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OG11 PS 240

by T_A_M » Mon Apr 20, 2009 7:47 am
What are your thoughts on rates and work questions?
Best approach to solve this?


Seed Mixture X is 40% ryegrass and 60% bluegrass by weight. Seed mixture Y is 25% ryegrass and 75% fescue. If a mixture of X and Y contains 30% ryegrass what % of the weight of the mixture is X?

T_A_M
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by bluementor » Mon Apr 20, 2009 8:13 am
You only have to consider the composition of ryegrass in all three mixtures, since this is the only common element.

Mixture X : 40% ryegrass
Mixture Y: 25% ryegrass
Final Mixture X+Y: 30% ryegrass

Lets say the fraction of X in the final mixture is A. So the fraction of Y in the final mixture is (1-A), since the final mixture is made up of only X and Y.

You can now use weighted average to solve this:

A(0.4) + (1-A)(0.25) = (0.3)(A + 1 - A)
0.4A + 0.25 - 0.25A = 0.3
0.15A = 0.05
A = 0.05/0.15 = 1/3

So mixture X makes up 1/3 of the final mixture:

1/3 * 100% = 33.33%.

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by T_A_M » Mon Apr 20, 2009 8:17 am
Hi -
Thanks,
I managed to get to the same answer following the solution provided in OG. Do you have any good approaches / strategies to solve any mixture problems?
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by T_A_M » Mon Apr 20, 2009 8:18 am
Appreciate the solution, my question is more around good techniques / tools to solve the question in 2 mins?
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by cool_sni » Wed Apr 22, 2009 8:35 am
Here's another way to approach this problem
final mixture contain 30% of Ryegrass.
X contributes 40-30=10% (clearly you need less amount from X)
Y contributes 30-25 = 5%

So contribution by X = 5/15=1/3=33%
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by figs » Wed Apr 22, 2009 10:04 am
Here's another way to approach this problem
final mixture contain 30% of Ryegrass.
X contributes 40-30=10% (clearly you need less amount from X)
Y contributes 30-25 = 5%

So contribution by X = 5/15=1/3=33%

Can you explain last step??
5/15 ..is not the contribution by Y??
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by cool_sni » Wed Apr 22, 2009 11:00 am
As you can see X needs to contribute less than Y.
Amount X needs to contribute is the difference Y has from final %.
So,
X -> 5/(5+10)
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