## Seed mixture $$X$$ is $$40$$ percent ryegrass and $$60$$ percent bluegrass by weight; seed mixture $$Y$$ is $$25$$ perce

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### Seed mixture $$X$$ is $$40$$ percent ryegrass and $$60$$ percent bluegrass by weight; seed mixture $$Y$$ is $$25$$ perce

by Vincen » Fri Aug 06, 2021 7:38 am

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## Global Stats

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. $$66\frac23\%$$

Source: Official Guide

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### Re: Seed mixture $$X$$ is $$40$$ percent ryegrass and $$60$$ percent bluegrass by weight; seed mixture $$Y$$ is $$25$$ p

by [email protected] » Wed Aug 11, 2021 5:09 am
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. $$66\frac23\%$$

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.

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### Re: Seed mixture $$X$$ is $$40$$ percent ryegrass and $$60$$ percent bluegrass by weight; seed mixture $$Y$$ is $$25$$ p

by swerve » Thu Aug 12, 2021 5:52 am
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. $$66\frac23\%$$

Source: Official Guide
Let $$M = x + y$$

$$M =$$ New mixture
$$x =$$ Mixture $$X$$
$$y =$$ Mixture $$Y$$

What do we need to find? $$\Rightarrow \dfrac{x}{M}\ast 100$$

Equating Ryegrass in the mixture

$$0.4x + 0.25y = 0.3M$$
$$0.4x + 0.25(M-x) = 0.3M$$
$$0.4x + 0.25M - 0.25x = 0.3M$$
$$0.15x = 0.05M$$
$$\dfrac{x}{M} = \dfrac{1}{3}$$

Hence, $$33.33\%.$$

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