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
