seed mixture

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by GMATinsight » Wed Nov 26, 2014 9:28 pm
simba12123 wrote:Seed mixture x is 40 percent ryegrass and 60 percent bluegrass by weight, seed mixture y is 25 percent ryegrass and 75 percent fescue. If a mixture of x and y contains 30 percent ryegrass. WHat percent of the weight of the mixture is x?


a. 10 %
b.33.3%
c.40%
d.50%
e.66.67%

qa is b


although the answer explanation is nice in the back of the og 11 book, I am looking for a faster and less algebraic approach. ANy help?
Mixture of 40% Ryegrass and 25% Ryegrass is 30% Ryegrass

If Mixture of 40% Ryegrass and 25% Ryegrass were (40+25)/2 = 32.5% Ryegrass then both would have been equal in quantity

But Since Mixture of 40% Ryegrass and 25% Ryegrass is 30% Ryegrass so it's more influenced by 25% Ryegrass and therefore

40------(40-30)=10-------30------(30-25)=5-------25
i.e. the ratio must be 10:5 = 2:1 = y : x

i.e. x = (1/3)*100 = 33.33%

Answer: Option B
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by Danny@GMATAcademy » Tue Jul 05, 2016 8:24 am
A couple of ways to solve this problem:

Algebra:https://youtu.be/obHR1MewMhA
Balance Method:https://youtu.be/EkkOwiyrhhQ

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by 800_or_bust » Tue Jul 05, 2016 11:11 am
simba12123 wrote:Seed mixture x is 40 percent ryegrass and 60 percent bluegrass by weight, seed mixture y is 25 percent ryegrass and 75 percent fescue. If a mixture of x and y contains 30 percent ryegrass. WHat percent of the weight of the mixture is x?


a. 10 %
b.33.3%
c.40%
d.50%
e.66.67%

qa is b


although the answer explanation is nice in the back of the og 11 book, I am looking for a faster and less algebraic approach. ANy help?
I know this is very old, but I saw it had been bumped up today by Danny. A quicker way to solve this is as follows:

(1) Notice that the resultant mixture is closer to the composition of Y than X, so you know there must be more of Y than X. Therefore, choices (D) and (E) can be eliminated right off the bat and ignored altogether.

(2) Test answer choice B by seeing what the composition of the resulting mixture would be if we added 1/3 of X to 2/3 of Y. In this case, B works and B is the correct answer. However, had the resulting mixture contained too much ryegrass, then we would have added too much of X, and thus the correct answer would have been A. If the resulting mixture would have contained too little ryegrass, then we would not have added enough X, and thus the correct answer would have been C.

Hence, it would be possible to come to a correct solution on this question merely by checking choice (B).
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by Jeff@TargetTestPrep » Fri Jan 05, 2018 9:31 am
Seed mixture x is 40 percent ryegrass and 60 percent bluegrass by weight, seed mixture y is 25 percent ryegrass and 75 percent fescue. If a mixture of x and y contains 30 percent ryegrass. WHat percent of the weight of the mixture is x?


a. 10 %
b.33.3%
c.40%
d.50%
e.66.67%
We are given information about ryegrass, bluegrass, and fescue. However, we are told the mixture of X and Y is 30 percent ryegrass. Thus, we only care about the ryegrass, so we can ignore fescue and bluegrass.

We are given that seed mixture X is 40% ryegrass and that seed mixture Y is 25% ryegrass. We are also given that the weight of the combined mixture is 30% ryegrass. With x representing the total weight of mixture X, and y representing the total weight of mixture Y, we can create the following equation:

0.4x + 0.25y = 0.3(x + y)

We can multiply the entire equation by 100:

40x + 25y = 30x + 30y

10x = 5y

2x = y

The question asks what percentage of the weight of the mixture is x.

We can create an expression for this:

x/(x+y) * 100 = ?

Since we know that 2x = y, we can substitute in 2x for y in our expression. So, we have:

x/(x+2x) * 100 = x/(3x) * 100 = 1/3 * 100 = 33.33%

Answer: B

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by Brent@GMATPrepNow » Fri Jan 05, 2018 12:29 pm
Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of this mixture is X ?

(A) 10%
(B) 33 1/3%
(C) 40%
(D) 50%
(E) 66 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

For more information on weighted averages, you can watch this video: https://www.gmatprepnow.com/module/gmat- ... ics?id=805
Here are some additional practice questions related to weighted averages:
- https://www.beatthegmat.com/weighted-ave ... 17237.html
- https://www.beatthegmat.com/weighted-ave ... 14506.html
- https://www.beatthegmat.com/average-weig ... 57853.html
- https://www.beatthegmat.com/averages-que ... 87118.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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