OG11 PS 240

[T_A_M]

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

[bluementor]

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%.

-BM-

[T_A_M]

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?

[T_A_M]

Appreciate the solution, my question is more around good techniques / tools to solve the question in 2 mins?

[cool_sni]

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%

[figs]

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??

[cool_sni]

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)