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

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Manager
Joined: 01 Jan 2017
Posts: 50
WE: General Management (Consulting)
Many fertilizers are given an NPK rating based on the percentages of t  [#permalink]

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18 Oct 2019, 04:10
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45% (medium)

Question Stats:

71% (02:55) correct 29% (03:02) wrong based on 49 sessions

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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

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Joined: 22 Feb 2014
Posts: 46
Re: Many fertilizers are given an NPK rating based on the percentages of t  [#permalink]

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18 Oct 2019, 09:32
C

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Joined: 13 Jan 2019
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Re: Many fertilizers are given an NPK rating based on the percentages of t  [#permalink]

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18 Oct 2019, 15:02
Hi,

Can anyone explain how to solve this question?

thanks!
Manager
Joined: 01 Jan 2017
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19 Oct 2019, 00:33
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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, $$P+K=11+12=23$$.

Hope it helps.
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Re: Many fertilizers are given an NPK rating based on the percentages of t  [#permalink]

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19 Oct 2019, 00:35
1
100 gms of Fertilizer A contains 20 gms (N), 10 gms (P) and 10 gms (K)
100 gms of Fertilizer B contains 50 gms (N), 13 gms (P) and 16 gms (K)
Let's assume 100x gms of A need to be mixed with 100y gms of B so that the resultant fertilizer (let's call it C) contains 30% of (N). Then the quantity of (N) in C will be (20x+50y) which should be equal to 30% (3/10th) of (100x+100y). So:
20x+50y = (3/10)(100x+100y).....> x=2y. So A and B need to be mixed in the ratio of 2:1. Let's mix 200 gms of A and 100 gms of B to give us 300 gms of C which will contain (20*2+50*1)=90 gms, (10*2+13*1)=33gms and (10*2+16*1)=36 gms of (N),(P) and (K) respectively.
The respective percentages of (N), (P) and (K) will thus be 90/3=30, 33/3=11 and 36/3=12.
Sum of percentages of (P) and (K) will be (11+12)=23. ANS:C
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Posts: 8164
Re: Many fertilizers are given an NPK rating based on the percentages of t  [#permalink]

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19 Oct 2019, 00:43
DealMakerOne wrote:
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

A very simple method would be

A - 20-10-10, so 20% nitrogen
B - 50-13-16, so 50% nitrogen
Final mix has 30% nitrogen

Therefore, proportion of A by weighted average method = $$\frac{50-30}{50-20}=\frac{2}{3}$$ and B will be $$1-\frac{2}{3}=\frac{1}{3}$$

Thus, sum of the percentages of phosphorous and potassium in the mixture is $$(10+10)*\frac{2}{3}+(13+16)*\frac{1}{3}=\frac{40}{3}+\frac{29}{3}=\frac{69}{3}=23$$

C
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Re: Many fertilizers are given an NPK rating based on the percentages of t   [#permalink] 19 Oct 2019, 00:43
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