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# In a restaraunt, there are 3 popular dishes A, B and C . On

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Senior Manager
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In a restaraunt, there are 3 popular dishes A, B and C . On [#permalink]

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02 Jul 2005, 11:07
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In a restaraunt, there are 3 popular dishes A, B and C . On a given evening, 71% of the diners ate Recipe A, 84% ate B and 86% ate C. The percentage of diners who ate all the dishes must be greater than what what number?

A. 47%

B. 42%

C. 41%

D. 40%

E. 49%
Manager
Joined: 28 Aug 2004
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02 Jul 2005, 11:23
AJB77 wrote:
In a restaraunt, there are 3 popular dishes A, B and C . On a given evening, 71% of the diners ate Recipe A, 84% ate B and 86% ate C. The percentage of diners who ate all the dishes must be greater than what what number?

A. 47%

B. 42%

C. 41%

D. 40%

E. 49%

Let's look at it this way, what's the least pssible number of eaters who ate 3 dishes.

we have 71+84+86=241

for the least of 3 dishes eaters, we have maximum eaters of 2 dishes, that's 200

we're left with 41. so it must be greater than 40, D.
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02 Jul 2005, 17:25
Dan wrote:
AJB77 wrote:
In a restaraunt, there are 3 popular dishes A, B and C . On a given evening, 71% of the diners ate Recipe A, 84% ate B and 86% ate C. The percentage of diners who ate all the dishes must be greater than what what number?

A. 47%

B. 42%

C. 41%

D. 40%

E. 49%

Let's look at it this way, what's the least pssible number of eaters who ate 3 dishes.

we have 71+84+86=241

for the least of 3 dishes eaters, we have maximum eaters of 2 dishes, that's 200

we're left with 41. so it must be greater than 40, D.

plz Dan
how do u know that ""we have maximum eaters of 2 dishes, that's 200""
thanks
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03 Jul 2005, 12:22
that's an assumption and not an actual figure since the stem asks for the minimum number...

suppose you've 100 people (100%) and each eats out 2 dishes ==> that's 200 for a total...

I am assuming this is the OA, but who knows!?

Director
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03 Jul 2005, 12:48
ABC = AUBUC - (A + B + C) + (AB + BC + AC) = 100 - (71 + 84 + 86) + (AB + BC + AC)

ABC = (AB + BC + AC) - 141

now we need to do more transformations (for example AC~B means intersection of AC excluding B)

ABC = (AB~C + BC~A + AC~B + 3ABC) - 141

(AB~C + BC~A + AC~B) = 141 - 2ABC

now the greates value (AB~C + BC~A + AC~B) can take is 100 - ABC (draw a Venn diagram when everyone had at least 2 recipes )

so (AB~C + BC~A + AC~B) = 141 - 2ABC =< 100 - ABC

=> 141-100 =< ABC
or ABC >=41

Director
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03 Jul 2005, 12:51
Actually one doesn't have to solve the problem at all since the lowest number is 40%, so the correct answer will always be greater than 40%.
so the answer must be D.
03 Jul 2005, 12:51
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