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# Orange Computers is breaking up its conference attendees ...

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Current Student
Joined: 06 Sep 2013
Posts: 1978

Kudos [?]: 719 [0], given: 355

Concentration: Finance
Orange Computers is breaking up its conference attendees ... [#permalink]

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18 Sep 2013, 17:10
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Question Stats:

100% (01:27) correct 0% (00:00) wrong based on 8 sessions

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Orange Computers is breaking up its conference attendees into groups. Each group must have exactly one person from Division A, two people from Division B, and three people from Division C. There are 20 people from Division A, 30 people from Division B, and 40 people from Division C at the conference. What is the smallest number of people who will not be able to be assigned to a group?

Kudos [?]: 719 [0], given: 355

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7676

Kudos [?]: 17381 [1], given: 232

Location: Pune, India
Re: Orange Computers is breaking up its conference attendees ... [#permalink]

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18 Sep 2013, 19:31
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Expert's post
jlgdr wrote:
Orange Computers is breaking up its conference attendees into groups. Each group must have exactly one person from Division A, two people from Division B, and three people from Division C. There are 20 people from Division A, 30 people from Division B, and 40 people from Division C at the conference. What is the smallest number of people who will not be able to be assigned to a group?

Each Group - 1 from Division A, 2 from Division B, 3 from Division C

Division A has 20 people so we can make 20 groups.
Division B has 30 people so we can make only 30/2 = 15 groups (Each group must have 2 people from B)
Division C has 40 people so we can make only 40/3 = 13 groups (Each group must have 3 people from C)

We can make only 13 groups since division C will run out of people after that.

Leftover people:
1 from C
30 - 13*2 = 4 from B
20 - 13 = 7 from A
A total of 12 people.
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Kudos [?]: 17381 [1], given: 232

Re: Orange Computers is breaking up its conference attendees ...   [#permalink] 18 Sep 2013, 19:31
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