Question

A fertilizer has the analysis of 12-24-20. How many pounds of N, P, and K are there in the 3000 lbs. of fertilizer?

Answer 1

Answer:

$W_N=642.857ibs$

$W_P=1285.714ibs$

$W_K=1071.43ibs$

Step-by-step explanation:

From the question we are told that:

Ratio of NPK $R=12-24-20.$

Weight $W=3000 lbs$

Generally the sum of the ratio is given as

 $\sum R=12+24+20$

 $\sum R=56$

Generally the Weight of N is mathematically given by

 $W_N=\frac{12}{56}*3000$

 $W_N=642.857ibs$

Generally the Weight of P is mathematically given by

 $W_P=\frac{24}{56}*3000$

 $W_P=1285.714ibs$

Generally the Weight of K is mathematically given by

 $W_K=\frac{20}{56}*3000$

  $W_K=1071.43ibs$