My Math Forum I'm stumped! Mixing Liquid Concentrates

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 March 14th, 2017, 08:42 PM #1 Newbie     Joined: Mar 2017 From: Canada Posts: 2 Thanks: 0 Math Focus: Fuzzy Logic I'm stumped! Mixing Liquid Concentrates This SHOULD be easy, but I don't know why I can't solve it. I've tried Excel and still can't do it. I've spent countless hours on this and am stumped! Suppose the concentrates below are all different chemicals. They also have the same consistency as water and can be used together to dilute each other with no additional water needed. Each concentrate below has its own bottle because they are a DIFFERENT type of chemical from each other. The label on each specifies an amount of liquid that MUST be diluted to a specific ratio of another liquid. Water can be added if needed as long as they're diluted to the ratio. Liguid Concentrate to Liquid ratio: c1 needs 1:2 (one part liquid concentrate, 2 parts liquid (i.e. water or another concentrate) c2 needs 1:3 c3 needs 1:4 c4 needs 1:8 PROBLEM: I need a MIXTURE of all the above concentrates that will fill a 1000ml vessel. The original dilution ratios MUST still be preserved for EACH concentrate in the final mixture. I thought this would be as simple as adding fractions I just don't know why my mind can't figure it out. Thanks for any help; it's REALLY appreciated! Last edited by skipjack; March 15th, 2017 at 12:28 AM.
 March 15th, 2017, 12:42 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,310 Thanks: 1981 For, say, c2, the volume of c2 must be 250ml so that the remaining 750ml of other liquids gives the desired 1:3 ratio for the c2. Similarly, the volume of c3 must be 200ml. Adding the volumes for all four concentrates gives a total that is less than 1000ml, so the rest of the 1000ml is water. Thanks from romsek
 March 15th, 2017, 12:55 AM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,325 Thanks: 1233 ok.. Suppose we have volume $V$ $c1 = \dfrac{V}{3}$ $c2 = \dfrac{V}{4}$ $c3 = \dfrac{V}{5}$ $c4 = \dfrac{V}{9}$ $c1 + c2 + c3 + c4 = V \dfrac {161}{180}$ Now... let's add $\dfrac {19V}{180}$ of water now the concentrations add to 1 and we simply multiply by $V=1000$ to get the amounts (in ml) $c1 = 333 ~ \dfrac 1 3$ $c2 = 250$ $c3 = 200$ $c4 = 111 ~\dfrac 1 9$ $water = 105 ~\dfrac 5 9$ You can check that the concentration ratios are all correct and that these amount sum to 1000 ml.

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