Many fertilizers are given an NPK rating based on the percentages of t

Many fertilizers are given an NPK rating based on the percentages of the three major plant nutrients they contain: nitrogen (N), phosphorous (P), and potassium (K). For example, a fertilizer with an NPK rating of 5-7-3 contains 5 percent nitrogen, 7 percent phosphorous, and 3 percent potassium. A farmer has two fertilizers: fertilizer A, with an NPK rating of 20-10-10, and fertilizer B, with an NPK rating of 50-13-16. If the farmer mixes the two fertilizers such that the mixture contains 30 percent nitrogen, what is the sum of the percentages of phosphorous and potassium in the mixture?

(A) 14.5
(B) 18
(C) 23
(D) 28
(E) 56

Ratio of nutrients in fertilizer A: \(N:P:K=20:10:10\)
Ratio of nutrients in fertilizer B: \(N:P:K=50:13:16\)

After the farmer mixed two fertilizers, it is known that the resulted (or mixed) fertilizer contains 30% of nitrogen (N). So we are asked about the sum of the percentages of the rest two nutrients (P and K). In other words, first, we need to find a ratio of nutrients in the new (mixed) fertilizer.

Let \(X\) be the weight of fertilizer A and \(Y\) be the weight of fertilizer B. Using the mixture rule, we know that the new fertilizer contains 30% of nitrogen (N):

\(\frac{N}{100}\) = \(\frac{(0.2*x+0.5*y)}{(x + y)}\)

\(30X+30Y=20X+50Y\)

\(10X=20Y\)

\(X=2Y\)

So now we know that the ratio of weight of fertilizer A to the weight of fertilizer B is 1 to 2. After this, we can easily apply the same technique to find the % of P and K.

\(\frac{P}{100}\) = \(\frac{(0.1*2+0.13*1)}{(2 + 1)}\)

\(20+13=3P\)

\(P=11\)

\(\frac{K}{100}\) = \(\frac{(0.1*2+0.16*1)}{(2 + 1)}\)

\(20+16=3K\)

\(K=12\)

Finally, \(P+K=11+12=23\).

Answer: C.

Hope it helps.