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Junior | Next Rank: 30 Posts
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by joinn » Wed Jul 06, 2011 8:49 pm
if a drink is made by mixing 2 shots of a liquor containing 15% alcohol
with 3 shots of a liquor containing 20% alcohol, what is the content of the mixed alcohol drink. How to solve this problem. This problem is from weighted averages chapter from manhattan math book series no 4.

pals no options available, explain anyone..
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by krishnasty » Wed Jul 06, 2011 9:57 pm
IMO [spoiler]18%[/spoiler]

wat's the OA?
i have never been good in mixture, but here's my approach..
(15%)(2) + (20%)(3) = (x%)(5)
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by Jim@Knewton » Wed Jul 06, 2011 10:40 pm
Let 1 shot = s (quantity)
=> New mix overall quantity = 2s + 3s = 5s
=> New mix alcohol quantity = 0.15*2s + 0.20*3s = 0.9s
=> New content [spoiler]= 0.9s/5s = 0.18 = 18%[/spoiler] alcohol
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by amit2k9 » Thu Jul 07, 2011 12:39 am
you can use the allegation formula -

2/3 = (20-x)/(x-15)

this gives x=18.

also 2 units of 15 to 3 units of 20 means it will be closer to 20 using weighted average.

or

taking 15 = 0 means 20 = 5.
using weighted average -

[2*(0)+ 3*(5)]/(2+3) = 3
thus 15+3 = 18.
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