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Math help?

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forgive me if this has been discussed previously, i cannot seem to find the answer..

When I have a 80 proof whiskey, I would typically add about 5mL of water to a 60mL aliquot to obtain my desired whiskey:water ratio

However when I have a "cask strength" bottle of whiskey say e.g. 120 proof I tend to either over or under dilute my mix. ( i eyeball it )

Here is the question:

If i wish to end up with both the same quantity (at the end of the mix) and same %alchohol:water ratio, how much water should I add to how much "cask strength" whiskey to get the same ratio as:

80 proof 60mL + 3mL

80 proof 60mL + 5mL

80 proof 60mL + 10mL

(you can use 90 proof base whiskey if you care to do the math twice)

in other words if if i have add the 5 ml to 60mL i wind up with 65mL of a now 37% or 78 proof mix(assuming my math is correct) I want the same 65 mL of a mix of water and 120 proof whiskey...

so? is there a simple algebraic formula to use to plug in any two "proofs" and water amounts?

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I believe it's:

60mL 80pf + 3mL water = 40mL 120pf + 23mL water

60mL 80pf + 5mL water = 40mL 120pf + 25mL water

60mL 80pf + 10mL water = 40mL 120pf + 30mL water

The procedure is to figure out the amount of pure alcohol in the mixture (in your case, 24mL, or volume booze times percentage alcohol). Then divide that number (24mL) by the alcohol fraction of the new booze (.6). This gives you the volume of the new booze to use (40mL), and just fill to the final total volume with water.

So for 90pf stuff, you would use 53.3mL and fill with water.

Edited by Penwu (log)

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