Using the "trick" for weighted averages / mixtures

Hey everyone,
I just watched Greg’s concept series on “Mixtures” where he teaches us the “trick” calculating the differences between the averages / mixtures.

My question is, I’ve sometimes come across questions where the final mixture doesnt specify the % of just ONE component, but both.

For example, this was a question covered in the practice class:

“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?”

What if the question was:

“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 needs to contain 30% ryegrass and 40% bluegrass , what percent of the weight of the mixture is X?”

How would we use the trick to solve a 2-tier question like this?

The problem is that your variation need not be solvable. When you’re mixing two items, the composition of X and Y in the mixture are dependent on each other - it seems to be that you’re assuming that it is independent. In other words, if there was an answer to your second part, it would be the exact same as the original question, and if that answer does not result in a mixture of 40% bluegrass, the question is faulty.

Hi, thanks for your response! I guess the attached is a question which represents my doubt better (i.e. two variables / ratios to solve for). Can we use the “trick” to solve this?

Two tins A and B contain mixtures of wheat and rice. In A, the weights of wheat and rice are in
the ratio 2 : 3 and in B they are in the ratio 3 : 7. What quantities must be taken from A and B to
form a mixture containing 5 kg of wheat and 11 kg of rice?

Here’s a hint: you know the proportion of wheat in the final mixture. Can you reduce this problem to a standard mixture problem that way? The only problem is that you will need to multiply the final result to get a 16 kg total mixture.

Sorry, I didnt catch that. Could you please elaborate how to set up the “trick” with the “distances” in a case like this?

Hint: the proportion of wheat in the final mixture is about 31.25%. The proportion of wheat in A is 40% and about 43% in B.

Does that help?

Ahhh, understood now - I got it! Works using the trick!!
Thank you so much

Getting stuck on another one with 2 ratios. Any hints for the “trick” in this?

Two vessels contain mixtures of spirit and water. In the first vessel the ratio of spirit to water is 8 : 3 and in the second vessel the ratio is 5 : 1. A 35–litre cask is filled from these vessels so as to contain a mixture of spirit and water in the ratio of 4 : 1. How many litres are taken from the first vessel?

In this, whats tripping me out is they are asking how much of the entire first vessel we need (as opposed to how much water / spirit we need). If the question was the latter, I could’ve set up the trick and distances, but am getting confused given the question is set up a bit differently.

So you have a 8/11 proportion of spirit in vessel 1, 5/6 proportion of spirit in vessel 2, and a 80% proportion of spirit in the end. Can you find the ratio of the volume taken out from vessel 1 and 2?