How many pounds of 10-20-10 should be applied to an acre if the fertilizer recommendation is 26 lbs of P?

How many pounds of 10-20-10 should be applied to an acre if the fertilizer recommendation is 26 lbs of P?

• A. 300
• B. 400
• C. 500
• D. 600

Answer: (A)

To determine how many pounds of the fertilizer formulation 10-20-10 should be applied to provide 26 pounds of phosphorus (P), it is essential to analyze the nutrient content of the fertilizer. The numbers in the fertilizer formulation represent the percentage by weight of nitrogen (N), phosphorus (P), and potassium (K) respectively. This means that in 10-20-10 fertilizer, there is: - 10% nitrogen - 20% phosphorus - 10% potassium To calculate how much of the 10-20-10 fertilizer is needed to supply 26 pounds of phosphorus, we first determine how much phosphorus is contained in one pound of the fertilizer. Since the fertilizer is 20% phosphorus, it means there are 0.2 pounds of phosphorus in a 1-pound application of 10-20-10. To find out how many pounds of the fertilizer are needed to yield 26 pounds of phosphorus, you would set up the following equation: Let \( x \) be the pounds of 10-20-10 needed: \[ 0.2 \times x = 26 \] To solve for \( x \): \[ x = \frac{26}{0.2} =

To determine how many pounds of the fertilizer formulation 10-20-10 should be applied to provide 26 pounds of phosphorus (P), it is essential to analyze the nutrient content of the fertilizer.

The numbers in the fertilizer formulation represent the percentage by weight of nitrogen (N), phosphorus (P), and potassium (K) respectively. This means that in 10-20-10 fertilizer, there is:

• 10% nitrogen

• 20% phosphorus

• 10% potassium

To calculate how much of the 10-20-10 fertilizer is needed to supply 26 pounds of phosphorus, we first determine how much phosphorus is contained in one pound of the fertilizer. Since the fertilizer is 20% phosphorus, it means there are 0.2 pounds of phosphorus in a 1-pound application of 10-20-10.

To find out how many pounds of the fertilizer are needed to yield 26 pounds of phosphorus, you would set up the following equation:

Let ( x ) be the pounds of 10-20-10 needed:

[ 0.2 \times x = 26 ]

To solve for ( x ):

[ x = \frac{26}{0.2} =