Question

Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of the mixture is X ?

Answer 1

Answer: The weight of X is
`33(1)/(3)\%` of weight of mixture.

Explanation:

Since we have given that

Percentage of seed mixture X for ryegrass = 40%

Percentage of seed mixture Y for ryegrass = 25%

If a mixture of X and Y contains 30 percent ryegrass,

Let total seed mixture be 100

So, for seed X = x

For seed Y = 100-x

So, According to question,

`0.4x+0.25(100-x)=30\\\\0.4x+25-0.25x=30\\\\0.15x=30-25\\\\0.15x=5\\\\x=(5)/(0.15)\\\\x=(100)/(3)`

So, weight of mixture X is given by

`\frac{\text{Weight of X}}{\text{Weight of mixture}}* 100\\\\=((100)/(3))/(100)* 100\\\\=(100)/(3)\%\\\\=33(1)/(3)\%`

Hence, the weight of X is
`33(1)/(3)\%` of weight of mixture.