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qhoc0010
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Mixture A is 15 percent alcohol, and mixture B is 50 percent 08 Feb 2005, 22:16
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Mixture A is 15 percent alcohol, and mixture B is 50 percent 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) 3.0

I can solve this problem. It is no big deal. But my method is too long because it involved in 4 equations with 4 variables. What is the quickest way?



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Mixture A is 15 percent alcohol, and mixture B is 50 percent 08 Feb 2005, 22:35
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Hmmm I'm using two equations.

15A+50B=30(A+B)
A+B=4

From the first one you get 3A=4B
substitute A into the second one
4A+4B=16
4A+3A=16
Z=16/7=2.3
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qhoc0010
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Mixture A is 15 percent alcohol, and mixture B is 50 percent 08 Feb 2005, 22:43
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HongHu
Hmmm I'm using two equations.

15A+50B=30(A+B)
A+B=4

From the first one you get 3A=4B
substitute A into the second one
4A+4B=16
4A+3A=16
Z=16/7=2.3
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Well, that is one way.
Mine is lengthly.

a/x = 15/100
b/y = 50/100
(a+b)/(x+y) = 30/100
x+y = 4
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MA
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Mixture A is 15 percent alcohol, and mixture B is 50 percent 08 Feb 2005, 22:45
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W.15+(4-W).5=.3(4)
solving for W=.80/.35=2.7 approx.
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Folaa3
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Mixture A is 15 percent alcohol, and mixture B is 50 percent 09 Feb 2005, 05:29
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qhoc0010
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W.15+(4-W).5=.3(4)
solving for W=.80/.35=2.7 approx.

OK, that is a nice one.
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Ghocoo10, but what is the OA? I notice 2.7 is not among the ans choices.
And seriously, what is the quickest way to get this done folks :lol:
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Mixture A is 15 percent alcohol, and mixture B is 50 percent 09 Feb 2005, 06:56
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I HAVE ANOTHER APPROACH...

(30-15).A = (50-30).(4-A)
15A = 20(4-A)
15A = 80-20A
35A = 80
A =80/35 =16/7 =2.3

THE FORMULA HERE IS

(RESULTING SOLN - SOLN A) * A's VOL = (SOLN B - RESULTING SOLN) * B's VOL
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Mixture A is 15 percent alcohol, and mixture B is 50 percent 09 Feb 2005, 07:22
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great formula. Where is the reference from?
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qhoc0010
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Mixture A is 15 percent alcohol, and mixture B is 50 percent 09 Feb 2005, 08:55
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OA is (C)
Thanks all.
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banerjeea_98
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Mixture A is 15 percent alcohol, and mixture B is 50 percent 09 Feb 2005, 09:55
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"C"

Let X be vol or "A", so B is 4-X

X*15/100 + (4-X)*50/100 = 4*30/100


Solve for X = 2.28 ~ 2.3
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mbafarbe
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Mixture A is 15 percent alcohol, and mixture B is 50 percent 09 Feb 2005, 12:10
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I did it the same way as HongHu and it did not take too much time.
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HongHu
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Mixture A is 15 percent alcohol, and mixture B is 50 percent 09 Feb 2005, 12:31
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Sometimes two equations can be simpler than one equation. It may be easier to check the validity, and simpler for the calculations too.
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Folaa3
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Mixture A is 15 percent alcohol, and mixture B is 50 percent 09 Feb 2005, 21:08
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banerjeea_98
"C"

Let X be vol or "A", so B is 4-X

X*15/100 + (4-X)*50/100 = 4*30/100


Solve for X = 2.28 ~ 2.3
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Wow this approach is simple and short. Thanks banerjeea
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MA
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Mixture A is 15 percent alcohol, and mixture B is 50 percent 09 Feb 2005, 21:34
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Folaa3
qhoc0010
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W.15+(4-W).5=.3(4)
solving for W=.80/.35=2.7 approx.

OK, that is a nice one.

Ghocoo10, but what is the OA? I notice 2.7 is not among the ans choices.
And seriously, what is the quickest way to get this done folks :lol:
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Thanx Folaa,

i am editing my answer...
W.15+(4-W).5=.3(4)
solving for W=.80/.35=2.3 approx.
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Mixture A is 15 percent alcohol, and mixture B is 50 percent 11 Feb 2005, 11:55
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Use "Poor Man's Logic"

If we just averaged the two (equal concentrations) we would have 32.5%. Thus, we need a little more of the 15% solution and a little less of the 50% solution. Accoringly, 2.3 looks like a viable answer >>> verify 2.3 with plug'n'chug.



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