Iowa State University Crops
Competition
Math exam - College Division
Instructions
Round final answers to one decimal place, unless otherwise stated.
Write your final answer in the box provided, remember to include the units. No credit will be given
for answers outside of the boxes.
1. Farmer Sam is trying to calculate his fertilizer needs. He is planning on producing 250 bushel per acre
corn this year. In order to do so, he plans to apply fertilizer at the removal rate. He assumes a removal
rate of 1.1 lbs. of N, 0.375 lbs. of P2O5, and 0.30 lbs. of K2O per bushel.
a. How much of each nutrient does he need to apply per acre at the removal rate for 250
bushels? (Answer in lbs of N, P2O5, and K2O)
250 bu
1A
1.1 lb N
1 bu
= 275 lb N/A
250 bu
1A
0.375 lb P2O5
1 bu
250 bu
1A
0.30 lb K2O
1 bu
Answer
275 lb N/A
93.8 lb P2O5/A
75 lb K2O/A
= 93.75 lb P2O5/A
=75 lb K2O/A
b. If he uses anhydrous ammonia (82-0-0), diammonium phosphate (18-46-0), and potash (0-060), how many pounds of each fertilizer does he need to fertilize one acre?
93.75 lb P2O5
1A
100 lb DAP
46 lb P2O5
275 lb-36.69lb N
1A
75 lb K2O
1A
= 203.8 lb DAP/A
100 lb AA
82 lb N
100 lb Potash
60 lb K2O

= 290.63 lb AA/A
203.8 lb DAP
1A
18 lb N
100 lb DAP
= 36.69 lb N/A
Answer
203.8 lb DAP/A
290.6 lb AA/A
125 lb Potash/A
= 125 lb Potash
c. If anhydrous ammonia is $850.00 per ton, diammonium phosphate is $575.00 per ton, and
potash is $600.00 per ton, how much will it cost him to fertilize one acre?
203.8 lb DAP
1 ton
2000 lb
$575.00
1 ton DAP
290.63 lb AA
1 ton
2000 lb
$850.00
1 ton AA
125 lb Potash
1 ton
2000 lb
= $58.59
$59.59 + $123.53 + $37.50 =
$219.61
= $123.52
$600.00
1 ton Potash
Answer
$219.61
= $37.50
d. If farmer Sam likes to cover his input purchases with grain sales, how many bushels of corn
per acre must he sell to cover all of his fertilizer costs if the current cash price of corn is $7.35 /
bushel?
$219.61 1 bu = 29.88 bu/A
1A
$7.35
Answer
29.9 bu/A
2. Farmer Brad is trying to calculate his seed order for the upcoming growing season. He farms 3,000
acres of prime Iowa farmland and uses a 50 / 50 corn and soybean rotation. In any given year, half of his
acres are in corn. Brad is trying to increase his corn yields, so he has been increasing his corn
population. This year, he is bumping his population up to 38,000 plants per acre.
a. If his seed test 95% germination (assume 100% purity), how much seed per acre should he
plant to account for the germination if he wants a final stand of 38,000?
38,000 plants 1 PLS
100% seed = 40,000 seed/A
1A
1 plant 95% PLS
Answer
40,000 seed/A
b. Brad also likes to account for other factors such as fungal diseases and insects. So he
estimates an additional 3% death loss. What population should he plant to achieve a final stand
of 38,000 plants per acre considering these additional factors? (Round up to the nearest seed)
40,000 seed ÷ 0.97 survival = 41,237.1 seed/A
Answer
41,238 seed/A
c. How many 80,000 kernel units does he need to purchase from his local seed dealer to meet
his planting needs? (Round up to the nearest unit)
3,000 acres total, 50/50 rotation: 1,500 acres corn
Answer
1500 A 41238 seed 1 unit
= 773.2 units
1A
80,000 seeds
774 units
d. After planting, farmer Brad is calculating stand counts in his field. His rows are on 24-inch
centers and he counts the plants in 100 foot of row. In his count he finds 175 plants. How many
plants per acre does he have?
175 plants
43560 ft2 = 38115 plants/A
(180 ft)(24/12 ft) 1 A
Answer
38,115 plants/A
3. Farmer Jeff loves to track his corn growth and development using growing degree days. He knows
that it takes approximately 90-120 growing degree days from planting to emergence for corn.
Date
May 4
May 5
May 6
May 7
May 8
May 9
May 10
May 11
May 12
May 13
May 14
May 15
Day, May day MAX
4
53
5
56
6
58
7
72
8
76
9
77
10
99
11
82
12
67
13
61
14
50
15
64
16
66
17
68
18
72
19
75
20
63
21
76
22
74
23
76
24
78
25
61
26
64
27
57
Daily Maximum
Temperature
53
56
58
72
76
77
99
82
67
61
50
64
day MIN adj-MAX**
43
53
48
56
46
58
47
72
50
76
57
77
59
86
64
82
60
67
54
61
42
50
42
64
36
66
36
68
39
72
48
75
59
63
59
76
57
74
55
76
55
78
49
61
44
64
54
57
Daily Minimum
Temperature
43
48
46
47
50
57
59
64
60
54
42
42
adj-MIN** GDD
50
1.5
50
3
50
4
50
11
50
13
57
17
59 22.5
64
23
60 13.5
54
7.5
50
0
50
7
50
8
50
9
50
11
50 12.5
59
11
59 17.5
57 15.5
55 15.5
55 16.5
50
5.5
50
7
54
5.5
Date
May 16
May 17
May 18
May 19
May 20
May 21
May 22
May 23
May 24
May 25
May 26
May 27
cumlative GDD
1.5
4.5
8.5
19.5
32.5
49.5
72
95
108.5
116
116
123
131
140
151
163.5
174.5
192
207.5
223
239.5
245
252
257.5
Daily Maximum
Daily Minimum
Temperature
Temperature
66
36
68
36
72
39
75
48
63
59
76
59
74
57
76
55
78
55
61
49
64
44
57
54
**remember to adjust for critical temperature range of corn
day of emergence (1st leaf)
2nd leaf
a. If he plants corn on May 4th, what day will his corn emerge if he assumes that it takes 100
growing degree days to emerge?
GGD = [Tmax + Tmin]/2 – Tbase
Answer
Tbase = 50 for corn
May 12th
86  Tmax  Tmin  50
b. If a corn plant puts on a new leaf every 84 growing degree days after emergence, on what day
will the second leaf appear?
Answer
May 21st
4. Farmer Frank has a center pivot irrigation system with a diameter of 2,640 feet.
a. How many acres covered by the irrigation system?
Area = r2 = (2640/2 ft)2 1 A
= 125.66 A
43560 ft2
Answer
125.7 A
b. How many total gallons of water is applied to the field if he applies 2 inches of water over the
whole area? (Round to the nearest 10 thousand gallons)
125.66 A 43560 ft2 (1 ft2)(2/12 ft H2O) 1 gl
= 6,820,000 gl
2
3
1A
1 ft
.1337 ft
Or
(27,154 gl/A/in)(125.66A)(2in) = 6,820,000 gl
Answer
6,820,000 gl
5. Farmer John harvested a quarter section of corn which yielded a bumper crop of 250 bushels per
acre. The corn had a fairly high moisture content of 25%. How many total bushels of corn has he
harvested, adjusted to 15.5% moisture, from the whole quarter section?
¼ section 1 mi2
640 A 250 bu corn = 40,000 bu corn
1 section 1 mi2 1 A
(40,000 bu corn)(25% H2O) = 10,000 bu H2O
Answer
35,503.0 bu
40,000 bu corn – 10,000 bu H2O = 30,000 bu dry corn
30,000 bu dry corn = X bu corn @15.5%  X = 35,503.0 bu corn @15.5%
100%-15.5%
100%
6. Farmer Meaghan was calibrating her sprayer. She collected the following information when
calibrating her field sprayer:
Measured individual nozzle delivery rate = 28 fluid ounces per 30 seconds
Speed of travel = 6 mph
Nozzles = 54
Pressure = 40 psi
Tank volume = 300 gallons
Boom width = 90 feet = (90ft)(12in/ft) = 1080 in
a. Calculate the spray volume delivery rate in gallons per acre.
GPM = 28 fl oz 60 sec 1 gl
= .4375 gl/min
30 sec 1 min 128 fl oz
Nozzle width = W = 1080 in ÷ 54 = 20 in
Answer
21.7 gl/A
GPA=(GPM x mph x W)/5940
GPA = (.4375 gl/min)(6 mph)(20 in)/5940 = 21.66 gl/A
b. Traveling at 6 mph, what is the maximum number of acres can she spray in one hour (not
accounting for turning, etc.)?
1 hr 6 mi 5280 ft (1 ft)(90 ft) 1 A
= 65.45 A
1 hr 1 mi
1 ft
43560 ft2
Answer
65.5 A
c. If she applies Trifluralin at 1.8 pounds active ingredient per hectare using Treflan EC, how
many gallons of product is required to mix one full tank (Treflan EC contains 480 grams active
ingredient per liter of product)?
300 gl 1 A
1 hA
1.8 lb a.i. 454 g a.i. 1 L Treflan 1 gl
= 2.52 gl Treflan/tank
1 tank 21.66 gl 2.471 A 1 hA
1 lb a.i.
480 g a.i.
3.78 L
Answer
2.5 gl Treflan/tank
7. A cylinder water tank is 12 feet tall and has a diameter of 4 feet.
a. How many cubic feet is the tank’s capacity?
Volume = h x r2 = (12ft)(4/2)2 = 150.80 ft3
Answer
150.80 ft3
b. How many gallons will it take to fill the tank to half capacity?
(½)150.80 ft3 1 gl
= 563.94 gl
.1337 ft3
Answer
563.9 gl
8 [Tiebreaker]. The tiebreaker will only be scored in the event of a tie between competitors in the top
five rankings. "Extra credit" will not be rewarded. Do not solve the tiebreaker problem unless you
have completed the entire exam.
Agronomist Erik has developed a new product for insect control. Below are information for Erik’s
product and a common generic.
Name
Erik
Generic
Active ingredient (a.i.) 0.2 ounces a.i. / fluid ounce 180 g a.i. / Liter
Product weight
9 pounds / gal
2.3 pounds / Liter
Product cost
$4.25 / quart
$0.05 / mL
a. If a farmer applies the active ingredient at 3 pounds per hectare using Erik’s product, how
many gallons of product is required to spray 10 acres?
Answer
7.59 gl
b. If a farmer applies the active ingredient at 3 pounds per hectare using the generic product,
how many gallons of product is required to spray 10 acres?
Answer
8.10 gl
c. Which product has a lower cost per unit of active ingredient?
$0.66/oz Erik
$7.88/oz Generic
Answer
Erik's