I do not know how OG solves this Q but the way I would solve it is by creating the following two equations

X+Y = 100 (assuming that the total mixture is 100 since this is a % question this makes it simple)

0.4X + 0.25Y = 30 (this comes from the fact that 40% of X + 25% of Y will give you 30% of total)

solve for X from the two equations to get X = 33.333 = 33 1/3

Is B the correct answer?

vinviper1 wrote:Seed nuxtyre X us 40% ryegrass and 60 % blue grewass by weight; seed mixture Y is 25% ryegrass and 75% fescue. if a mixture of X and Y contains 30 % ryegrass, what percent of the weight of the mixture is X?
10
33 1/3
40
50
66 2/3

Wanted to see if there is a better way than OG.

THANKS!

I want to start by focusing on:

Wanted to see if there is a better way than OG.

There's pretty much ALWAYS a better way than the OG.

The OG is a good source of questions. The OG is a horrible source of explanations.

Now on to the question.

If we use common sense, this question requires almost 0 math and takes very little time.

We can look at this problem as one involving weighted averages.

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

The first thing we note is that the % in the final mixture is closer to Y than to X. Therefore, there must be more Y than X in the mixture. Quickly eliminate (d) and (e).

If we draw a number line we can see the exact relationships:

Y ---5---- Mix ------10--------X

X is twice as far from the average than Y. Therefore, Y has twice as much weight as X does in the final mixture.

So, the ratio of Y:X is 2:1, which means that X makes up 1/3 of the mixture: choose (b).

* * *

A few extra examples to clarify how this works:

A --- 5---- Mix ---------15------- B

B is 3 times as far from the average as A; A has 3 times as much weight as B. Ratio of A:B = 3:1

C ----- 10 ------ Mix ----------15-----------D

D is 1.5 times as far from the average as C; C has 1.5 times as much weight as D. Ratio of C:D = 1.5:1 = 3:2

E ---3--- Mix -----5----- F

F is 5/3 as far from the average as E; E has 5/3 times as much weight as F. Ratio of E:F = 5/3:1 = 5:3

In fact, this last example demonstrates that we can just flip the actual distances to get the ratio.

Actual question after flipping distances... Y:X = 10:5 = 2:1

From the examples above, after flipping distances:

A:B = 15:5 = 3:1
C:D = 15:10 = 3:2
E:F = 5:3

I appreciate the time on this Stu. IT has really cleared it up and wow is that better than the OGs explanation

Stu -

didn't quite understand the Number Line...approach.

Y----5----mix-------10-----X ????

Can you PLEASE explain.

Stu -

didn't quite understand the Number Line...approach.

Y----5----mix-------10-----X ????

Can you PLEASE explain.

GMAT_crusher wrote:Stu -

didn't quite understand the Number Line...approach.

Y----5----mix-------10-----X ????

Can you PLEASE explain.

The concentration of Y was 5% below that of the mixture. The concentration of X was 10% above that of the mixture.

If anything else needs clarification, let me know.