AnaMariaF wrote:
Hi,
Can anyone explain how to solve this question?
thanks!
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.
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