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Three pounds of 05 grass seed contain 5 percent herbicide. A different [#permalink]
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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
*\((Concen_A)(Vol_A) + (Con_B)(Vol_B) = (Con_{A+B})(Vol_{A+B})\)
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Three pounds of 05 grass seed contain 5 percent herbicide. A different [#permalink]
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


We have, 3 of 5%
Let x of 20%

Then,
5%(3) + 20%(x) = 15%(3+x)
\(=> 0.15 + 0.2x = 0.45 + 0.15x\)
\(=> 0.05x = 0.3\)
\(=> x = 6\)

Hence, OA is (D).
GMAT Club Bot
Three pounds of 05 grass seed contain 5 percent herbicide. A different [#permalink]
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