1,000 kg of carbonaceous materials with a C:N ratio of 12:1 are applied to soil. 25% of the 400 kg of carbon in the materials is assimilated by microorganisms, reducing the C:N ratio to 8:1. This leaves 20.8 kg of nitrogen available for crops. An additional 39.2 kg of nitrogen is needed. Applying 85.2 kg of urea fertilizer which is 46% nitrogen would provide this amount. Fertilizer A, with 15% nitrogen content and costing RM8/kg of nitrogen, is a better buy than Fertilizer B at RM11.25/kg of nitrogen. One liter of liquid fertilizer containing 20

1. 1,000 kg of carbonaceous materials with a C:N ratio of 12:1 are applied to a
soilcontaining mixed population of microorganisms. The amount of C in
carbonaceous materials is 40%, 2 5% of C in th e sub st ra te is assimila t e d b y
micro o rg an isms, an d t he C: N rat io of microorganisms is 8:1.
a)Calculate the amount of N available for crop use.
b)If the crop requires 60 kg N, how much urea must be applied?
(Hint: urea contains 46% N)
(10 marks)
QUESTION 3
a) NPK content of fertilizer A is 15 - 15 - 15 and sells for RM60 per 50
kg bag; whereas NPK content of fertilizer B is 10 - 8 - 8 and costs
RM22.50 per 20 kg bag. Which is the best buy on the basis of
nitrogen?
(7 marks)
b) A liter of liquid fertilizer weighs 1.5 kg. NPK analysis of the
fertilizer is 20 - 4 -10. How much N, P, and K are in one liter?
(3 marks)
Amt of C in carbonaceous materials
0 . 4 x 1 0 0 0 kg = 400 kg
Amt of N in carbonaceous materials
400 kg - 12 = 33.3 kg C:N ratio 12:1)
Amt of C assimilated by microbes
0.25 x 400 kg = 100 kg
Amt of N in carbonaceous materials
100 kg ÷ 8 = 12.5 kg (C:N ratio 8:1)
Amt of available N for crop use
3.3 kg – 12.5 kg = 20.8 kg
Amt of additional N required by crop
60 k g - 2 0 . 8 kg = 39.2 kg
Amt of urea to be applied
(1 kg urea= 0.46 kg N) x 39.2 kg N 85.2 kg

2. Fertilizer A Fertilizer B
N content 0.15 x 50 kg 0.10 x 20 kg
= 7.5 kg = 2 kg
Price of N RM 60 ÷ 7.5 kg RM
22.50 ÷ 2
=RM 8 / kg =RM 11.25 / kg
Fertilizer A is the best buy.
Amt of NPK in liquid fertilizer
0.20 x 1.5 kg = 0.3 kg N
0.04 x 1.5 kg = 0.06 kg P
0.10 x 1.5 kg = 0.15 kg K

3. SCHEMA
NO ANSWERS MARKS
la Amt of C in carbonaceous materials 1½
0 . 4 x 1 0 0 0 kg = 400 kg
Amt of N in carbonaceous materials
400 kg - 12 = 33.3 kg C:N ratio 12:1) 1½
Amt of C assimilated by microbes
0.25 x 400 kg = 100 kg 1½
Amt of N in carbonaceous materials
100 kg ÷ 8 = 12.5 kg (C:N ratio 8:1) 1½
Amt of available N for crop use 1
3.3 kg – 12.5 kg = 20.8 kg
b Amt of additional N required by crop
60 k g - 2 0 . 8 kg = 39.2 kg 1
Amt of urea to be applied
(1 kg urea= 0.46 kg N) x 39.2 kg N 85.2 kg 2

4. Fertilizer A Fertilizer B 3
N content 0.15 x 3
50 kg 0.10 x 1
20 kg 1
= 7.5 kg = 2 kg
Price of N RM 60
÷ 7.5 kg RM
3a 22.50 ÷ 2
=RM 8 / kg =RM
11.25 / kg
Fertilizer A is the best buy.
Amt of NPK in liquid fertilizer
0.20 x 1.5 kg = 0.3 kg N
b
0.04 x 1.5 kg = 0.06 kg P 1
0.10 x 1.5 kg = 0.15 kg K 1