilizer | single_nutrient_rate | numeric | kg/ha | WSU extension fertilizer calculations | easy | inverse-recompute | product rate pushed back through the label analysis | 0.02 | true | true |
f_oxide_0000920 | A recommendation is 63 kg/ha of P2O5 across 30 ha. How much elemental phosphorus (P) is that per hectare? | Elemental P = 0.437 x P2O5 (the oxide-to-element factor for phosphorus).
P = 0.437 x 63 = 27.53 kg/ha.
The final answer is $\boxed{27.53 kg/ha}$. | 27.53 kg/ha | fertilizer | oxide_to_element | numeric | kg P/ha | Cornell CSS412 | easy | closed-form | single published conversion factor | 0.01 | true | true |
f_refusal_0001175 | It is V6 and the forecast is dry. Soil is a sandy clay. The nearest weather station is 25 km away. When should I apply manure to a 12 ha centre pivot? | Manure timing is governed by local nutrient-management regulation and site conditions such as soil temperature, frozen ground and proximity to water. This is a regulated question and the jurisdiction is not given.
The details given do not substitute for it — none of them determine the value asked for.
The missing input... | cannot be determined without local extension guidance and the relevant regulations | refusal | insufficient_input | text | null | n/a — refusal by construction | medium | by-construction | the question deliberately omits an input the calculation requires | null | false | true |
f_refusal_0000015 | The field was in sugarbeet last season. How many days can I graze the home paddock? | Grazing days need the forage dry matter available per hectare and a utilisation assumption. Neither is provided, and standing forage cannot be estimated from field size alone.
The details given do not substitute for it — none of them determine the value asked for.
The missing input is the current forage dry matter on o... | cannot be determined without the current forage dry matter on offer | refusal | insufficient_input | text | null | n/a — refusal by construction | medium | by-construction | the question deliberately omits an input the calculation requires | null | false | true |
f_irrigation_0000963 | Maize (grain) is at mid-season on 6 ha. Reference evapotranspiration is 5.2 mm/day and the FAO-56 mid-season crop coefficient is 1.2. If you replace 10 days of crop water use with a system delivering 180 m3/h, how long must it run? | FAO-56: ETc = ETo x Kc = 5.2 x 1.2 = 6.24 mm/day.
Over 10 days the crop uses 6.24 x 10 = 62.4 mm.
1 mm applied over 1 ha is 10 m3, so 62.4 mm over 6 ha = 62.4 x 10 x 6 = 3,744 m3.
Run time = 3,744 / 180 = 20.8 hours.
The final answer is $\boxed{20.8 hours}$. | 20.8 hours | irrigation | etc_runtime | numeric | hours | FAO Irrigation & Drainage Paper 56 | hard | dimensional-roundtrip | depth reconstructed from the computed volume | 0.02 | true | true |
f_fert_rate_0001799 | A soil test for soybean on the north field calls for 150 kg N/ha. You are using ammonium nitrate (34-0-0). How much product do you need per hectare, and how much in total for 12 ha? | The label analysis ammonium nitrate (34-0-0) means the product is 34% N by weight.
Product rate = N required / (%N / 100) = 150 / (34/100) = 441.18 kg/ha.
Total for 12 ha = 441.18 x 12 = 5,294.12 kg.
Check: 441.18 kg/ha x 34% = 150 kg N/ha, which is the rate asked for.
The final answer is $\boxed{441.18 kg/ha (5,294.1... | 441.18 kg/ha (5,294.12 kg total) | fertilizer | single_nutrient_rate | numeric | kg/ha | WSU extension fertilizer calculations | easy | inverse-recompute | product rate pushed back through the label analysis | 0.02 | true | true |
f_irrigation_0001722 | Wheat (winter) is at mid-season on 12 ha. Reference evapotranspiration is 7.4 mm/day and the FAO-56 mid-season crop coefficient is 1.15. If you replace 5 days of crop water use with a system delivering 100 m3/h, how long must it run? | FAO-56: ETc = ETo x Kc = 7.4 x 1.15 = 8.51 mm/day.
Over 5 days the crop uses 8.51 x 5 = 42.55 mm.
1 mm applied over 1 ha is 10 m3, so 42.55 mm over 12 ha = 42.55 x 10 x 12 = 5,106 m3.
Run time = 5,106 / 100 = 51.06 hours.
The final answer is $\boxed{51.06 hours}$. | 51.06 hours | irrigation | etc_runtime | numeric | hours | FAO Irrigation & Drainage Paper 56 | hard | dimensional-roundtrip | depth reconstructed from the computed volume | 0.02 | true | true |
f_grain_0002725 | A weigh wagon shows 5,500 lb of corn from one acre at 25.6% moisture. What is the yield in bushels per acre at the 15.5% standard? | Dry matter is conserved, so weight at 15.5% = 5,500 x (100 - 25.6) / (100 - 15.5)
= 5,500 x 74.4 / 84.5 = 4,842.6 lb.
Corn is 56 lb per bushel, so yield = 4,842.6 / 56 = 86.48 bu/ac.
The final answer is $\boxed{86.48 bu/ac}$. | 86.48 bu/ac | grain | yield_at_standard | numeric | bu/ac | OSU — bushels, test weights and calculations | medium | dry-matter-conservation | dry weight re-derived and re-divided | 0.02 | true | true |
f_npk_blend_0000804 | Your recommendation is 200-85-60 kg/ha of N-P2O5-K2O. Using DAP (18-46-0), potash (0-0-60) and urea (46-0-0), what rate of each do you apply? Meet the phosphate with DAP first and credit the nitrogen it carries. | Phosphate first. DAP is 46% P2O5, so DAP = 85 / 0.46 = 184.78 kg/ha.
That DAP also carries nitrogen: 184.78 x 18% = 33.26 kg N/ha.
Potash is 60% K2O, so potash = 60 / 0.60 = 100 kg/ha.
Remaining N = 200 - 33.26 = 166.74 kg/ha.
Urea is 46% N, so urea = 166.74 / 0.46 = 362.48 kg/ha.
Check: N 200, P2O5 85, K2O 60 — matche... | DAP 184.78 kg/ha, potash 100 kg/ha, urea 362.48 kg/ha | fertilizer | npk_blend | numeric | kg/ha | WSU extension fertilizer calculations | hard | nutrient-balance | all three nutrient totals re-summed from the product rates | 0.02 | true | true |
f_economics_0001396 | UAN 32 (32-0-0) costs 705 per tonne. What is the cost per kg of actual nitrogen? | One tonne is 1000 kg of product carrying 32% N = 320 kg N.
Cost per kg N = 705 / 320 = 2.2 per kg N.
The final answer is $\boxed{2.2}$. | 2.2 | economics | cost_per_kg_nutrient | numeric | currency/kg N | input costing | easy | closed-form | unit-rate identity | 0.02 | true | true |
f_livestock_0000000 | 25 head averaging 650 kg graze a 40 ha block of 12 ha carrying 2,600 kg DM/ha. At 2.2% of bodyweight intake and 55% utilisation, how many grazing days does the paddock provide? | Intake per head = 650 x 2.2% = 14.3 kg DM/day.
Herd demand = 25 x 14.3 = 357.5 kg DM/day.
Usable forage = 2,600 x 12 x 55% = 17,160 kg DM.
Grazing days = 17,160 / 357.5 = 48 days.
The final answer is $\boxed{48 days}$. | 48 days | livestock | grazing_days | numeric | days | standard forage budgeting | medium | demand-supply-roundtrip | usable forage and herd demand recomputed separately | 0.02 | true | true |
f_fert_rate_0001716 | A soil test for canola on a 12 ha centre pivot calls for 120 kg N/ha. You are using ammonium nitrate (34-0-0). How much product do you need per hectare, and how much in total for 65 ha? | The label analysis ammonium nitrate (34-0-0) means the product is 34% N by weight.
Product rate = N required / (%N / 100) = 120 / (34/100) = 352.94 kg/ha.
Total for 65 ha = 352.94 x 65 = 22,941.18 kg.
Check: 352.94 kg/ha x 34% = 120 kg N/ha, which is the rate asked for.
The final answer is $\boxed{352.94 kg/ha (22,941... | 352.94 kg/ha (22,941.18 kg total) | fertilizer | single_nutrient_rate | numeric | kg/ha | WSU extension fertilizer calculations | easy | inverse-recompute | product rate pushed back through the label analysis | 0.02 | true | true |
f_grain_0001827 | A weigh wagon shows 5,675 lb of corn from one acre at 19.8% moisture. What is the yield in bushels per acre at the 15.5% standard? | Dry matter is conserved, so weight at 15.5% = 5,675 x (100 - 19.8) / (100 - 15.5)
= 5,675 x 80.2 / 84.5 = 5,386.21 lb.
Corn is 56 lb per bushel, so yield = 5,386.21 / 56 = 96.18 bu/ac.
The final answer is $\boxed{96.18 bu/ac}$. | 96.18 bu/ac | grain | yield_at_standard | numeric | bu/ac | OSU — bushels, test weights and calculations | medium | dry-matter-conservation | dry weight re-derived and re-divided | 0.02 | true | true |
f_seeding_0001481 | You want 400,000 established barley plants per hectare. Thousand-kernel weight is 25.2 g, germination 88% and expected field emergence 85%. What seeding rate in kg/ha? | Establishment = 88% x 85% = 74.8% of seeds sown.
Seeds to sow = 400,000 / 0.748 = 534,759 seeds/ha.
A thousand-kernel weight of 25.2 g means one seed weighs 25.2/1000 g.
Rate = 534,759 x 25.2 / 1000 / 1000 = 13.48 kg/ha.
Check: 13.48 kg/ha at that TKW and establishment gives back 400,000 plants/ha.
The final answer is... | 13.48 kg/ha | seeding | rate_from_population | numeric | kg/ha | standard extension seeding-rate arithmetic | medium | inverse-recompute | population re-derived from the computed seed rate | 0.02 | true | true |
f_irrigation_0001791 | Maize (grain) is at mid-season on 12 ha. Reference evapotranspiration is 6.8 mm/day and the FAO-56 mid-season crop coefficient is 1.2. If you replace 10 days of crop water use with a system delivering 180 m3/h, how long must it run? | FAO-56: ETc = ETo x Kc = 6.8 x 1.2 = 8.16 mm/day.
Over 10 days the crop uses 8.16 x 10 = 81.6 mm.
1 mm applied over 1 ha is 10 m3, so 81.6 mm over 12 ha = 81.6 x 10 x 12 = 9,792 m3.
Run time = 9,792 / 180 = 54.4 hours.
The final answer is $\boxed{54.4 hours}$. | 54.4 hours | irrigation | etc_runtime | numeric | hours | FAO Irrigation & Drainage Paper 56 | hard | dimensional-roundtrip | depth reconstructed from the computed volume | 0.02 | true | true |
f_grain_0000668 | A weigh wagon shows 7,725 lb of corn from one acre at 23.4% moisture. What is the yield in bushels per acre at the 15.5% standard? | Dry matter is conserved, so weight at 15.5% = 7,725 x (100 - 23.4) / (100 - 15.5)
= 7,725 x 76.6 / 84.5 = 7,002.78 lb.
Corn is 56 lb per bushel, so yield = 7,002.78 / 56 = 125.05 bu/ac.
The final answer is $\boxed{125.05 bu/ac}$. | 125.05 bu/ac | grain | yield_at_standard | numeric | bu/ac | OSU — bushels, test weights and calculations | medium | dry-matter-conservation | dry weight re-derived and re-divided | 0.02 | true | true |
f_gdd_0001290 | Accumulate growing degree days for soybean over these 5 days (max/min air temperature in F): day 1 65/45F; day 2 82/68F; day 3 65/41F; day 4 72/61F; day 5 64/39F. What is the total GDD? | GDD per day = ((Tmax + Tmin)/2) - 50F, the base temperature for soybean.
Before averaging, highs above 86F are counted as 86F, and lows below 50F are counted as 50F. A negative daily value is recorded as 0.
Day 1: 65/45 -> 7.5 GDD (running 7.5).
Day 2: 82/68 -> 25 GDD (running 32.5).
Day 3: 65/41 -> 7.5 GDD (runn... | 63.5 | degree_days | accumulate | numeric | GDD (F) | NDSU NDAWN | medium | per-day-resum | daily contributions re-summed independently of the accumulator | 0.01 | true | true |
f_refusal_0001791 | The field was in maize last season. It is V6 and the forecast is dry. How much product do I put in the tank for a 12 ha centre pivot? | Product per tank depends on the sprayer's actual output in L/ha, which comes from nozzle flow, travel speed and nozzle spacing. None of those are given.
The details given do not substitute for it — none of them determine the value asked for.
The missing input is the nozzle output and travel speed.
The final answer is ... | cannot be determined without the nozzle output and travel speed | refusal | insufficient_input | text | null | n/a — refusal by construction | medium | by-construction | the question deliberately omits an input the calculation requires | null | false | true |
f_gdd_0001606 | Accumulate growing degree days for soybean over these 10 days (max/min air temperature in F): day 1 70/41F; day 2 95/68F; day 3 84/54F; day 4 66/45F; day 5 81/64F; day 6 75/46F; day 7 75/67F; day 8 69/44F; day 9 58/42F; day 10 72/58F. What is the total GDD? | GDD per day = ((Tmax + Tmin)/2) - 50F, the base temperature for soybean.
Before averaging, highs above 86F are counted as 86F, and lows below 50F are counted as 50F. A negative daily value is recorded as 0.
Day 1: 70/41 -> 10 GDD (running 10).
Day 2: 95/68 -> 27 GDD (running 37).
Day 3: 84/54 -> 19 GDD (running 5... | 148.5 | degree_days | accumulate | numeric | GDD (F) | NDSU NDAWN | medium | per-day-resum | daily contributions re-summed independently of the accumulator | 0.01 | true | true |
f_seeding_0000873 | You want 300,000 established maize plants per hectare. Thousand-kernel weight is 5.3 g, germination 94% and expected field emergence 90%. What seeding rate in kg/ha? | Establishment = 94% x 90% = 84.6% of seeds sown.
Seeds to sow = 300,000 / 0.846 = 354,610 seeds/ha.
A thousand-kernel weight of 5.3 g means one seed weighs 5.3/1000 g.
Rate = 354,610 x 5.3 / 1000 / 1000 = 1.88 kg/ha.
Check: 1.88 kg/ha at that TKW and establishment gives back 300,000 plants/ha.
The final answer is $\bo... | 1.88 kg/ha | seeding | rate_from_population | numeric | kg/ha | standard extension seeding-rate arithmetic | medium | inverse-recompute | population re-derived from the computed seed rate | 0.02 | true | true |
f_grain_0000770 | Corn tests 57.8 lb/bu at 16.1% moisture. What is the expected test weight once it is dried to 15.5%? | Adjusted test weight = ((100 - MC_dry) / (100 - MC_wet)) x TW_wet.
= ((100 - 15.5) / (100 - 16.1)) x 57.8
= (84.5 / 83.9) x 57.8 = 58.21 lb/bu.
The final answer is $\boxed{58.21 lb/bu}$. | 58.21 lb/bu | grain | test_weight | numeric | lb/bu | Purdue — grain test weight considerations for corn | medium | closed-form | published test-weight identity | 0.02 | true | true |
f_fert_rate_0000457 | A soil test for potato on the north field calls for 75 kg N/ha. You are using UAN 32 (32-0-0). How much product do you need per hectare, and how much in total for 50 ha? | The label analysis UAN 32 (32-0-0) means the product is 32% N by weight.
Product rate = N required / (%N / 100) = 75 / (32/100) = 234.38 kg/ha.
Total for 50 ha = 234.38 x 50 = 11,718.75 kg.
Check: 234.38 kg/ha x 32% = 75 kg N/ha, which is the rate asked for.
The final answer is $\boxed{234.38 kg/ha (11,718.75 kg total... | 234.38 kg/ha (11,718.75 kg total) | fertilizer | single_nutrient_rate | numeric | kg/ha | WSU extension fertilizer calculations | easy | inverse-recompute | product rate pushed back through the label analysis | 0.02 | true | true |
f_units_0000041 | Convert 287.6 lb/acre to kg/ha. | 1 lb = 1/2.20462 kg and 1 ac = 1/2.47105 ha.
287.6 x 2.47105 / 2.20462 = 322.36 kg/ha.
The final answer is $\boxed{322.36 kg/ha}$. | 322.36 kg/ha | units | lbac_kgha | numeric | kg/ha | SI / US customary conversion factors | easy | closed-form | exact conversion constant | 0.01 | true | true |
f_livestock_0001052 | 30 head averaging 450 kg graze an 8 ha trial block of 8 ha carrying 3,400 kg DM/ha. At 2.8% of bodyweight intake and 40% utilisation, how many grazing days does the paddock provide? | Intake per head = 450 x 2.8% = 12.6 kg DM/day.
Herd demand = 30 x 12.6 = 378 kg DM/day.
Usable forage = 3,400 x 8 x 40% = 10,880 kg DM.
Grazing days = 10,880 / 378 = 28.78 days.
The final answer is $\boxed{28.78 days}$. | 28.78 days | livestock | grazing_days | numeric | days | standard forage budgeting | medium | demand-supply-roundtrip | usable forage and herd demand recomputed separately | 0.02 | true | true |
f_seeding_0000060 | You want 250,000 established canola plants per hectare. Thousand-kernel weight is 26.4 g, germination 85% and expected field emergence 92%. What seeding rate in kg/ha? | Establishment = 85% x 92% = 78.2% of seeds sown.
Seeds to sow = 250,000 / 0.782 = 319,693 seeds/ha.
A thousand-kernel weight of 26.4 g means one seed weighs 26.4/1000 g.
Rate = 319,693 x 26.4 / 1000 / 1000 = 8.44 kg/ha.
Check: 8.44 kg/ha at that TKW and establishment gives back 250,000 plants/ha.
The final answer is $... | 8.44 kg/ha | seeding | rate_from_population | numeric | kg/ha | standard extension seeding-rate arithmetic | medium | inverse-recompute | population re-derived from the computed seed rate | 0.02 | true | true |
f_grain_0002478 | Corn tests 52.8 lb/bu at 17.5% moisture. What is the expected test weight once it is dried to 15.5%? | Adjusted test weight = ((100 - MC_dry) / (100 - MC_wet)) x TW_wet.
= ((100 - 15.5) / (100 - 17.5)) x 52.8
= (84.5 / 82.5) x 52.8 = 54.08 lb/bu.
The final answer is $\boxed{54.08 lb/bu}$. | 54.08 lb/bu | grain | test_weight | numeric | lb/bu | Purdue — grain test weight considerations for corn | medium | closed-form | published test-weight identity | 0.02 | true | true |
f_npk_blend_0000790 | Your recommendation is 65-20-85 kg/ha of N-P2O5-K2O. Using DAP (18-46-0), potash (0-0-60) and urea (46-0-0), what rate of each do you apply? Meet the phosphate with DAP first and credit the nitrogen it carries. | Phosphate first. DAP is 46% P2O5, so DAP = 20 / 0.46 = 43.48 kg/ha.
That DAP also carries nitrogen: 43.48 x 18% = 7.83 kg N/ha.
Potash is 60% K2O, so potash = 85 / 0.60 = 141.67 kg/ha.
Remaining N = 65 - 7.83 = 57.17 kg/ha.
Urea is 46% N, so urea = 57.17 / 0.46 = 124.29 kg/ha.
Check: N 65, P2O5 20, K2O 85 — matches the... | DAP 43.48 kg/ha, potash 141.67 kg/ha, urea 124.29 kg/ha | fertilizer | npk_blend | numeric | kg/ha | WSU extension fertilizer calculations | hard | nutrient-balance | all three nutrient totals re-summed from the product rates | 0.02 | true | true |
f_grain_0001177 | A weigh wagon shows 4,550 lb of corn from one acre at 21.4% moisture. What is the yield in bushels per acre at the 15.5% standard? | Dry matter is conserved, so weight at 15.5% = 4,550 x (100 - 21.4) / (100 - 15.5)
= 4,550 x 78.6 / 84.5 = 4,232.31 lb.
Corn is 56 lb per bushel, so yield = 4,232.31 / 56 = 75.58 bu/ac.
The final answer is $\boxed{75.58 bu/ac}$. | 75.58 bu/ac | grain | yield_at_standard | numeric | bu/ac | OSU — bushels, test weights and calculations | medium | dry-matter-conservation | dry weight re-derived and re-divided | 0.02 | true | true |
f_gdd_0000505 | Accumulate growing degree days for corn over these 10 days (max/min air temperature in F): day 1 60/43F; day 2 87/63F; day 3 53/44F; day 4 68/39F; day 5 81/51F; day 6 73/49F; day 7 62/50F; day 8 82/57F; day 9 66/57F; day 10 84/68F. What is the total GDD? | GDD per day = ((Tmax + Tmin)/2) - 50F, the base temperature for corn.
Before averaging, highs above 86F are counted as 86F, and lows below 50F are counted as 50F. A negative daily value is recorded as 0.
Day 1: 60/43 -> 5 GDD (running 5).
Day 2: 87/63 -> 24.5 GDD (running 29.5).
Day 3: 53/44 -> 1.5 GDD (running 3... | 130.5 | degree_days | accumulate | numeric | GDD (F) | NDSU NDAWN | medium | per-day-resum | daily contributions re-summed independently of the accumulator | 0.01 | true | true |
f_oxide_0000193 | A recommendation is 161 kg/ha of P2O5 across 80 ha. How much elemental phosphorus (P) is that per hectare? | Elemental P = 0.437 x P2O5 (the oxide-to-element factor for phosphorus).
P = 0.437 x 161 = 70.36 kg/ha.
The final answer is $\boxed{70.36 kg/ha}$. | 70.36 kg/ha | fertilizer | oxide_to_element | numeric | kg P/ha | Cornell CSS412 | easy | closed-form | single published conversion factor | 0.01 | true | true |
f_seeding_0000774 | You want 220,000 established barley plants per hectare. Thousand-kernel weight is 15.1 g, germination 92% and expected field emergence 92%. What seeding rate in kg/ha? | Establishment = 92% x 92% = 84.64% of seeds sown.
Seeds to sow = 220,000 / 0.8464 = 259,924 seeds/ha.
A thousand-kernel weight of 15.1 g means one seed weighs 15.1/1000 g.
Rate = 259,924 x 15.1 / 1000 / 1000 = 3.92 kg/ha.
Check: 3.92 kg/ha at that TKW and establishment gives back 220,000 plants/ha.
The final answer is... | 3.92 kg/ha | seeding | rate_from_population | numeric | kg/ha | standard extension seeding-rate arithmetic | medium | inverse-recompute | population re-derived from the computed seed rate | 0.02 | true | true |
f_irrigation_0001209 | Soybean is at mid-season on 15 ha. Reference evapotranspiration is 6.8 mm/day and the FAO-56 mid-season crop coefficient is 1.15. If you replace 5 days of crop water use with a system delivering 80 m3/h, how long must it run? | FAO-56: ETc = ETo x Kc = 6.8 x 1.15 = 7.82 mm/day.
Over 5 days the crop uses 7.82 x 5 = 39.1 mm.
1 mm applied over 1 ha is 10 m3, so 39.1 mm over 15 ha = 39.1 x 10 x 15 = 5,865 m3.
Run time = 5,865 / 80 = 73.31 hours.
The final answer is $\boxed{73.31 hours}$. | 73.31 hours | irrigation | etc_runtime | numeric | hours | FAO Irrigation & Drainage Paper 56 | hard | dimensional-roundtrip | depth reconstructed from the computed volume | 0.02 | true | true |
f_grain_0001080 | Corn tests 54.9 lb/bu at 18.1% moisture. What is the expected test weight once it is dried to 15.5%? | Adjusted test weight = ((100 - MC_dry) / (100 - MC_wet)) x TW_wet.
= ((100 - 15.5) / (100 - 18.1)) x 54.9
= (84.5 / 81.9) x 54.9 = 56.64 lb/bu.
The final answer is $\boxed{56.64 lb/bu}$. | 56.64 lb/bu | grain | test_weight | numeric | lb/bu | Purdue — grain test weight considerations for corn | medium | closed-form | published test-weight identity | 0.02 | true | true |
f_fert_rate_0001465 | A soil test for maize on a 65 ha dryland field calls for 60 kg N/ha. You are using NPK blend (15-15-15). How much product do you need per hectare, and how much in total for 12 ha? | The label analysis NPK blend (15-15-15) means the product is 15% N by weight.
Product rate = N required / (%N / 100) = 60 / (15/100) = 400 kg/ha.
Total for 12 ha = 400 x 12 = 4,800 kg.
Check: 400 kg/ha x 15% = 60 kg N/ha, which is the rate asked for.
The final answer is $\boxed{400 kg/ha (4,800 kg total)}$. | 400 kg/ha (4,800 kg total) | fertilizer | single_nutrient_rate | numeric | kg/ha | WSU extension fertilizer calculations | easy | inverse-recompute | product rate pushed back through the label analysis | 0.02 | true | true |
f_refusal_0001534 | Soil is a peaty loam. The field was in winter wheat last season. How much herbicide should I spray on sugarbeet? | Application rate is set by the registered product label, which is product- and jurisdiction-specific and legally binding. It cannot be inferred.
The details given do not substitute for it — none of them determine the value asked for.
The missing input is the label rate for the specific product.
The final answer is $\b... | cannot be determined without the label rate for the specific product | refusal | insufficient_input | text | null | n/a — refusal by construction | medium | by-construction | the question deliberately omits an input the calculation requires | null | false | true |
f_irrigation_0000071 | Onion (dry) is at mid-season on 40 ha. Reference evapotranspiration is 5 mm/day and the FAO-56 mid-season crop coefficient is 1.05. If you replace 7 days of crop water use with a system delivering 180 m3/h, how long must it run? | FAO-56: ETc = ETo x Kc = 5 x 1.05 = 5.25 mm/day.
Over 7 days the crop uses 5.25 x 7 = 36.75 mm.
1 mm applied over 1 ha is 10 m3, so 36.75 mm over 40 ha = 36.75 x 10 x 40 = 14,700 m3.
Run time = 14,700 / 180 = 81.67 hours.
The final answer is $\boxed{81.67 hours}$. | 81.67 hours | irrigation | etc_runtime | numeric | hours | FAO Irrigation & Drainage Paper 56 | hard | dimensional-roundtrip | depth reconstructed from the computed volume | 0.02 | true | true |
f_livestock_0000710 | 100 head averaging 600 kg graze a 65 ha dryland field of 12 ha carrying 1,800 kg DM/ha. At 3.0% of bodyweight intake and 65% utilisation, how many grazing days does the paddock provide? | Intake per head = 600 x 3.0% = 18 kg DM/day.
Herd demand = 100 x 18 = 1,800 kg DM/day.
Usable forage = 1,800 x 12 x 65% = 14,040 kg DM.
Grazing days = 14,040 / 1,800 = 7.8 days.
The final answer is $\boxed{7.8 days}$. | 7.8 days | livestock | grazing_days | numeric | days | standard forage budgeting | medium | demand-supply-roundtrip | usable forage and herd demand recomputed separately | 0.02 | true | true |
f_irrigation_0000175 | Onion (dry) is at mid-season on 10 ha. Reference evapotranspiration is 8.6 mm/day and the FAO-56 mid-season crop coefficient is 1.05. If you replace 5 days of crop water use with a system delivering 250 m3/h, how long must it run? | FAO-56: ETc = ETo x Kc = 8.6 x 1.05 = 9.03 mm/day.
Over 5 days the crop uses 9.03 x 5 = 45.15 mm.
1 mm applied over 1 ha is 10 m3, so 45.15 mm over 10 ha = 45.15 x 10 x 10 = 4,515 m3.
Run time = 4,515 / 250 = 18.06 hours.
The final answer is $\boxed{18.06 hours}$. | 18.06 hours | irrigation | etc_runtime | numeric | hours | FAO Irrigation & Drainage Paper 56 | hard | dimensional-roundtrip | depth reconstructed from the computed volume | 0.02 | true | true |
f_irrigation_0000364 | Alfalfa is at mid-season on 10 ha. Reference evapotranspiration is 7.1 mm/day and the FAO-56 mid-season crop coefficient is 1.2. If you replace 3 days of crop water use with a system delivering 150 m3/h, how long must it run? | FAO-56: ETc = ETo x Kc = 7.1 x 1.2 = 8.52 mm/day.
Over 3 days the crop uses 8.52 x 3 = 25.56 mm.
1 mm applied over 1 ha is 10 m3, so 25.56 mm over 10 ha = 25.56 x 10 x 10 = 2,556 m3.
Run time = 2,556 / 150 = 17.04 hours.
The final answer is $\boxed{17.04 hours}$. | 17.04 hours | irrigation | etc_runtime | numeric | hours | FAO Irrigation & Drainage Paper 56 | hard | dimensional-roundtrip | depth reconstructed from the computed volume | 0.02 | true | true |
f_economics_0000179 | ammonium sulphate (21-0-0) costs 495 per tonne. What is the cost per kg of actual nitrogen? | One tonne is 1000 kg of product carrying 21% N = 210 kg N.
Cost per kg N = 495 / 210 = 2.36 per kg N.
The final answer is $\boxed{2.36}$. | 2.36 | economics | cost_per_kg_nutrient | numeric | currency/kg N | input costing | easy | closed-form | unit-rate identity | 0.02 | true | true |
f_gdd_0001651 | Accumulate growing degree days for wheat over these 5 days (max/min air temperature in F): day 1 71/46F; day 2 69/59F; day 3 75/45F; day 4 66/39F; day 5 59/39F. What is the total GDD? | GDD per day = ((Tmax + Tmin)/2) - 32F, the base temperature for wheat.
Before averaging, and lows below 32F are counted as 32F. A negative daily value is recorded as 0.
Day 1: 71/46 -> 26.5 GDD (running 26.5).
Day 2: 69/59 -> 32 GDD (running 58.5).
Day 3: 75/45 -> 28 GDD (running 86.5).
Day 4: 66/39 -> 20.5 GDD... | 124 | degree_days | accumulate | numeric | GDD (F) | NDSU NDAWN | medium | per-day-resum | daily contributions re-summed independently of the accumulator | 0.01 | true | true |
f_irrigation_0001307 | Maize (grain) is at mid-season on 6 ha. Reference evapotranspiration is 6.5 mm/day and the FAO-56 mid-season crop coefficient is 1.2. If you replace 5 days of crop water use with a system delivering 120 m3/h, how long must it run? | FAO-56: ETc = ETo x Kc = 6.5 x 1.2 = 7.8 mm/day.
Over 5 days the crop uses 7.8 x 5 = 39 mm.
1 mm applied over 1 ha is 10 m3, so 39 mm over 6 ha = 39 x 10 x 6 = 2,340 m3.
Run time = 2,340 / 120 = 19.5 hours.
The final answer is $\boxed{19.5 hours}$. | 19.5 hours | irrigation | etc_runtime | numeric | hours | FAO Irrigation & Drainage Paper 56 | hard | dimensional-roundtrip | depth reconstructed from the computed volume | 0.02 | true | true |
f_economics_0000056 | Variable costs are 700 per hectare and the crop sells for 305 per tonne. What yield do you need just to cover variable costs? | Breakeven yield = variable cost / price = 700 / 305 = 2.3 t/ha.
The final answer is $\boxed{2.3 t/ha}$. | 2.3 t/ha | economics | breakeven_yield | numeric | t/ha | gross margin budgeting | easy | closed-form | division identity | 0.02 | true | true |
f_refusal_0000760 | We are 80 days after sowing. Soil is a silty clay loam. The field was in maize last season. When should I apply manure to the home paddock? | Manure timing is governed by local nutrient-management regulation and site conditions such as soil temperature, frozen ground and proximity to water. This is a regulated question and the jurisdiction is not given.
The details given do not substitute for it — none of them determine the value asked for.
The missing input... | cannot be determined without local extension guidance and the relevant regulations | refusal | insufficient_input | text | null | n/a — refusal by construction | medium | by-construction | the question deliberately omits an input the calculation requires | null | false | true |
f_units_0000196 | Convert 297.3 kg/ha to lb/acre. | 1 kg = 2.20462 lb and 1 ha = 2.47105 ac.
297.3 x 2.20462 / 2.47105 = 265.24 lb/ac.
The final answer is $\boxed{265.24 lb/ac}$. | 265.24 lb/ac | units | kgha_lbac | numeric | lb/ac | SI / US customary conversion factors | easy | closed-form | exact conversion constant | 0.01 | true | true |
f_irrigation_0000889 | Barley is at mid-season on 15 ha. Reference evapotranspiration is 4.6 mm/day and the FAO-56 mid-season crop coefficient is 1.15. If you replace 7 days of crop water use with a system delivering 200 m3/h, how long must it run? | FAO-56: ETc = ETo x Kc = 4.6 x 1.15 = 5.29 mm/day.
Over 7 days the crop uses 5.29 x 7 = 37.03 mm.
1 mm applied over 1 ha is 10 m3, so 37.03 mm over 15 ha = 37.03 x 10 x 15 = 5,554.5 m3.
Run time = 5,554.5 / 200 = 27.77 hours.
The final answer is $\boxed{27.77 hours}$. | 27.77 hours | irrigation | etc_runtime | numeric | hours | FAO Irrigation & Drainage Paper 56 | hard | dimensional-roundtrip | depth reconstructed from the computed volume | 0.02 | true | true |
f_fert_rate_0002227 | A soil test for maize on the home paddock calls for 80 kg N/ha. You are using NPK blend (15-15-15). How much product do you need per hectare, and how much in total for 20 ha? | The label analysis NPK blend (15-15-15) means the product is 15% N by weight.
Product rate = N required / (%N / 100) = 80 / (15/100) = 533.33 kg/ha.
Total for 20 ha = 533.33 x 20 = 10,666.67 kg.
Check: 533.33 kg/ha x 15% = 80 kg N/ha, which is the rate asked for.
The final answer is $\boxed{533.33 kg/ha (10,666.67 kg ... | 533.33 kg/ha (10,666.67 kg total) | fertilizer | single_nutrient_rate | numeric | kg/ha | WSU extension fertilizer calculations | easy | inverse-recompute | product rate pushed back through the label analysis | 0.02 | true | true |
f_grain_0000623 | Corn tests 48.5 lb/bu at 23.5% moisture. What is the expected test weight once it is dried to 15.5%? | Adjusted test weight = ((100 - MC_dry) / (100 - MC_wet)) x TW_wet.
= ((100 - 15.5) / (100 - 23.5)) x 48.5
= (84.5 / 76.5) x 48.5 = 53.57 lb/bu.
The final answer is $\boxed{53.57 lb/bu}$. | 53.57 lb/bu | grain | test_weight | numeric | lb/bu | Purdue — grain test weight considerations for corn | medium | closed-form | published test-weight identity | 0.02 | true | true |
f_irrigation_0001074 | Cotton is at mid-season on 10 ha. Reference evapotranspiration is 5.5 mm/day and the FAO-56 mid-season crop coefficient is 1.18. If you replace 10 days of crop water use with a system delivering 80 m3/h, how long must it run? | FAO-56: ETc = ETo x Kc = 5.5 x 1.18 = 6.49 mm/day.
Over 10 days the crop uses 6.49 x 10 = 64.9 mm.
1 mm applied over 1 ha is 10 m3, so 64.9 mm over 10 ha = 64.9 x 10 x 10 = 6,490 m3.
Run time = 6,490 / 80 = 81.12 hours.
The final answer is $\boxed{81.12 hours}$. | 81.12 hours | irrigation | etc_runtime | numeric | hours | FAO Irrigation & Drainage Paper 56 | hard | dimensional-roundtrip | depth reconstructed from the computed volume | 0.02 | true | true |
f_gdd_0000836 | Accumulate growing degree days for corn over these 7 days (max/min air temperature in F): day 1 86/59F; day 2 69/51F; day 3 59/43F; day 4 58/45F; day 5 63/48F; day 6 80/66F; day 7 69/50F. What is the total GDD? | GDD per day = ((Tmax + Tmin)/2) - 50F, the base temperature for corn.
Before averaging, highs above 86F are counted as 86F, and lows below 50F are counted as 50F. A negative daily value is recorded as 0.
Day 1: 86/59 -> 22.5 GDD (running 22.5).
Day 2: 69/51 -> 10 GDD (running 32.5).
Day 3: 59/43 -> 4.5 GDD (runni... | 80 | degree_days | accumulate | numeric | GDD (F) | NDSU NDAWN | medium | per-day-resum | daily contributions re-summed independently of the accumulator | 0.01 | true | true |
f_gdd_0001706 | Accumulate growing degree days for wheat over these 7 days (max/min air temperature in F): day 1 78/61F; day 2 68/40F; day 3 70/61F; day 4 57/43F; day 5 81/63F; day 6 77/59F; day 7 73/59F. What is the total GDD? | GDD per day = ((Tmax + Tmin)/2) - 32F, the base temperature for wheat.
Before averaging, and lows below 32F are counted as 32F. A negative daily value is recorded as 0.
Day 1: 78/61 -> 37.5 GDD (running 37.5).
Day 2: 68/40 -> 22 GDD (running 59.5).
Day 3: 70/61 -> 33.5 GDD (running 93).
Day 4: 57/43 -> 18 GDD (... | 221 | degree_days | accumulate | numeric | GDD (F) | NDSU NDAWN | medium | per-day-resum | daily contributions re-summed independently of the accumulator | 0.01 | true | true |
f_spray_0000841 | Your sprayer runs 2.16 L/min per nozzle at 12 km/h with nozzles 0.375 m apart. The label rate is 3.58 L/ha and the tank holds 600 L. What is the output in L/ha, and how much product goes in one tank? | Output L/ha = (L/min x 600) / (km/h x nozzle spacing in m).
= (2.16 x 600) / (12 x 0.375) = 288 L/ha.
One tank covers 600 / 288 = 2.08 ha.
Product per tank = 3.58 x 2.08 = 7.46 L.
The final answer is $\boxed{288 L/ha; 7.46 L per tank}$. | 288 L/ha; 7.46 L per tank | spraying | calibration | numeric | L/ha | sprayer calibration 600-rule | hard | area-roundtrip | product per tank cross-checked against hectares per tank | 0.02 | true | true |
f_irrigation_0000192 | Soybean is at mid-season on 40 ha. Reference evapotranspiration is 8.5 mm/day and the FAO-56 mid-season crop coefficient is 1.15. If you replace 5 days of crop water use with a system delivering 100 m3/h, how long must it run? | FAO-56: ETc = ETo x Kc = 8.5 x 1.15 = 9.77 mm/day.
Over 5 days the crop uses 9.77 x 5 = 48.87 mm.
1 mm applied over 1 ha is 10 m3, so 48.87 mm over 40 ha = 48.87 x 10 x 40 = 19,550 m3.
Run time = 19,550 / 100 = 195.5 hours.
The final answer is $\boxed{195.5 hours}$. | 195.5 hours | irrigation | etc_runtime | numeric | hours | FAO Irrigation & Drainage Paper 56 | hard | dimensional-roundtrip | depth reconstructed from the computed volume | 0.02 | true | true |
f_irrigation_0001458 | Rice is at mid-season on 15 ha. Reference evapotranspiration is 3.3 mm/day and the FAO-56 mid-season crop coefficient is 1.2. If you replace 5 days of crop water use with a system delivering 180 m3/h, how long must it run? | FAO-56: ETc = ETo x Kc = 3.3 x 1.2 = 3.96 mm/day.
Over 5 days the crop uses 3.96 x 5 = 19.8 mm.
1 mm applied over 1 ha is 10 m3, so 19.8 mm over 15 ha = 19.8 x 10 x 15 = 2,970 m3.
Run time = 2,970 / 180 = 16.5 hours.
The final answer is $\boxed{16.5 hours}$. | 16.5 hours | irrigation | etc_runtime | numeric | hours | FAO Irrigation & Drainage Paper 56 | hard | dimensional-roundtrip | depth reconstructed from the computed volume | 0.02 | true | true |
f_gdd_0000666 | Accumulate growing degree days for soybean over these 5 days (max/min air temperature in F): day 1 80/61F; day 2 62/42F; day 3 85/61F; day 4 61/51F; day 5 75/49F. What is the total GDD? | GDD per day = ((Tmax + Tmin)/2) - 50F, the base temperature for soybean.
Before averaging, highs above 86F are counted as 86F, and lows below 50F are counted as 50F. A negative daily value is recorded as 0.
Day 1: 80/61 -> 20.5 GDD (running 20.5).
Day 2: 62/42 -> 6 GDD (running 26.5).
Day 3: 85/61 -> 23 GDD (runn... | 68 | degree_days | accumulate | numeric | GDD (F) | NDSU NDAWN | medium | per-day-resum | daily contributions re-summed independently of the accumulator | 0.01 | true | true |
f_gdd_0000920 | Accumulate growing degree days for soybean over these 5 days (max/min air temperature in F): day 1 77/50F; day 2 74/62F; day 3 82/62F; day 4 63/44F; day 5 75/50F. What is the total GDD? | GDD per day = ((Tmax + Tmin)/2) - 50F, the base temperature for soybean.
Before averaging, highs above 86F are counted as 86F, and lows below 50F are counted as 50F. A negative daily value is recorded as 0.
Day 1: 77/50 -> 13.5 GDD (running 13.5).
Day 2: 74/62 -> 18 GDD (running 31.5).
Day 3: 82/62 -> 22 GDD (run... | 72.5 | degree_days | accumulate | numeric | GDD (F) | NDSU NDAWN | medium | per-day-resum | daily contributions re-summed independently of the accumulator | 0.01 | true | true |
f_seeding_0000830 | You want 400,000 established soybean plants per hectare. Thousand-kernel weight is 13.7 g, germination 90% and expected field emergence 85%. What seeding rate in kg/ha? | Establishment = 90% x 85% = 76.5% of seeds sown.
Seeds to sow = 400,000 / 0.765 = 522,876 seeds/ha.
A thousand-kernel weight of 13.7 g means one seed weighs 13.7/1000 g.
Rate = 522,876 x 13.7 / 1000 / 1000 = 7.16 kg/ha.
Check: 7.16 kg/ha at that TKW and establishment gives back 400,000 plants/ha.
The final answer is $... | 7.16 kg/ha | seeding | rate_from_population | numeric | kg/ha | standard extension seeding-rate arithmetic | medium | inverse-recompute | population re-derived from the computed seed rate | 0.02 | true | true |
f_irrigation_0000938 | Wheat (winter) is at mid-season on 10 ha. Reference evapotranspiration is 6 mm/day and the FAO-56 mid-season crop coefficient is 1.15. If you replace 10 days of crop water use with a system delivering 200 m3/h, how long must it run? | FAO-56: ETc = ETo x Kc = 6 x 1.15 = 6.9 mm/day.
Over 10 days the crop uses 6.9 x 10 = 69 mm.
1 mm applied over 1 ha is 10 m3, so 69 mm over 10 ha = 69 x 10 x 10 = 6,900 m3.
Run time = 6,900 / 200 = 34.5 hours.
The final answer is $\boxed{34.5 hours}$. | 34.5 hours | irrigation | etc_runtime | numeric | hours | FAO Irrigation & Drainage Paper 56 | hard | dimensional-roundtrip | depth reconstructed from the computed volume | 0.02 | true | true |
f_units_0000071 | Convert 382.6 lb/acre to kg/ha. | 1 lb = 1/2.20462 kg and 1 ac = 1/2.47105 ha.
382.6 x 2.47105 / 2.20462 = 428.84 kg/ha.
The final answer is $\boxed{428.84 kg/ha}$. | 428.84 kg/ha | units | lbac_kgha | numeric | kg/ha | SI / US customary conversion factors | easy | closed-form | exact conversion constant | 0.01 | true | true |
f_refusal_0000531 | Soil is a sandy loam. We are 100 days after sowing. How many days can I graze an 8 ha trial block? | Grazing days need the forage dry matter available per hectare and a utilisation assumption. Neither is provided, and standing forage cannot be estimated from field size alone.
The details given do not substitute for it — none of them determine the value asked for.
The missing input is the current forage dry matter on o... | cannot be determined without the current forage dry matter on offer | refusal | insufficient_input | text | null | n/a — refusal by construction | medium | by-construction | the question deliberately omits an input the calculation requires | null | false | true |
f_irrigation_0001919 | Alfalfa is at mid-season on 6 ha. Reference evapotranspiration is 3.2 mm/day and the FAO-56 mid-season crop coefficient is 1.2. If you replace 10 days of crop water use with a system delivering 150 m3/h, how long must it run? | FAO-56: ETc = ETo x Kc = 3.2 x 1.2 = 3.84 mm/day.
Over 10 days the crop uses 3.84 x 10 = 38.4 mm.
1 mm applied over 1 ha is 10 m3, so 38.4 mm over 6 ha = 38.4 x 10 x 6 = 2,304 m3.
Run time = 2,304 / 150 = 15.36 hours.
The final answer is $\boxed{15.36 hours}$. | 15.36 hours | irrigation | etc_runtime | numeric | hours | FAO Irrigation & Drainage Paper 56 | hard | dimensional-roundtrip | depth reconstructed from the computed volume | 0.02 | true | true |
f_refusal_0000881 | Soil is a chalky loam. What will my sugarbeet weigh once it is dried down? | Dried weight follows from the starting moisture and the target moisture. The starting moisture is not given, and dry-matter conservation cannot be applied without it.
The details given do not substitute for it — none of them determine the value asked for.
The missing input is the grain moisture at harvest.
The final a... | cannot be determined without the grain moisture at harvest | refusal | insufficient_input | text | null | n/a — refusal by construction | medium | by-construction | the question deliberately omits an input the calculation requires | null | false | true |
f_units_0000042 | How many acres is 23 hectares? | 1 ha = 2.47105 ac.
23 x 2.47105 = 56.83 ac.
The final answer is $\boxed{56.83 ac}$. | 56.83 ac | units | ha_ac | numeric | ac | SI / US customary conversion factors | easy | closed-form | exact conversion constant | 0.01 | true | true |
f_fert_rate_0000981 | A soil test for barley on a 12 ha centre pivot calls for 140 kg N/ha. You are using urea (46-0-0). How much product do you need per hectare, and how much in total for 40 ha? | The label analysis urea (46-0-0) means the product is 46% N by weight.
Product rate = N required / (%N / 100) = 140 / (46/100) = 304.35 kg/ha.
Total for 40 ha = 304.35 x 40 = 12,173.91 kg.
Check: 304.35 kg/ha x 46% = 140 kg N/ha, which is the rate asked for.
The final answer is $\boxed{304.35 kg/ha (12,173.91 kg total... | 304.35 kg/ha (12,173.91 kg total) | fertilizer | single_nutrient_rate | numeric | kg/ha | WSU extension fertilizer calculations | easy | inverse-recompute | product rate pushed back through the label analysis | 0.02 | true | true |
f_grain_0000371 | A weigh wagon shows 7,675 lb of corn from one acre at 22.7% moisture. What is the yield in bushels per acre at the 15.5% standard? | Dry matter is conserved, so weight at 15.5% = 7,675 x (100 - 22.7) / (100 - 15.5)
= 7,675 x 77.3 / 84.5 = 7,021.04 lb.
Corn is 56 lb per bushel, so yield = 7,021.04 / 56 = 125.38 bu/ac.
The final answer is $\boxed{125.38 bu/ac}$. | 125.38 bu/ac | grain | yield_at_standard | numeric | bu/ac | OSU — bushels, test weights and calculations | medium | dry-matter-conservation | dry weight re-derived and re-divided | 0.02 | true | true |
f_irrigation_0001597 | Tomato is at mid-season on 40 ha. Reference evapotranspiration is 5.8 mm/day and the FAO-56 mid-season crop coefficient is 1.15. If you replace 7 days of crop water use with a system delivering 200 m3/h, how long must it run? | FAO-56: ETc = ETo x Kc = 5.8 x 1.15 = 6.67 mm/day.
Over 7 days the crop uses 6.67 x 7 = 46.69 mm.
1 mm applied over 1 ha is 10 m3, so 46.69 mm over 40 ha = 46.69 x 10 x 40 = 18,676 m3.
Run time = 18,676 / 200 = 93.38 hours.
The final answer is $\boxed{93.38 hours}$. | 93.38 hours | irrigation | etc_runtime | numeric | hours | FAO Irrigation & Drainage Paper 56 | hard | dimensional-roundtrip | depth reconstructed from the computed volume | 0.02 | true | true |
f_irrigation_0000478 | Maize (grain) is at mid-season on 6 ha. Reference evapotranspiration is 3.7 mm/day and the FAO-56 mid-season crop coefficient is 1.2. If you replace 5 days of crop water use with a system delivering 100 m3/h, how long must it run? | FAO-56: ETc = ETo x Kc = 3.7 x 1.2 = 4.44 mm/day.
Over 5 days the crop uses 4.44 x 5 = 22.2 mm.
1 mm applied over 1 ha is 10 m3, so 22.2 mm over 6 ha = 22.2 x 10 x 6 = 1,332 m3.
Run time = 1,332 / 100 = 13.32 hours.
The final answer is $\boxed{13.32 hours}$. | 13.32 hours | irrigation | etc_runtime | numeric | hours | FAO Irrigation & Drainage Paper 56 | hard | dimensional-roundtrip | depth reconstructed from the computed volume | 0.02 | true | true |
f_npk_blend_0000582 | Your recommendation is 165-60-80 kg/ha of N-P2O5-K2O. Using DAP (18-46-0), potash (0-0-60) and urea (46-0-0), what rate of each do you apply? Meet the phosphate with DAP first and credit the nitrogen it carries. | Phosphate first. DAP is 46% P2O5, so DAP = 60 / 0.46 = 130.43 kg/ha.
That DAP also carries nitrogen: 130.43 x 18% = 23.48 kg N/ha.
Potash is 60% K2O, so potash = 80 / 0.60 = 133.33 kg/ha.
Remaining N = 165 - 23.48 = 141.52 kg/ha.
Urea is 46% N, so urea = 141.52 / 0.46 = 307.66 kg/ha.
Check: N 165, P2O5 60, K2O 80 — mat... | DAP 130.43 kg/ha, potash 133.33 kg/ha, urea 307.66 kg/ha | fertilizer | npk_blend | numeric | kg/ha | WSU extension fertilizer calculations | hard | nutrient-balance | all three nutrient totals re-summed from the product rates | 0.02 | true | true |
f_economics_0000309 | urea (46-0-0) costs 390 per tonne. What is the cost per kg of actual nitrogen? | One tonne is 1000 kg of product carrying 46% N = 460 kg N.
Cost per kg N = 390 / 460 = 0.85 per kg N.
The final answer is $\boxed{0.85}$. | 0.85 | economics | cost_per_kg_nutrient | numeric | currency/kg N | input costing | easy | closed-form | unit-rate identity | 0.02 | true | true |
f_grain_0002542 | A weigh wagon shows 8,350 lb of corn from one acre at 24.9% moisture. What is the yield in bushels per acre at the 15.5% standard? | Dry matter is conserved, so weight at 15.5% = 8,350 x (100 - 24.9) / (100 - 15.5)
= 8,350 x 75.1 / 84.5 = 7,421.12 lb.
Corn is 56 lb per bushel, so yield = 7,421.12 / 56 = 132.52 bu/ac.
The final answer is $\boxed{132.52 bu/ac}$. | 132.52 bu/ac | grain | yield_at_standard | numeric | bu/ac | OSU — bushels, test weights and calculations | medium | dry-matter-conservation | dry weight re-derived and re-divided | 0.02 | true | true |
f_economics_0000567 | Variable costs are 1280 per hectare and the crop sells for 155 per tonne. What yield do you need just to cover variable costs? | Breakeven yield = variable cost / price = 1280 / 155 = 8.26 t/ha.
The final answer is $\boxed{8.26 t/ha}$. | 8.26 t/ha | economics | breakeven_yield | numeric | t/ha | gross margin budgeting | easy | closed-form | division identity | 0.02 | true | true |
f_economics_0000012 | A crop yields 5.2 t/ha and sells for 290 per tonne. Variable costs are 1160 per hectare. What is the gross margin per hectare? | Revenue = 5.2 x 290 = 1,508 per ha.
Gross margin = revenue - variable costs = 1,508 - 1160 = 348 per ha.
The final answer is $\boxed{348}$. | 348 | economics | gross_margin | numeric | currency/ha | gross margin budgeting | easy | closed-form | revenue re-multiplied | 0.02 | true | true |
f_grain_0002129 | Corn tests 55.9 lb/bu at 22.9% moisture. What is the expected test weight once it is dried to 15.5%? | Adjusted test weight = ((100 - MC_dry) / (100 - MC_wet)) x TW_wet.
= ((100 - 15.5) / (100 - 22.9)) x 55.9
= (84.5 / 77.1) x 55.9 = 61.27 lb/bu.
The final answer is $\boxed{61.27 lb/bu}$. | 61.27 lb/bu | grain | test_weight | numeric | lb/bu | Purdue — grain test weight considerations for corn | medium | closed-form | published test-weight identity | 0.02 | true | true |
f_refusal_0001277 | Last year the same field yielded 4.6 t/ha. It is flowering and the forecast is dry. The block is 6 ha. What will my sunflower weigh once it is dried down? | Dried weight follows from the starting moisture and the target moisture. The starting moisture is not given, and dry-matter conservation cannot be applied without it.
The details given do not substitute for it — none of them determine the value asked for.
The missing input is the grain moisture at harvest.
The final a... | cannot be determined without the grain moisture at harvest | refusal | insufficient_input | text | null | n/a — refusal by construction | medium | by-construction | the question deliberately omits an input the calculation requires | null | false | true |
f_economics_0001177 | Variable costs are 1480 per hectare and the crop sells for 330 per tonne. What yield do you need just to cover variable costs? | Breakeven yield = variable cost / price = 1480 / 330 = 4.48 t/ha.
The final answer is $\boxed{4.48 t/ha}$. | 4.48 t/ha | economics | breakeven_yield | numeric | t/ha | gross margin budgeting | easy | closed-form | division identity | 0.02 | true | true |
Verified Agronomy, 17,199 rows
The arithmetic a working agronomist actually does: fertiliser rates, growing degree days, irrigation scheduling, seeding rates, grain moisture, sprayer calibration, forage budgeting. Every formula traced to a citable published source, and every number checked.
And 1,796 rows (10.4%) that cannot be answered at all, where the correct response is to say so and name the missing input.
Verify this card
Do not take my word for any of this. Here is how to check it yourself, from a terminal, with no account and no token.
The short way. One clone, one command, 22 checks.
git clone https://github.com/A1VARA5/verified-agronomy-17k
cd verified-agronomy-17k
python verify.py
Standard library only, no install step, Python 3.9 or newer. It prints a PASS or FAIL line per
check and re-derives every number on this page: the row count from the Hub's own parquet index,
that the licence is CC0 and the field is populated rather than empty, the 26 calculator
assertions, the 25,324 arithmetic expressions, the 1,796 refusal rows, a per-row citation check,
the weights and their exact byte size, the base model resolving, both Kaggle mirrors and the live
interface. Use --offline for the eight checks that need no network. The same script runs in
GitHub Actions on every push and once a day, which is what the badge above reports.
The long way, step by step. The expected results are in the comments.
# 1. Row count, read from the Hub's own parquet index instead of from this page.
# Expect "num_rows":17199
curl -s "https://datasets-server.huggingface.co/size?dataset=manifesta/verified-agronomy-17k" \
| grep -o '"num_rows":[0-9]*' | head -1
# 2. Do the formulas match the sources they cite? Expect "RESULT: 0 failed", 26 PASS lines.
B=https://huggingface.co/datasets/manifesta/verified-agronomy-17k/resolve/main/build
for f in agri_calc.py test_agri_calc.py check_agri.py; do curl -sLO $B/$f; done
python test_agri_calc.py
# 3. Does the arithmetic in the published solutions actually compute?
# Expect "25,324 expressions checked, 0 inconsistent" and "RESULT: 0 failed".
pip install pyarrow
curl -sLO https://huggingface.co/datasets/manifesta/verified-agronomy-17k/resolve/main/data/train-00000-of-00001.parquet
python check_agri.py train-00000-of-00001.parquet
If step 1 comes back with "the server is busier than usual", that is the Hub warming its cache. Run it again.
Step 2 runs every calculator against a worked example published by the source it cites, not against itself. 26 assertions, 0 failures. Anyone can open the referenced page and see the same number. This is the check that caught the P₂O₅ to P factor being 0.437 in the Cornell reference and 0.436 in my head, before that digit reached thousands of rows.
Step 3 re-parses the published parquet, pulls out every chained arithmetic expression written
into the worked solutions, evaluates each one independently and compares. It also re-checks that
the boxed answer equals the answer field on all 17,199 rows, that ids are unique, that no two
questions are identical, that all 1,796 refusal rows actually refuse and name their missing input,
and that no answerable row refuses. The download is 1,192,585 bytes and it finishes in seconds.
Two numbers I corrected on this card
The arithmetic count is 25,324, not 30,710. The 30,710 figure was real, but it was measured on the 18,900-row candidate pool before near-duplicate removal. The published file is 17,199 rows and gives 25,324. I would rather the card match what the script prints on the file you can download.
The calculator count is 26, not 25. test_agri_calc.py runs 26 assertions. Count the PASS
lines.
The weights, and the checks that go with them
The AutoScientist run has reported and the adapter is published:
- Weights on Hugging Face:
manifesta/adaption_agronomy_calc_problems - Weights on Kaggle: adaption_agro-b3684c92
# 4. Do the weights download, and are they the size the card claims?
# Expect HTTP 200 and content-length 119273568.
curl -sIL https://huggingface.co/manifesta/adaption_agronomy_calc_problems/resolve/main/adapter_model.safetensors \
| grep -iE "^HTTP|^content-length" | tail -2
# 5. Does the declared base model resolve? Expect 200, then 401 for the string the export wrote.
curl -s -o /dev/null -w "%{http_code}\n" https://huggingface.co/api/models/google/gemma-3-4b-it
curl -s -o /dev/null -w "%{http_code}\n" https://huggingface.co/api/models/togethercomputer/gemma-3-4b-it
# 6. Is the live interface up? Expect 200.
curl -s -o /dev/null -w "%{http_code}\n" https://manifestaagriculture.adaptionlabs.app/
The 401 in step 5 is the point, not a mistake. adapter_config.json shipped from the platform
export declaring togethercomputer/gemma-3-4b-it, a repository that does not exist, so nothing
could resolve a base model from the artifact. I corrected it to google/gemma-3-4b-it on
2026-08-13 and kept the original string verbatim in the same file as
original_base_model_name_or_path. No weight changed. Both values are readable in the published
config, so you can see what shipped first and decide for yourself whether the change was right.
The training result, and it is not a good one
The adapted model scored 46 against the base model's 54. It went backwards. That is a regression, not a tie and not a win, and it belongs at the top of this section rather than after the data-quality number.
| Base model | google/gemma-3-4b-it |
| Trained model | adaption_gemma_3_4b_it_agronomy_calc_problems_587e58e3 |
| Weights | Hugging Face and Kaggle |
| Rows generated, ingested, after expansion | 17,199 / 17,028 / about 28,069 |
| AutoScientist | 46 adapted against 54 base |
| Adaptive Data quality | 5.0 to 8.4, +68.0% relative, grade C to B, percentile 7.2 to 31.5 |
Two numbers, two different measurements. The +68.0% is the improvement in the data. The 46 against 54 is the model. The data got substantially better, the largest lift of the three entries in this series, and the model still got worse. Conflating those would be the easiest way to make this entry read better than it is.
The mechanism, out of trainer_state.json
I edited the recipe by hand instead of accepting the platform default: n_epochs from 1 to 3,
and learning_rate from 1e-5 to 1e-4, on a Tiny AutoScientist (sub-10B) base with domain and
diversity expansion enabled. The hypothesis was that the two earlier runs in this series had been
starved of optimizer steps.
| Optimizer steps | 66 |
| Epochs | 3 |
| Peak learning rate | 1.0e-4, reached at step 8 |
| Gradient norm range | 0.0795 to 2.6548 |
max_grad_norm |
2 |
| Steps that exceeded the threshold | 5 of 66, and they are steps 1, 2, 3, 4, 5 |
| Training loss | 1.6436 at step 1, 0.9727 at step 66 |
| Eval loss across 5 evaluations | 1.2917, 1.1639, 1.1239, 1.0999, 1.0919 |
Five clipped steps, all five in the warmup before the learning rate reached its peak, and from step 6 onward the gradient never went above 2 again. That is a mild start, not a broken run.
The loss curve after it is textbook: eval loss fell at every single evaluation, and the model still lost the head-to-head. That is the useful finding here. A monotonic eval curve is not evidence the model got better at the task, it is evidence it got better at reproducing the training distribution. This corpus has one deliberately rigid output contract, every row closing with the same boxed line, and three epochs at 1e-4 is enough to fit that shape hard. A preference judge scores the answer, not the shape.
The hypothesis was that more steps at a higher learning rate would help. It did not. I am recording that rather than reporting only the +68.0%, because a card that documents a hypothesis that failed is more useful than one that hides it. The full diagnosis, the per-step numbers and a cross-run comparison against the three other adapters are on the model card.
What this means for the dataset. The corpus is verified, cited and internally consistent, and none of that is in question here. What has not been shown is that it improves a 4B model under this recipe. If you train on it, start from one epoch at 1e-5 rather than what I ran, and score against the judge rather than against loss.
Why this exists
Agronomy questions look like general knowledge and are actually arithmetic with domain constants. Get the constant wrong and the answer is confidently, precisely wrong. A fertiliser rate that is out by the P₂O₅ to P factor is still a plausible-looking number.
So this dataset takes the same stance as a verified math corpus: a row earns its place by passing a check. Two checks, in fact. The formula is checked against its source, and the arithmetic is checked against itself.
The two-layer verification
Layer 1, the formulas match their published sources. build/test_agri_calc.py checks every
calculator against a worked example published by the source it cites. 26 checks, 0 failures, including:
| Check | Source | Published value | Ours |
|---|---|---|---|
| 50 kg N/ha via urea 46-0-0 | WSU extension | 108.7 | 108.7 |
| Moisture shrink factor to 15.5% | Pioneer | 1.1834 | 1.1834 |
| P from P₂O₅ | Cornell CSS412 | ×0.437 | ×0.437 |
| Corn GDD, 95°F/45°F day | NDSU NDAWN | 18.0 (both clamps) | 18.0 |
| ETc = ETo × Kc | FAO-56 | identity | identity |
Citing a formula is not evidence you implemented it correctly. Matching the source's own numbers is. This caught a real error: P₂O₅→P is 0.437, not the 0.436 that would otherwise have propagated into thousands of rows.
Layer 2, the arithmetic in every solution actually computes. build/check_agri.py re-parses
the finished dataset, extracts every arithmetic expression written into the worked solutions,
evaluates it independently, and compares. 25,324 expressions checked on the published file, 0 inconsistent. If a step
says 31.8 x 10 x 20 = 6,360, that has been confirmed.
Each row also passed a generator-time check appropriate to its family: an inverse recomputation, a nutrient balance, a dimensional round-trip, or a dry-matter conservation check.
What is in it
| Family | Rows | Share | Example |
|---|---|---|---|
| Fertiliser | 3,772 | 21.9% | product rate from a soil-test recommendation; N-P-K blends crediting DAP's nitrogen |
| Growing degree days | 1,980 | 11.5% | accumulation with the published caps and floors |
| Grain | 1,845 | 10.7% | moisture shrink, bushels at 15.5%, test-weight adjustment |
| Unanswerable | 1,796 | 10.4% | cannot be determined without <input> |
| Seeding | 1,603 | 9.3% | kg/ha from target population, TKW, germination, emergence |
| Irrigation | 1,577 | 9.2% | ETc = ETo × Kc → depth → volume → run time |
| Spraying | 1,428 | 8.3% | 600-rule calibration, product per tank |
| Economics | 1,240 | 7.2% | gross margin, breakeven yield, cost per kg nutrient |
| Livestock | 1,238 | 7.2% | intake, utilisation, grazing days |
| Units | 720 | 4.2% | kg/ha ↔ lb/ac, ha ↔ ac |
15,403 rows (89.6%) are exactly gradable, see exactly_gradable. Difficulty: 3,784 hard,
8,462 medium, 4,953 easy.
The refusal slice
Fertiliser rate needs a current soil test. Irrigation run time needs site-specific reference evapotranspiration. Seeding rate in kg/ha needs the thousand-kernel weight of the actual seed lot. Product rates come from the registered label. Manure timing is governed by local regulation.
These questions are unanswerable by construction, and many include real but irrelevant context (field size, previous crop, soil texture, growth stage) precisely to test whether plausible detail gets mistaken for the input the calculation requires:
"The block is 18 ha. How many days can I graze the home paddock?" →
cannot be determined without the current forage dry matter on offer
Field size is genuinely relevant to grazing days and still does not determine them without forage on offer. That is the trap, on purpose.
Output contract
Every row closes with one uniform line:
The final answer is $\boxed{ANSWER}$.
Unit inside the box. This is deliberate and makes the corpus machine-gradable without an LLM judge.
Verified on 100% of rows, and the boxed content matches the answer field on 100% of rows.
Loading
from datasets import load_dataset
ds = load_dataset("manifesta/verified-agronomy-17k", split="train")
print(ds[0]["question"], ds[0]["worked_solution"])
refusals = ds.filter(lambda r: r["family"] == "refusal") # 1,796
gradable = ds.filter(lambda r: r["exactly_gradable"]) # 15,403
| Field | Notes |
|---|---|
question |
The scenario. Carries the numeric inputs the verified answer depends on, so do not paraphrase. |
worked_solution |
Stepwise calculation ending in the boxed final answer |
answer |
The final answer alone |
family and subtask |
See the table above |
answer_type, answer_unit, numeric_tolerance |
numeric or text; tolerance is relative (0.02 = ±2%) |
exactly_gradable |
False for refusal rows (free text) |
citation |
The published source for that row's formula |
verified_by, verification_note |
Which check that row passed, and what it proved |
difficulty |
easy, medium, hard |
Provenance and licensing, CC0
There is one source, and it is a script in this repo. The same four scripts, byte identical, plus
the build manifest and the training state, are in the
GitHub repository, where verify.py SHA-256s
both copies and fails if they ever stop matching.
| Source | Rows | Licence | Link |
|---|---|---|---|
| Generated by my own engine | 17,199 | CC0-1.0 | build/gen_agri.py |
| Third-party text, scraped content or images | 0 | not applicable | |
| Total | 17,199 | CC0-1.0 |
No third-party corpus was read, so there is no attribution chain to inherit and no share-alike term to propagate. Released CC0-1.0, no restrictions.
The formulas are a different question from the text. Formulas are facts, and facts are not
copyrightable, but a fact still has to come from somewhere and be checkable. Every row names the
source its formula came from in the citation column. These are the counts:
| Citation | Rows | Source |
|---|---|---|
| WSU extension fertilizer calculations | 2,880 | |
| NDSU NDAWN | 1,980 | corn GDD, wheat GDD |
| Refusal by construction, no formula | 1,796 | not applicable |
| Standard extension seeding-rate arithmetic | 1,603 | target population, TKW, germination, emergence |
| FAO Irrigation and Drainage Paper 56 | 1,577 | FAO-56 |
| Sprayer calibration, 600-rule | 1,428 | L/ha = L/min x 600 / (km/h x m) |
| Standard forage budgeting | 1,238 | intake, utilisation, grazing days |
| Gross margin budgeting | 1,043 | yield x price minus cost |
| OSU bushels, test weights and calculations | 932 | fact sheet |
| Cornell CSS412 | 892 | fertilizer module |
| Purdue, grain test weight for corn | 805 | newsletter |
| SI and US customary conversion factors | 720 | definitional |
| Input costing | 197 | cost per kg of nutrient |
| Pioneer agronomy, moisture shrink | 108 | article |
| Total | 17,199 |
Count them from the shipped file rather than trusting the table:
from datasets import load_dataset
from collections import Counter
ds = load_dataset("manifesta/verified-agronomy-17k", split="train")
print(Counter(ds["citation"]))
build/test_agri_calc.py is the part that matters. It checks each calculator against a worked
example published by the source in that table, so the citation is not decoration.
Limitations, read these
- It is arithmetic, not agronomy judgement. This teaches a model to compute correctly and to refuse honestly. It does not teach what rate to recommend. That is a soil-test, variety, season and jurisdiction specific decision, which is exactly what the refusal slice says.
- 100% generated. Realistic and internally verified, but not drawn from field records. There is no observational data here.
- Constants are largely North-American and FAO conventions: 56 lb/bu corn, 15.5% market moisture, °F degree-day bases alongside metric rates. A model trained on this inherits those conventions.
- Kc values are FAO-56 mid-season single coefficients. Real scheduling uses stage-specific and often dual coefficients; this is the simplified standard case.
- The refusal share (10.4%) is a design choice. A model trained on this may skew toward refusal; downsample that family if you want a more answer-eager model.
- Verification proves the arithmetic, not the agronomy. Every number computes and every formula matches its source. Whether a given scenario is agronomically sensible was spot-checked by hand, not reviewed at 17,199-row scale.
- No public agronomy benchmark exists to decontaminate against, so there is no contamination
check to show. The set is internally exact- and near-deduplicated (MinHash 0.85, 1,701 removed).
Every one of those 1,701 removed row ids is listed in
evidence/neardup_removed_ids.json. I did not preserve the one-off script that ran that pass, so the id list is the audit record instead of the code. That is a real gap and it is better said than left out. - The trained model lost to its base, 46 to 54. The corpus has not been shown to improve a 4B model under the recipe I ran.
Citation
@misc{verified_agronomy_17k,
title = {Verified Agronomy: 17,199 source-cited agricultural calculations with a refusal slice},
author = {Aivaras Navardauskas},
year = {2026},
url = {https://huggingface.co/datasets/manifesta/verified-agronomy-17k}
}
Formula sources: FAO-56 · NDSU NDAWN · Cornell CSS412 · WSU extension · Purdue · Ohio State · Pioneer.
Build code, evidence files and 22 runnable checks: https://github.com/A1VARA5/verified-agronomy-17k Weights: https://huggingface.co/manifesta/adaption_agronomy_calc_problems Live interface: https://manifestaagriculture.adaptionlabs.app/
Built with Adaptive Data by Adaption.
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