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Mixture A is 15% alcohol, and mixture B is 50% alcohol. If

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Mixture A is 15% alcohol, and mixture B is 50% alcohol. If [#permalink] New post 04 May 2006, 08:31
Mixture A is 15% alcohol, and mixture B is 50% alcohol. If the two are poured together to create a 4 gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?

A. 1.5
B. 1.7
C. 2.3
D. 2.5
E. 3

please explain your solution.
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Re: Mixture [#permalink] New post 04 May 2006, 08:49
kuristar wrote:
Mixture A is 15% alcohol, and mixture B is 50% alcohol. If the two are poured together to create a 4 gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?

A. 1.5
B. 1.7
C. 2.3
D. 2.5
E. 3

please explain your solution.

I ll go with C
I used the approach that ancient merchants used for mixtures))
Hope this helps

here we go
15mix A 20parts of mixture A
/
/
30 final mix
\
\
50 mix B 15parts of mixture B
just draw the following scheme
4 gallons of final mixture contains of 35 parts so 1 part of mixture is 4/35
we have 20 parts of mixture A hence (4/35)*20=80/35 or 2.3
C for me
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 [#permalink] New post 04 May 2006, 08:53
How did you get the parts per mixture (A, B respectively)?

Thanks!
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 [#permalink] New post 04 May 2006, 08:58
let A be x gallons, B is (4-x) gallons

0.15x + 0.5(4-x) = 0.3 * 4
0.15x + 2 - 0.50x = 1.2

Solving for x,
x = 2.3

Answer is C
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 [#permalink] New post 04 May 2006, 09:07
Thanks, now it seems obvious (of course!).
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 [#permalink] New post 04 May 2006, 09:10
kuristar wrote:
How did you get the parts per mixture (A, B respectively)?

Thanks!

we have 30 percent final mixture that contains mixture A and B
to figure out the parts per mixture use the scheme above as follows
we subtract the smaller from larger 50-30 =20 this is parts of mixture A
and for mixture B we subtract 15 from 30 cause it is smaller now we have 15parts of mixture B
Not sure that it is the good explanation but English is not my firt language))

try this approach it really works

lets check
for example we need to get 45 percent mixture of alchohol
we have mixt A 20 percent alc
and mixt B 65 percent alc

using scheme
20--- 20 parts of mixt A
/
45
\
65--- 25 parts of mixt B in final mixture
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 [#permalink] New post 04 May 2006, 09:29
Got it. thanks, would have never thought of it that way.
  [#permalink] 04 May 2006, 09:29
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