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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegra [#permalink]
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Carcass wrote:
Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 % fescue. If a mixture of X and Y contains 30% ryegrass, what percent of the weight of the mixture is X?


A. 10%

B. \(33\frac{1}{3}\)%

C. 40%

D. 50%

E. \(66\frac{2}{3}\)%



This looks like a job for weighted averages!

Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

Mixture X is 40 percent ryegrass
Mixture Y is 25 percent ryegrass

Let x = the PERCENT of mixture X needed (in other words, x/100 = the proportion of mixture X needed)
So, 100-x = the PERCENT of mixture Y needed (in other words, (100-x)/100 = the proportion of mixture Y needed)

Weighted average of groups combined = 30%

Now take the above formula and plug in the values to get:
30 = (x/100)(40) + [(100-x)/100](25)
Multiply both sides by 100 to get: 30 = 40x + (100-x)(25)
Expand: 3000 = 40x + 2500 - 25x
Simplify: 3000 = 15x + 2500
So: 500 = 15x
Solve: x = 500/15
= 100/3
= 33 1/3

So, mixture X is 33 1/3 % of the COMBINED mix.

Answer: B

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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegra [#permalink]
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First notice that X and Y both contain ryegrass, and their resultant mixture X+Y does as well, so we can set up the equation with the information given in the question:

\(\frac{4}{10}X + \frac{1}{4}Y = \frac{3}{10}(X+Y)\)

Now we solve to get the ratio of X to Y:

\(\frac{4}{10}X + \frac{1}{4}Y = \frac{3}{10}X + \frac{3}{10}Y\)

\(\frac{1}{10}X + \frac{1}{4}Y = \frac{3}{10}Y\)

\(\frac{1}{10}X + \frac{5}{20}Y = \frac{6}{20}Y\)

\(\frac{1}{10}X = \frac{1}{20}Y\)

\(\frac{X}{Y} = \frac{1}{2}\)

So we have: \(X:Y = 1:2\)

Which translates to: 1 part X for every 2 parts of Y.

If you sum both parts (X and Y), then you get 3 parts.

One popular way to write this is: 1z : 2z = 3z.

So X is \(\frac{1}{3}\) of the new mix.

The answer is therefore B.
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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegra [#permalink]
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Take Ryegrass into account only, as the resulting mixture is 30% ryegrass

Use the concept of Weighted Average;
40X + 25Y = 30 (X+Y)
10X = 5Y
2X = Y

Now, Total mixture = (X+Y) = X + 2X = 3X
Therefore, %X of total = (X/3X) x 100 = 33.33%

Hence, option B
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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegra [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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