Enginpasa1 wrote:Is there an easier way to do this?
seed mixture X is 40% ryegrass and 60% bluegrass. Seed mixture Y is 25% 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.33.33%
c.40%
d.50%
e.66.66 %
I am stumped
I am not very good at explaining questions over the net.. so bear with me.
This question is based on the concept of weighted average. Let me explain weighted average with an example.
Consider 2 classes. Class A has 3 students and Class B has 5 students. Suppose the average marks in English for class A is 60 and that for Class B is 80. What is the average marks in English for Class A and Class B combined ?. It is not 60 + 80 = 140 / 2 = 70 .. Why??, because the weights for the 2 marks are different. So the average in this case is (60*3+80*5 )/(3+5) = 72.5.
Suppose that Class A and Class B both had 4 students in that case what would be the average. It would be (60*4+80*4)/(4+4)=70, which is the same as (60 +80)/2 = 70, this is because both the marks have the same weight.
Using the same concept of weighted average in the above question . Take "a" liters of mixture X and "b" liters of mixture Y. So the new mixture created out of these 2 will be (a+b) liters in volume. Now the new mixture has 30% of ryegrass. This 30% is basically the weighted average of the ryegrass in the 2 mixtures X and Y mixed together in quantity "a"" and "b" respectively.
So we have (40*a+25*b)/(a+b)=30, solving this equation we have..
40a+25b=30a +30b, 10a=5b, therefore a/b=1/2, so a/(a+b)= 1/3, so the percentage of X in the new mixture is 33%.
There are many ways to solve this question, one way has already been illustrated in the above post by codesnooker. You can also solve this question by allegation, which if used can help solve such questions in a few seconds. My post is to basically explain the concept of weighted average. Hope this helps you.
Regards.
