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Thanks All of the coffee mixtures sold in a certain store

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Thanks All of the coffee mixtures sold in a certain store [#permalink] New post 22 Mar 2004, 17:17
Please explain your answer. Thanks

All of the coffee mixtures sold in a certain store contain either Columbian (C), Jamaican (J), or Brazilian coffee, or some combination of those. Of all the coffee mixtures 33 contain C, 43 contain J, and 42 contain B. Of those, 16 contain at least C & J, 18 at least J & B, and 8 contain at least B & C, and 5 contain all three. How many different mixtures are sold at the store?

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 [#permalink] New post 22 Mar 2004, 17:44
Halle,

Please explain your answer, which is by the way wrong.

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 [#permalink] New post 22 Mar 2004, 17:45
is it 56 ..if right... will explain

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 [#permalink] New post 22 Mar 2004, 17:51
CJ, CB, JB, CJB

Just 4 mixtures.
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 [#permalink] New post 22 Mar 2004, 17:56
81? How?
Also, even if there are four mixtures Halle, the answer is off. I was pretty confused by this problem, thats why I think its important to see how any of us would approach this problem.

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 [#permalink] New post 22 Mar 2004, 18:09
CJB+CJ+BJ+CB+C+J+B = X

CJB = 5
CJ = 16 - 5 = 11
BJ = 18 - 5 = 13
CB = 8 - 5 = 3
C = 33 - 11 - 5 - 3 = 14
J = 43 - 13 - 5 - 11 = 14
B = 42 - 3 - 5 - 13 = 21

X = 5 + 11 + 13 + 3 + 14 + 14 + 21 = 81

It is much easier to show with a Venn Diagram but I don't know how to do it. Maybe Gmatblast can. Also, I don't like the wording of the question too much. ie. "16 contain at least C & J" Why "at least"? Why not just "16 contain C & J"

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 [#permalink] New post 22 Mar 2004, 19:17
Paul wrote:
CJB+CJ+BJ+CB+C+J+B = X

CJB = 5
CJ = 16 - 5 = 11
BJ = 18 - 5 = 13
CB = 8 - 5 = 3
C = 33 - 11 - 5 - 3 = 14
J = 43 - 13 - 5 - 11 = 14
B = 42 - 3 - 5 - 13 = 21

X = 5 + 11 + 13 + 3 + 14 + 14 + 21 = 81

It is much easier to show with a Venn Diagram but I don't know how to do it. Maybe Gmatblast can. Also, I don't like the wording of the question too much. ie. "16 contain at least C & J" Why "at least"? Why not just "16 contain C & J"


Paul, your approach is absolutely correct. The reason it states at least is to show that some of those 16 can contain more then just two blends. If the problem stated "16 contain C & J" then C & J is not =16-5, but rather 21-5. In other words, the area of two circles in Venn digram that contains C & J only would be 16 and not 11, and the total would change.

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 [#permalink] New post 22 Mar 2004, 19:24
Paul wrote:
CJB+CJ+BJ+CB+C+J+B = X

CJB = 5
CJ = 16 - 5 = 11
BJ = 18 - 5 = 13
CB = 8 - 5 = 3
C = 33 - 11 - 5 - 3 = 14
J = 43 - 13 - 5 - 11 = 14
B = 42 - 3 - 5 - 13 = 21

X = 5 + 11 + 13 + 3 + 14 + 14 + 21 = 81

It is much easier to show with a Venn Diagram but I don't know how to do it. Maybe Gmatblast can. Also, I don't like the wording of the question too much. ie. "16 contain at least C & J" Why "at least"? Why not just "16 contain C & J"


Paul, i accord your solution, I arrived at the same juncture using Ven Diagram.
BUT, the stem asks for "MIOXTURES OF COFFEES" and not toal no of coffee brands.
I think answer should be,
CJB + CJ + BJ + CB = 5 + 11 + 13 + 3 = 32.

PAUL, the word ATLEST has a subtle difference, it can't be ignored. ATLEAST tells you to, subtract n(J n B n C) from n(J n B) to count the no of coffee blends available with EXACT 2 Flavors in it.
I hope i am clear enough.

lvb9th, Please post official answer.

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 [#permalink] New post 22 Mar 2004, 19:36
Dharmin wrote:
Paul wrote:
CJB+CJ+BJ+CB+C+J+B = X

CJB = 5
CJ = 16 - 5 = 11
BJ = 18 - 5 = 13
CB = 8 - 5 = 3
C = 33 - 11 - 5 - 3 = 14
J = 43 - 13 - 5 - 11 = 14
B = 42 - 3 - 5 - 13 = 21

X = 5 + 11 + 13 + 3 + 14 + 14 + 21 = 81

It is much easier to show with a Venn Diagram but I don't know how to do it. Maybe Gmatblast can. Also, I don't like the wording of the question too much. ie. "16 contain at least C & J" Why "at least"? Why not just "16 contain C & J"


Paul, i accord your solution, I arrived at the same juncture using Ven Diagram.
BUT, the stem asks for "MIOXTURES OF COFFEES" and not toal no of coffee brands.
I think answer should be,
CJB + CJ + BJ + CB = 5 + 11 + 13 + 3 = 32.

PAUL, the word ATLEST has a subtle difference, it can't be ignored. ATLEAST tells you to, subtract n(J n B n C) from n(J n B) to count the no of coffee blends available with EXACT 2 Flavors in it.
I hope i am clear enough.

lvb9th, Please post official answer.

Dharmin


I refer you to the problem: All of the coffee mixtures sold in a certain store contain either Columbian (C), Jamaican (J), or Brazilian coffee, or some combination of those.
This means that any of C, J or B alone will be considered a mixture.

And yes guys, "at least" is crucial. Just figured it out :)

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 [#permalink] New post 22 Mar 2004, 19:45
Dharmin,

Pauls answer is BEST. I also agree with you that the question stem is phrased incorrectly. But, the main issue here is to recognize the AT LEAST part and arrive at the right conclusion. If in fact we ignored the at least part, we would've arrived at the wrong answer. The most important fact is, that the wrong answer might be included in the real GMAT.

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 [#permalink] New post 23 Mar 2004, 09:27
Thats was excellent Paul...If I recall it correctly...this is a Kaplan problem...and one of the hardest...you solved it very well...way to go Paul..!

And yeah... I think sometime back even Anand had suggested to change ur signature to 750+ quest...I guess it time you do that.


Vivek.

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 [#permalink] New post 24 Mar 2004, 01:38
shubhangi wrote:
is it 56 ..if right... will explain
:roll: :roll: :roll: :roll: :roll: :roll:

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  [#permalink] 24 Mar 2004, 01:38
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