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DS: Mixed Cocktail

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XxxyyY
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DS: Mixed Cocktail [#permalink] New post   08 Jun 2006, 09:03
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A
B
C
D
E

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Solvent A contains 24% of benzene. What is the benzene concentration of the mixed cocktail of solvent A and B?

(1) The mixing ratio of solvent A and B is 1:4

(2) The benzene concentration of solvent B is 1.5 times greater than benzene concentration of the mixed cocktail.



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Re: DS: Mixed Cocktail [#permalink] New post   08 Jun 2006, 11:49
Yurik79 wrote:
X & Y wrote:
Solvent A contains 24% of benzene. What is the benzene concentration of the mixed cocktail of solvent A and B?

(1) The mixing ratio of solvent A and B is 1:4

(2) The benzene concentration of solvent B is 1.5 times greater than benzene concentration of the mixed cocktail.

D for me


Can you show how, using (1) you are getting the answer
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Re: DS: Mixed Cocktail [#permalink] New post   08 Jun 2006, 13:54
I get C as well.

Without knowing the amount of B and A that are mixed together (st. 1) and the concentration of B (st.2), I don't see how you can determine the concentration of the mixed cocktail.
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Re: DS: Mixed Cocktail [#permalink] New post   09 Jun 2006, 01:31
C for me too.

1) we know the ratios the solvents are mixed but don't have a idea if solvent has benzene or not.

2) we know the benzene concentration in solvent B. But don't know anything of the mixture.

Both: the required information is there.
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Re: DS: Mixed Cocktail [#permalink] New post   09 Jun 2006, 09:32
gmatmba wrote:
Yurik79 wrote:
X & Y wrote:
Solvent A contains 24% of benzene. What is the benzene concentration of the mixed cocktail of solvent A and B?

(1) The mixing ratio of solvent A and B is 1:4

(2) The benzene concentration of solvent B is 1.5 times greater than benzene concentration of the mixed cocktail.

D for me


Can you show how, using (1) you are getting the answer

Oops sorry my fault )))agree with C
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Re: DS: Mixed Cocktail [#permalink] New post   30 Dec 2006, 22:27
Can someone explain this please?

Thanks!
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Re: DS: Mixed Cocktail [#permalink] New post   30 Dec 2006, 23:45
C.

cond 1: you don't know the concentration of B, so no way you can solve.

cond 2:
Let total mixed volume be 1.
and A has volume r, then B has 1-r.
you get this equation:

x = r*24% + (1-r)*x*(1+1.5).

you solve x = 24%*r/(2.5*r - 1.5).

but u don't know r. so 2 is Insuff.

Together. since r /1-r = 1/4. u can solve r. So, both sufficient
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Re: DS: Mixed Cocktail [#permalink] New post   30 Dec 2006, 23:48
X & Y wrote:
Solvent A contains 24% of benzene. What is the benzene concentration of the mixed cocktail of solvent A and B?

(1) The mixing ratio of solvent A and B is 1:4

(2) The benzene concentration of solvent B is 1.5 times greater than benzene concentration of the mixed cocktail.


Let´s see:

To know the concentration of A+B, we need to know both the proportion in which A and B mix AND B´s % of benzene. While (1) gives us only the former, (2) gives us only the latter. Only if we join (1) and (2) we have sufficient data to solve the problem => the answer is C.
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Re: DS: Mixed Cocktail [#permalink] New post   31 Dec 2006, 07:35
tennis_ball wrote:
C.

cond 1: you don't know the concentration of B, so no way you can solve.

cond 2:
Let total mixed volume be 1.
and A has volume r, then B has 1-r.
you get this equation:

x = r*24% + (1-r)*x*(1+1.5).

you solve x = 24%*r/(2.5*r - 1.5).

but u don't know r. so 2 is Insuff.

Together. since r /1-r = 1/4. u can solve r. So, both sufficient


x here ix total volume 2 liquids?
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Re: DS: Mixed Cocktail [#permalink] New post   01 Jan 2007, 01:29
x is the concentration of the mixed.



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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Re: DS: Mixed Cocktail [#permalink]
01 Jan 2007, 01:29
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