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stolyar, paul, stoolfi, praetorian and akamaibrah are

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anandnk
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stolyar, paul, stoolfi, praetorian and akamaibrah are [#permalink] New post   11 Feb 2004, 17:21
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stolyar, paul, stoolfi, praetorian and akamaibrah are friends and hosting individual newyear parties. Folks from GMAT club are invited to join.
Stolyar can accomodate 25
Paul can accomodate 30
Stoolfi can acomodate 40
Praetorian can accomodate 15
Akamaibrah can accomodate 10

They are yet to send out personal invitations. Every person who receives invitation will attend 100%.

They together (5 of them ) consider it as successfull year only if atleast 3 of them can attract as many members as they can accomodate. What is the minimum number of invitations that should be sent out.

Show your work fellas.



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Re: stolyar, paul, stoolfi, praetorian and akamaibrah are [#permalink] New post   11 Feb 2004, 18:16
Hmmm, I wonder if this question has some kind of trap. I also got 50 but is it really that simple? If yes then the explanation is that, provided that every card sent has 100% positive response, we only need to satisfy Akamaibrah, Praetorian and Stolyar's accomodation capacity. Therefore, 10+15+25=50. I would be very surprised if it were this simple though...
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Re: stolyar, paul, stoolfi, praetorian and akamaibrah are [#permalink] New post   Updated on: 11 Feb 2004, 20:35
I think I didnt state the problem properly.

A collective invitation is sent out. People start filling the houses randomly.

I am loving it! :smoke

Originally posted by anandnk on 11 Feb 2004, 18:48.
Last edited by anandnk on 11 Feb 2004, 20:35, edited 1 time in total.
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Re: stolyar, paul, stoolfi, praetorian and akamaibrah are [#permalink] New post   11 Feb 2004, 18:59
Oh, I see now. Then I think the answer is:
minimum # of people needed to make the event successful is Akamaibrah, Praetorian and Stolyar's accomodation capacity. Therefore, 10+15+25=50. Probability of anyone attending 1 of the 5 houses is 1/5. Then 50 / [1/5] = 250 invitations should be sent out at the minimum
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Re: stolyar, paul, stoolfi, praetorian and akamaibrah are [#permalink] New post   11 Feb 2004, 20:26
Paul wrote:
Oh, I see now. Then I think the answer is:
minimum # of people needed to make the event successful is Akamaibrah, Praetorian and Stolyar's accomodation capacity. Therefore, 10+15+25=50. Probability of anyone attending 1 of the 5 houses is 1/5. Then 50 / [1/5] = 250 invitations should be sent out at the minimum


maximum capacity (all 5 hosts) is 120.
So, it can't be 250.

Could it be 95.
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Re: stolyar, paul, stoolfi, praetorian and akamaibrah are [#permalink] New post   11 Feb 2004, 20:44
Yes, could not find out an answer right now but 250 is wrong. I again made a stupid mistake and interpreted the premise as probability of people accepting is 1 out of 5. Very bad. Sleep now. I'll think about it tomorrow unless someone comes up with the answer before that
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Re: stolyar, paul, stoolfi, praetorian and akamaibrah are [#permalink] New post   11 Feb 2004, 21:06
Is the answer 118?
What kind of problem is it?
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Re: stolyar, paul, stoolfi, praetorian and akamaibrah are [#permalink] New post   11 Feb 2004, 21:12
Gee, I'm back to this problem when I have to be sleeping now. I can't think of an answer but yet can't sleep without figuring it out. This is gonna haunt me tonight. :horror
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Re: stolyar, paul, stoolfi, praetorian and akamaibrah are [#permalink] New post   11 Feb 2004, 21:21
The question is asking us to think from worst case scenario.

Stolyar can accomodate 25
Paul can accomodate 30
Stoolfi can acomodate 40
Praetorian can accomodate 15
Akamaibrah can accomodate 10

Total = 120

Assume that every one got one less than their maximum capacity.
No one host got a full house.

Add one more guest , alteast one got full house.
Add anoither one, atleast two got full house.
Add a third guest, atleast three got full house.

Thus, minimum number of invitations = 118 should be to make sure
atleast three host got full house.
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Re: stolyar, paul, stoolfi, praetorian and akamaibrah are [#permalink] New post   11 Feb 2004, 21:40
I agree with kpadma, It should be 118 invitations.

First Stolyar's house is full - 25
Next Paul's house is full - 30

If we assume Stoolfi has only -39 guests
Praetorian has only - 14 guests
Akamaibrah has only - 9 guests

Total is 117 now. If we add one more invitation we should have at least three houses full in anyway.
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Re: stolyar, paul, stoolfi, praetorian and akamaibrah are [#permalink] New post   12 Feb 2004, 04:54
What Kpadma said makes sense. I slept with this problem in mind and came up with 10*5 + 4*5 + 3*10 = 100 invitations sent out provided that there is an even probability on the invitation cards of each one being chosen. But 118 seems right.
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Re: stolyar, paul, stoolfi, praetorian and akamaibrah are [#permalink] New post   12 Feb 2004, 06:22
looks like kpadma and geethu hit the first nail on the coffin

good show folks keep it up.



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Re: stolyar, paul, stoolfi, praetorian and akamaibrah are [#permalink]
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