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OG Mixtures Problem

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by aaggar7 » Mon Mar 11, 2013 7:53 pm

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Answers

A

B

C

D

E

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Difficulty—

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 1/3%
(C) 40%
(D) 50%
(E) 66 2/3%
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by Anju@Gurome » Mon Mar 11, 2013 9:06 pm
aaggar7 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 ?
This can be treated as an weighted average problem where 30 is the weighted average of 25 and 40 and we need to find out the weight of 40 in the average.

Say, the weight of 40 is x.
Hence, weight of 25 is (1 - x)

So, 40x + 25(1 - x) = 30
--> (40x - 25x) = (30 - 25)
--> 15x = 5
--> x = 1/3

Required percentage = 100/3 = 33â…“ %

The correct answer is B.
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by Anju@Gurome » Mon Mar 11, 2013 9:16 pm
Here is another different approach based on the same concept.

In any weighted average, thee weight of any end point in the average will be (Difference between the other end-point and the weighted average)/(Difference between the two end-points)

Hence, here weight of 40 in the average = |25 - 30|/|40 - 25| = 5/15 = 1/3

Required percentage = (1/3)*100 = 33â…“

The correct answer is B.[/quote]
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by aaggar7 » Tue Mar 12, 2013 12:26 am
Thanks Anju :)
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OG Mixtures Problem

by GMATGuruNY » Tue Mar 12, 2013 2:36 am
For a visual solution, check the second half of my post here:

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by [email protected] » Thu Jun 21, 2018 6:21 pm
Hi All,

We're told that seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight, seed mixture Y is 25 percent ryegrass and 75 percent fescue and a mixture of X and Y contains 30 percent ryegrass. We're asked for the percentage of the weight of the mixture that is seed X. This question can be solved in a couple of different ways, including by setting up a Weighted Average.

X = number of pounds of seed mixture X
Y = number of pounds of seed mixture Y

Since we're mixing together a certain amount of X and Y - and the respective percentages of ryegrass are 40% and 25%, respectively - we can create the following equation:

(.4X + .25Y)/(X+Y) = .3

.4X + .25Y = .3X + .3Y
.1X = .05Y
10X = 5Y
X/Y = 5/10 = 1/2

Thus, for every round of X, we have 2 pounds of Y and 3 pounds total. Thus, the percentage of the TOTAL mix that is seed mix X is 1/3 = 33 1/3%

Final Answer: B

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by swerve » Sun Jun 24, 2018 8:06 am
Let M = x + y

M = new mixture
x = Mixture X
y = Mixture Y

What do we need to find? (x/M)*100

Equating Ryegrass in the mixture,

0.4x + 0.25y = 0.3M
0.4x + 0.25(M-x) = 0.3M
0.4x + 0.25M - 0.25x = 0.3M
0.15x = 0.05M
x/M = 1/3

Hence, B is the correct answer. Regards!
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by Scott@TargetTestPrep » Mon Jun 25, 2018 11:11 am
aaggar7 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 1/3%
(C) 40%
(D) 50%
(E) 66 2/3%
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 care only 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 » Mon Jun 25, 2018 11:30 am
aaggar7 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 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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