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# Each member in a club with 90 members voted for exactly one

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Senior Manager
Joined: 01 May 2004
Posts: 337
Location: USA
Followers: 1

Kudos [?]: 20 [0], given: 0

Each member in a club with 90 members voted for exactly one [#permalink]  14 Jul 2004, 09:01
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Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Each member in a club with 90 members voted for exactly one of 4 candidates A, B, C or D. Did candidate A get the most votes?
Manager
Joined: 16 Jan 2004
Posts: 65
Location: NJ
Followers: 1

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

B.

B+c+D= 69

Now, evenif you make everything 21 (which is A) themax value is 63 hence A did not receive maximum votes. I remember seeing this question somewhere
Director
Joined: 01 Feb 2003
Posts: 854
Followers: 1

Kudos [?]: 35 [0], given: 0

Hmmm....that's called falling flat on my face (over confidence is attimes a killer)
Manager
Joined: 21 Jun 2004
Posts: 52
Followers: 1

Kudos [?]: 1 [0], given: 0

Can some one explain in littel more detail.

How did we come to the conluciosn that each could have received 21 votes. Is't there a possibility that one memebr received 0 votes, then what happens
CIO
Joined: 09 Mar 2003
Posts: 464
Followers: 2

Kudos [?]: 34 [0], given: 0

We don't know what each one got. But we want to know if A got the most, when there are 90 votes to go around. Number two says that A got 21, which leaves 69 votes for the other three candidates. With that many votes, someone in the other three has to be higher than 21, if not all of them, so the answer is no, which is an answer.

Intern
Joined: 07 Jun 2004
Posts: 29
Location: Sunnyvale
Followers: 0

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

Let's see if I got this straight. [#permalink]  15 Jul 2004, 10:47
Ok, so here is what I understand from what you guys are saying.

We know A=21
We therefore know B+C+D=69

The question never states how they could have voted.
Consequently, we could have seen a multitude of possibilities for B+C+D
For example,

1+1+67=69 ---> one is higher than 21, thus A is not highest
22+1+46=69 --->one is higher than 21, thus A is not highest
23+23+23=69 ---> all three are higher than 21, thus A is not higest.

I think the key to the answer is knowing
1) Disporportioned possibilities always leads to at least one higher than A
2) Evenly porportioned possibilities always leads to all three higher than A

I think one of the explanations someone gave explained the disporportioned possibility as a reason to choose B, but not taking into account the porportioned possibility could have still gotten the question wrong.

But since the porportioned result happens to correlate with what a disporportioned result is, we can still choose B as the answer.

This was a good question to think through. Thanks for question and dialogue guys. Amongst your conversation, I was able to conclude with what I just wrote. Very helpful. Thanks!
But this last option is the rea
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Davefor MBA

Intern
Joined: 13 Jul 2004
Posts: 14
Location: Melbourne
Followers: 0

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

my explanation is just simple like this:
the average of these 4 is 90/4 = 22.5, therefore 21 cant be the largest one. hence B answers the question straight away.
_________________

everyday is a new endeavor

Manager
Joined: 21 Jun 2004
Posts: 52
Followers: 1

Kudos [?]: 1 [0], given: 0

Thank you guys. It makes sense now
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