Problem Solving

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