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Of the 8 people having a meal, 5 are having fries, 4 are having burger

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Of the 8 people having a meal, 5 are having fries, 4 are having burger  [#permalink]

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New post 07 Jun 2017, 11:23
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Question Stats:

78% (01:47) correct 22% (01:50) wrong based on 94 sessions

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Of the 8 people having a meal, 5 are having fries, 4 are having burgers and 3 are having bacon. No other food is at the table. Each person is eating at least 1 item. If exactly 4 people are eating exactly 2 food items, how many people are eating all 3 items?

A- 0
B- 1
C- 2
D- 3
E- 4

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Re: Of the 8 people having a meal, 5 are having fries, 4 are having burger  [#permalink]

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New post 07 Jun 2017, 11:32
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Ans :A
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Re: Of the 8 people having a meal, 5 are having fries, 4 are having burger  [#permalink]

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New post 05 Sep 2018, 07:39
1
rohan2345 wrote:
Of the 8 people having a meal, 5 are having fries, 4 are having burgers and 3 are having bacon. No other food is at the table. Each person is eating at least 1 item. If exactly 4 people are eating exactly 2 food items, how many people are eating all 3 items?

A- 0
B- 1
C- 2
D- 3
E- 4



Dear Moderator,
Found this overlapping set problem in the combination section, hope you will look into this. Thank you.
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Re: Of the 8 people having a meal, 5 are having fries, 4 are having burger  [#permalink]

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New post 05 Sep 2018, 07:46
stne wrote:
rohan2345 wrote:
Of the 8 people having a meal, 5 are having fries, 4 are having burgers and 3 are having bacon. No other food is at the table. Each person is eating at least 1 item. If exactly 4 people are eating exactly 2 food items, how many people are eating all 3 items?

A- 0
B- 1
C- 2
D- 3
E- 4



Dear Moderator,
Found this overlapping set problem in the combination section, hope you will look into this. Thank you.

_______________
Edited. Thank you.
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Of the 8 people having a meal, 5 are having fries, 4 are having burger  [#permalink]

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New post 05 Sep 2018, 08:33
rohan2345 wrote:
Of the 8 people having a meal, 5 are having fries, 4 are having burgers and 3 are having bacon. No other food is at the table. Each person is eating at least 1 item. If exactly 4 people are eating exactly 2 food items, how many people are eating all 3 items?

A- 0
B- 1
C- 2
D- 3
E- 4


Immediate application of the simplifier:

\(A \cup B \cup C = A + B + C - \left[ {\sum {{\text{Exactly}}2} } \right] - 2 \cdot \left[ {{\text{Exactly3}}} \right]\)

In our case (see figure attached):

\(8 = {\text{F}} \cup {\text{Bu}} \cup {\text{Ba}}\mathop = \limits^{\left( * \right)} 12 - 4 - 2x\,\,\,\,\, \Rightarrow \,\,\,? = x = 0\)


(All numerical alternative choices - we may explore a particular case, as we did.)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
Attachments

05Set18_3w.gif
05Set18_3w.gif [ 27.81 KiB | Viewed 420 times ]


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Of the 8 people having a meal, 5 are having fries, 4 are having burger   [#permalink] 05 Sep 2018, 08:33
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