Suteja wrote:Seed mixture X is 40 percent ryegrass and 60 percent
bluegrass by weight; seed mixture Y is 25%
ryegrass and 7S 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
