Question

The amount of mixed fertilizer grade (12:32:16), urea and MoP required to supply 100 kg, Nitrogen, 50 kg Phosphorous and 50 kg Potash is _________.

• A. 50 kg Mixed Fertilizer (12:32:16), 90 kg Urea and 22 kg MoP
• B. 78 kg Mixed Fertilizer (12:32:16), 60 kg Urea and 32 kg MoP
• C. 156 kg Mixed Fertilizer (12:32:16), 177 kg Urea and 42 kg MoP
• D. 204 kg Mixed Fertilizer (12:32:16), 240 kg Urea and 54 kg MoP

Hint:

Always start calculating with the complex/mixed fertilizer to satisfy the most restricted nutrient (usually P or K first), then adjust the remaining deficits using single/straight fertilizers like Urea and MoP.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Fertilizer calculations involve determining the exact quantities of different single or mixed fertilizers required to meet a specific recommended dose of primary nutrients: Nitrogen (N), Phosphorus (\(\text{P}_2\text{O}_5\)), and Potassium (\(\text{K}_2\text{O}\)).
The calculations must account for nutrients supplied simultaneously by multi-nutrient complex fertilizers.
Key Formula or Approach:
1. Find the mixed fertilizer requirement first because it is the sole source of Phosphorus.
\[ \text{Mixed Fertilizer (kg)} = \frac{\text{Target Nutrient Weight (kg)}}{\text{Nutrient \% in Fertilizer}} \times 100 \]
2. Calculate the amounts of Nitrogen and Potassium supplied by this quantity of mixed fertilizer.
3. Subtract these supplied amounts from the total targets to find the remaining nutrient requirements.
4. Calculate the required single fertilizers (Urea for N, MoP for K) to fulfill the remaining deficits.

Step 2: Detailed Explanation:
Let us solve the problem systematically:
- Target Nutrient Dose: \(\text{N} = 100\text{ kg}\), \(\text{P}_2\text{O}_5 = 50\text{ kg}\), \(\text{K}_2\text{O} = 50\text{ kg}\).
- Available Fertilizers: Mixed Fertilizer (12:32:16), Urea (46% N), Muriate of Potash (60% \(\text{K}_2\text{O}\)).
Step 3.1: Calculate Mixed Fertilizer (12:32:16) required for Phosphorus:
Since the mixed fertilizer is the only source of \(\text{P}_2\text{O}_5\), and contains 32% \(\text{P}_2\text{O}_5\):
\[ \text{Mixed Fertilizer} = \frac{50}{32} \times 100 = 156.25\text{ kg} \approx 156\text{ kg} \]
Step 3.2: Calculate Nitrogen supplied and remaining deficit:
The 156.25 kg of mixed fertilizer contains 12% N:
\[ \text{N supplied} = 156.25 \times \frac{12}{100} = 18.75\text{ kg} \]
Remaining N needed from Urea:
\[ \text{N deficit} = 100 - 18.75 = 81.25\text{ kg} \]
Since Urea contains 46% N:
\[ \text{Urea required} = \frac{81.25}{46} \times 100 \approx 176.63\text{ kg} \approx 177\text{ kg} \]
Step 3.3: Calculate Potash supplied and remaining deficit:
The 156.25 kg of mixed fertilizer contains 16% \(\text{K}_2\text{O}\):
\[ \text{K}_2\text{O supplied} = 156.25 \times \frac{16}{100} = 25\text{ kg} \]
Remaining \(\text{K}_2\text{O}\) needed from MoP:
\[ \text{K}_2\text{O deficit} = 50 - 25 = 25\text{ kg} \]
Since MoP contains 60% \(\text{K}_2\text{O}\):
\[ \text{MoP required} = \frac{25}{60} \times 100 \approx 41.67\text{ kg} \approx 42\text{ kg} \]
Thus, the required amounts are: 156 kg of mixed fertilizer, 177 kg of Urea, and 42 kg of MoP.

Step 3: Final Answer:
The correct combination is 156 kg Mixed Fertilizer, 177 kg Urea, and 42 kg MoP (Option C).