Vincen wrote: ↑Fri Aug 06, 2021 7:38 am
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. \(33\frac13\%\)
C. \(40\%\)
D. \(50\%\)
E. \(66\frac23\%\)
Answer:
B
Source: Official Guide
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
Brent Hanneson - Creator of GMATPrepNow.com
