Firstly since 20% of Na

_{3}PO

_{4} mixture consists of contaminants, only 8g of Na

_{3}PO

_{4} is available. 164 g of it gives 31g P and thus 8g of it should give 1.512g P

Now 132 g of (NH

_{4})

_{2}SO

_{4} gives 28g N and hence 26.4 g of it should give 5.6g N.

Finally 101g of KNO

_{3} give 39 g K and hence 20g of it must give 7.72g of K.

So now to summarize what is available to us

P

1.512g

N

5.6g

K

7.72g

Thus by trial and error, we can observe that P is the limiting factor. Thus 1.512g P needs 1.5 times N and 2 time the K i.e. 2.268g N and 3.024g K.

These can be obtained from 10.69g of (NH

_{4})

_{2}SO

_{4}, 10g of the Na

_{3}PO

_{4} mixture given and 7.83g of KNO

_{3}.

Thus the net weight of fertilizer is

28.52g.