Dataset Viewer
Auto-converted to Parquet Duplicate
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
End of preview. Expand in Data Studio

Hugging Face Kaggle rows formulas arithmetic refusal license

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 — with 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.

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→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. 25/25 match, 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. 30,710 expressions checked, 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 — do not paraphrase.
worked_solution Stepwise calculation ending in the boxed final answer
answer The final answer alone
family / subtask See the table above
answer_type · answer_unit · numeric_tolerance numeric/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

Licensing — CC0

Every row was generated for this dataset. No third-party text, no scraped content, no images. The formulas are facts from public extension and FAO literature (facts are not copyrightable), and the sources are cited so you can check them. Released CC0-1.0 — no restrictions.

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 / 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).

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.

Built with Adaptive Data by Adaption.

Downloads last month
-