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

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Re: Many fertilizers are given an NPK rating based on the percentages of t [#permalink]
C

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Re: Many fertilizers are given an NPK rating based on the percentages of t [#permalink]
Hi,

Can anyone explain how to solve this question?

thanks!
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Re: Many fertilizers are given an NPK rating based on the percentages of t [#permalink]
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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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Re: Many fertilizers are given an NPK rating based on the percentages of t [#permalink]
DealMakerOne wrote:
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.

How did you go from one equation to the next (highlighted in red)?
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Re: Many fertilizers are given an NPK rating based on the percentages of t [#permalink]
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I did it as followed;

Fertilizer A + B = 70 / 23 / 26
Divided by 10, to get easier to the 30 percent nitrogen
7 / 2.3 / 2.6 times four = 28 / 9.2 / 10.4

Answer B and C are possible at this moment

Since the percentage of nitrogen has to be 30 it comes closer to a total percentage of 23
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Re: Many fertilizers are given an NPK rating based on the percentages of t [#permalink]
chetan2u wrote:
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

Why isn't the proportion 1/5 of A if we are taking 20%?
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Re: Many fertilizers are given an NPK rating based on the percentages of t [#permalink]
The 30 percent Nitrogen content in the overall mixture rests twice as close to Fertilizer A's Nitrogen percentage as it does to Mixture B's, so all of the respective percentages will land twice as close to Mixture A's percentages as Mixture B's. This means 11 percent for phosphorous and 12 percent for Potassium. 11 plus 12 is 23
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Re: Many fertilizers are given an NPK rating based on the percentages of t [#permalink]
chetan2u wrote:
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
Can you please explain the (50-30)/(50-20)? I didn't understand what 50, 20 and 30 mean here? Thanks
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Many fertilizers are given an NPK rating based on the percentages of t [#permalink]
Logic-based solution with few calculations:

The percentages of N in fertilizers A and B are 20% and 50% respectively.

The average of these two is 35%.

Since we know that the new mixture contains 30%, this means that there is more of fertilizer A than B in the mixture.

The percentages of P and K combined in fertilizers A and B are 20% and 29% respectively. Since there is more of fertilizer A, the total percentage of the two in the mixture must lie a little above 20 (cannot go above 29 or below 20). Only option C remains.­­
Many fertilizers are given an NPK rating based on the percentages of t [#permalink]
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