Question

What amount of Urea, SSP, and MOP is required for application in the wheat crop at the rate of 120, 64, 30 kg N, P₂O₅ and K₂O ha⁻¹?

• A. 261; 400 and 50 kg Urea, SSP and MOP
• B. 130; 200 and 50 kg Urea, SSP and MOP
• C. 261; 200 and 25 kg Urea, SSP and MOP
• D. 261; 100 and 25 kg Urea, SSP and MOP

Hint:

Fertilizer calculation:
- Fertilizer = Nutrient required Nutrient percentage × 100.
- Urea (46% N), SSP (16% P₂O₅), MOP (60% K₂O).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question tests the calculation of fertilizer doses based on nutrient content.

Step 2: Key Formula or Approach:
Amount of fertilizer = (Nutrient required Nutrient content of fertilizer) × 100.
Urea: 46% N.
SSP: 16% P₂O₅.
MOP: 60% K₂O.

Step 3: Detailed Explanation:
N required = 120 kg ha⁻¹.
Urea required = 120 0.46 = 260.87 ≈ 261 kg.
P₂O₅ required = 64 kg ha⁻¹.
SSP required = 64 0.16 = 400 kg.
K₂O required = 30 kg ha⁻¹.
MOP required = 30 0.60 = 50 kg.
Thus, the amounts are 261 kg Urea, 400 kg SSP, and 50 kg MOP.
Final Answer:
Thus, the correct answer is 261; 400 and 50 kg, which corresponds to option (A).
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