Nitrogen

One bag of urea containing 46 percent nitrogen weighs 25 kg. How many bags have to be purchased for fertilizing a field of 41003 square meters if the nitrogen dose is 50.0 kg per hectare?

Result

n =  18

Solution:

S=41003 m2=41003/10000 ha=4.1003 ha  m=50⋅ S=50⋅ 4.1003=205.015 kg 0.46 m1=m m1=m/0.46=205.015/0.46≐4100392≐445.6848 kg  n=⌈m1/25⌉=⌈445.6848/25⌉=18S=41003 \ m^2=41003 / 10000 \ ha=4.1003 \ ha \ \\ \ \\ m=50 \cdot \ S=50 \cdot \ 4.1003=205.015 \ \text{kg} \ \\ 0.46 \ m_{1}=m \ \\ m_{1}=m /0.46=205.015 /0.46 \doteq \dfrac{ 41003 }{ 92 } \doteq 445.6848 \ \text{kg} \ \\ \ \\ n=\lceil m_{1}/25 \rceil=\lceil 445.6848/25 \rceil=18

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