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DS: Mixed Cocktail 08 Jun 2006, 09:03
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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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gmatmba
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DS: Mixed Cocktail 08 Jun 2006, 11:49
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Yurik79
X & Y
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
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Can you show how, using (1) you are getting the answer
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tl372
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DS: Mixed Cocktail 08 Jun 2006, 13:54
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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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jaynayak
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DS: Mixed Cocktail 09 Jun 2006, 01:31
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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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Yurik79
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DS: Mixed Cocktail 09 Jun 2006, 09:32
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gmatmba
Yurik79
X & Y
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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Oops sorry my fault )))agree with C
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trivikram
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DS: Mixed Cocktail 30 Dec 2006, 22:27
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Can someone explain this please?

Thanks!
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tennis1ball
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DS: Mixed Cocktail 30 Dec 2006, 23:45
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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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Andr359
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DS: Mixed Cocktail 30 Dec 2006, 23:48
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X & Y
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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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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trivikram
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DS: Mixed Cocktail 31 Dec 2006, 07:35
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tennis_ball
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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x here ix total volume 2 liquids?
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tennis1ball
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DS: Mixed Cocktail 01 Jan 2007, 01:29
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x is the concentration of the mixed.



Archived Topic
Hi there,
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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