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
Cheers,
Brent