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# How many pounds of fertilizer that is 10 percent nitrogen must be adde

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Joined: 02 Sep 2009
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How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink]

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31 Oct 2017, 00:05
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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
[Reveal] Spoiler: OA

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Re: How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink]

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31 Oct 2017, 01:43
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Bunuel wrote:
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

10-------------20

--------18-------

2---------------8

Hence the ratios in which the mixtures are to be added are 1:4

Since we are adding 12 pounds of the 20 percent nitrogen mixture,
we will need 3 pounds of the 10 percent nitrogen mixture, so that
the resulting mixture has 18 percent nitrogen.

Therefore, we need 3 pounds of fertilizer mixture having 10 percent nitrogen(Option A)
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Joined: 30 Nov 2016
Posts: 37
Location: India
Concentration: Finance, Strategy
How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink]

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31 Oct 2017, 04:16
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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Re: How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink]

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31 Oct 2017, 04:44
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Bunuel wrote:
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.

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Re: How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink]

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01 Nov 2017, 17:05
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Bunuel wrote:
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

We add x pounds of fertilizer that is 10% nitrogen to 12 pounds of fertilizer that is 20% nitrogen, and the result is (x + 12) pounds of fertilizer that is 18% nitrogen. We can express this in the following equation:

0.1x + 0.2(12) = 0.18(x + 12)

10x + 20(12) = 18(x + 12)

10x + 240 = 18x + 216

24 = 8x

x = 3

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Re: How many pounds of fertilizer that is 10 percent nitrogen must be adde   [#permalink] 01 Nov 2017, 17:05
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