Question

The amount of mixed fertilizers grade (12:32:16), urea and MOP required to supply 100 kg N, 50 kg P\(_2\)O\(_5\) and 50 kg K\(_2\)O ha in rice:

• A. 50 kg (fertilizers grade): 90 kg (Urea): 22 kg (MOP)
• B. 78 kg (fertilizers grade): 60 kg (Urea): 32 kg (MOP)
• C. 156 kg (fertilizers grade): 177 kg (Urea): 42 kg (MOP)
• D. 204 kg (fertilizers grade): 240 kg (Urea): 54 kg (MOP)

Hint:

Fertilizer grade: N:P\(_2\)O\(_5\):K\(_2\)O percentages.
Urea: 46% N.
MOP: 60% K\(_2\)O.
Always balance the nutrients.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to calculate the amount of different fertilizers.
We have a mixed fertilizer grade and need to supplement with urea and MOP.

Step 2: Key Formula or Approach:
Calculate nutrient content from mixed fertilizer.
Calculate the remaining nutrients to be supplied by urea and MOP.

Step 3: Detailed Explanation:
Mixed fertilizer grade: 12:32:16 means 12% N, 32% P\(_2\)O\(_5\), 16% K\(_2\)O.
Let X = amount of mixed fertilizer.
P\(_2\)O\(_5\) from mixed fertilizer = 0.32X = 50 kg.
X = 50 0.32 = 156.25 kg.
N from mixed fertilizer = 0.12 \(\times\) 156.25 = 18.75 kg.
K\(_2\)O from mixed fertilizer = 0.16 \(\times\) 156.25 = 25 kg.
Remaining N = 100 - 18.75 = 81.25 kg.
Remaining K\(_2\)O = 50 - 25 = 25 kg.
Urea has 46% N.
Urea required = 81.25 0.46 = 176.63 kg (approx 177 kg).
MOP has 60% K\(_2\)O.
MOP required = 25 0.60 = 41.67 kg (approx 42 kg).
Thus, 156 kg mixed fertilizer, 177 kg urea, 42 kg MOP.

Step 4: Final Answer:
The correct amounts are 156 kg fertilizer grade, 177 kg Urea, 42 kg MOP.
Hence, the correct option is (C).