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Three pounds of 05 grass seed contain 5 percent herbicide. A different

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Three pounds of 05 grass seed contain 5 percent herbicide. A different  [#permalink]

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New post Updated on: 19 Oct 2014, 11:16
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Three pounds of 05 grass seed contain 5 percent herbicide. A different type of grass seed, 20, which contains 20 percent herbicide, will be mixed with three pounds of 05 grass seed. How much grass seed of type 20 should be added to the three pounds of 05 grass seed so that the mixture contains 15 percent herbicide?

A. 3
B. 3.75
C. 4.5
D. 6
E. 9

Originally posted by zadi on 19 Oct 2014, 11:07.
Last edited by Bunuel on 19 Oct 2014, 11:16, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Three pounds of 05 grass seed contain 5 percent herbicide. A different  [#permalink]

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New post 19 Oct 2014, 12:35
3
zadi wrote:
Three pounds of 05 grass seed contain 5 percent herbicide. A different type of grass seed, 20, which contains 20 percent herbicide, will be mixed with three pounds of 05 grass seed. How much grass seed of type 20 should be added to the three pounds of 05 grass seed so that the mixture contains 15 percent herbicide?

A. 3
B. 3.75
C. 4.5
D. 6
E. 9


05 grass seed contains 5% herbicide and its amount is 3 pound
20 grass seed contains 20% herbicide and its amount is x

when these two types of grass seeds are mixed, their average becomes 15%

thus we have
3(5)+x(20)/(x+3) = 15

15+20x=15x +45
5x=30
or x=6
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Re: Three pounds of 05 grass seed contain 5 percent herbicide. A different  [#permalink]

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New post 19 Oct 2014, 20:24
2
zadi wrote:
Three pounds of 05 grass seed contain 5 percent herbicide. A different type of grass seed, 20, which contains 20 percent herbicide, will be mixed with three pounds of 05 grass seed. How much grass seed of type 20 should be added to the three pounds of 05 grass seed so that the mixture contains 15 percent herbicide?

A. 3
B. 3.75
C. 4.5
D. 6
E. 9


Using weighted averages:

5% herbicide seed mixed with 20% herbicide seed to give 15% herbicide seed.

w1/w2 = (20 - 15)/(15 - 5) = 1/2

So 1 part of 5% was mixed with 2 parts of 20%.
Since 5% herbicide seed was 3 pounds, 20% herbicide seed must have been 6 pounds.

Answer (D)
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Three pounds of 05 grass seed contain 5 percent herbicide. A different  [#permalink]

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New post 27 Oct 2017, 12:31
1
zadi wrote:
Three pounds of 05 grass seed contain 5 percent herbicide. A different type of grass seed, 20, which contains 20 percent herbicide, will be mixed with three pounds of 05 grass seed. How much grass seed of type 20 should be added to the three pounds of 05 grass seed so that the mixture contains 15 percent herbicide?

A. 3
B. 3.75
C. 4.5
D. 6
E. 9

Weighted average as well, a slightly different version.*

The formula below makes it easier for me to see two mixtures with different concentrations and/or volumes are added to create a third, resultant, mixture with its concentration and volume.

A = mixture with .05 herbicide
x = mixture with .20 herbicide

We have 3 pounds of A.

.05(3) + .20(x) = .15(3 + x)
5(3) + 20x = 15(3 + x)
15 + 20x = 45 + 15x
5x = 30
x = 6

Answer D
*\((Concen_A)(Vol_A) + (Con_B)(Vol_B) = (Con_{A+B})(Vol_{A+B})\)
GMAT Club Bot
Three pounds of 05 grass seed contain 5 percent herbicide. A different &nbs [#permalink] 27 Oct 2017, 12:31
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