# Thread: Mixing ingredients where portion of one of ingredients is a fn of the final amount

1. ## Mixing ingredients where portion of one of ingredients is a fn of the final amount

I have two mixtures of flour and water.

Mixture 1 has equal parts flour and water

Mixture 2 has two parts flour to one part water

I also have supplies of flour and water

Problem (a) I'm going to mix flour and water and want to end up with mixture 1 -- equal parts flour and water. BUT in doing so I have to include an amount of Mixture 2 that is 1/6 of the total amount of mixture I end up with (i.e., 1/6 of the total of flour, water, and mixture 2).

Problem (b) is similar -- I'm going to mix flour and water and want to end up with mixture 2 -- two to one ratio of flour to water. BUT I have to include an amount of Mixture 1 that is 1/7 of the total end result.

This is my first post and not a school/college question

At 58 am a bit rusty on this stuff hence the question here.

My initial thoughts

F= flour
W= water
M1 = Mixture 1 = 0.5F +0.5W
M2 = Mixture 2 = 2.0F +1.0W

Ma = Mixture a = 0.5F + 0.5W at the end but must be made up of enough of M2 so as to constitute 1/6th of Ma

Ma = xF + yW + 1/6 (Ma)(M2).....
Thanks

2. The equation you wrote looks sensible, but you need more than 1 equation to solve the problem.

If you define:

F = the amount of flour you added
W = the amount of water you added
M2 = the amount of mixture 2 added

The total amount of water in your output is
$\frac{1}{3}M2 + W$

And the total amount of Flour is
$\frac{2}{3}M2 + F$

To end up with mixture1, you need to have the same amount of flour and water
$\frac{2}{3}M2 + F = \frac{1}{3}M2 + W$

You are also told to use M2 for 1/6 of your input, which means there should be 5 units of standalone (flour + water) for each 1 units of M2.
$(F + W) = 5M2$

Now all you need to do is decide how much mixture to make, to do this set an arbitrary amount of one of the ingredients eg
$M2 = 1$

Then solve the other two equations simultaneously.
$\frac{2}{3} + F = \frac{1}{3} + W$
$(F + W) = 5$

Which has a solution of
W=7/3
F = 8/3
M2 = 1

NB post was edited to fix a mistake earlier