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20% nitrogen in 12 pounds of fertilizer is 12/5.
Let us consider the new amount of 10% nitrogen to be x. so the total amount of nitrogen content will be 12/5 + x/10. And the total amount of fertilizer will be 12+x.
\(\frac{(2.4+ 0.x)}{(12+x)}\) =.18
X=3
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Bunuel
How many pounds of fertilizer that is 10 percent nitrogen must be added to 12 pounds of fertilizer that is 20 percent nitrogen so that the resulting mixture is 18 percent nitrogen?

(A) 3
(B) 6
(C) 12
(D) 24
(E) 48

Using weighted averages:

We need to mix 10% nitrogen fertiliser with 20% to get 18% nitrogen.

w1/w2 = (A2 - Aavg)/(Aavg - A1) = (20 - 18)/(18 - 10) = 1/4

For every 1 part of 10% nitrogen, we need 4 parts of 20% nitrogen.
Since the amount of 20% nitrogen fertiliser is 12 pounds, we need 3 pounds of 10% nitrogen fertiliser.

Answer (A)
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