LogFAQs > #892956778

Hello, friends. I'm into my second 24 hours of awakeness, spent the last far too long trying to figure this out, so please check my math for the following word problem.

I had $7. The store sold 12 packs of beer X for $6.09, with beer X being 5.5% alcohol and 12oz per can, but also sold 4 packs of beer X+! at $2.29. Beer X+! is 8% alcohol, but also 16 oz per beer.

I, being an alcoholic, want to maximize my ounces of alcohol per dollar spent. Sadly, my insomnia, combined with having purchased 3 of the X+! 4 packs and drinking some of them, has blown a gasket in my head. I want to know which option provides the more bang for the buck. My reasoning is as follows:

Conditions:
Y: 4 packs of beer, 16 oz/beer, at 8% alcohol (i.e. 0.08 oz alc/1 oz beer), $2.29/4 beers
Z: 12 packs of beer, 12 oz/beer, at 5.5% alcohol, $6.09/12 beers

Y = 12 beers * 16 oz beer/beer = 192oz beer
at 8% alcohol/beer, for Y :
Y = 192 oz beer * 0.08 oz alc/oz beer = .08 oz alc * 192 = 15.36 oz alc

Z = 12 beers 12 oz/beer = 144oz beer
at 5.5% alcohol/beer, for Z:
Z = 144 oz beer * 0.055 oz alc/oz beer = 144 * .055 oz alc = 7.92oz alc

Y also = $6.87 (3 for packs at $2.29 each)
Z also = $6.09 (1 12 pack at $6.09)

Y = $6.87 = 15.36 oz alc = $6.84/15.36 oz alc = $0.44/oz alc
Z = $6.09 = 7.92 oz /al = $0.77/oz alc

Accordingly, and as I said, I bought 3 of the X+!. Good? Yes or no, please show your work.
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This too shall pass.