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Averages

by Stuti567 » Mon Jul 04, 2016 3:03 am
55 people sit at a round table. Rs 220 is to be distributed among them so that each person receives an amount which is the average of the amounts with his immediate neighbours. In how many ways can this be achieved?

(a) 55C3 x 4!
(b) 0
(c) 4
(d) 1
(e) 55

Please answer with explanation.

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by [email protected] » Mon Jul 04, 2016 9:27 am
Hi Stuti567,

Is this a GMAT practice question? I ask because it doesn't have the 'style' of one (and if you're actually studying for the GMAT, then you might want to invest in more realistic practice materials). That having been said, this is more of a 'concept' question than anything else - and you can get to the correct answer by 'playing around' with the concept a little.

Given the details in the prompt, there is one obvious way to guarantee that each person has an amount that is the AVERAGE of the two people sitting next to him - if everyone has the exact SAME number of Rs (in this case, that would be 220/55 = 4). 4 is the average of 4 and 4.

Now, if you try to change ANY of those 4s to anything else, think about what would happen...

4 4 4 4 4 4 4 4 ....

If we make one of those 4s into a 3, then that 3 would need to be surrounded by a 2 and a 4 (since the average of 2 and 4 is 3).

_ _ _ 2 3 4 _ _ _

And there would have to be a 1 on the other side of that 2 (since the average of 1 and 3 is 2)

_ _ 1 2 3 4 _ _ _

And then there would have to be a 0 on the other side of the 1....

_ 0 1 2 3 4 _ _ _

But what's on the other side of the 0? -1. How can one of these people have negative Rs? It's not possible, thus you CANNOT change any of the 4s to any other number. That means there's only one possible solution to this question.

Final Answer: D

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by Stuti567 » Mon Jul 04, 2016 8:53 pm
[email protected] wrote:Hi Stuti567,

Is this a GMAT practice question? I ask because it doesn't have the 'style' of one (and if you're actually studying for the GMAT, then you might want to invest in more realistic practice materials). That having been said, this is more of a 'concept' question than anything else - and you can get to the correct answer by 'playing around' with the concept a little.

Given the details in the prompt, there is one obvious way to guarantee that each person has an amount that is the AVERAGE of the two people sitting next to him - if everyone has the exact SAME number of Rs (in this case, that would be 220/55 = 4). 4 is the average of 4 and 4.

Now, if you try to change ANY of those 4s to anything else, think about what would happen...

4 4 4 4 4 4 4 4 ....

If we make one of those 4s into a 3, then that 3 would need to be surrounded by a 2 and a 4 (since the average of 2 and 4 is 3).

_ _ _ 2 3 4 _ _ _

And there would have to be a 1 on the other side of that 2 (since the average of 1 and 3 is 2)

_ _ 1 2 3 4 _ _ _

And then there would have to be a 0 on the other side of the 1....

_ 0 1 2 3 4 _ _ _

But what's on the other side of the 0? -1. How can one of these people have negative Rs? It's not possible, thus you CANNOT change any of the 4s to any other number. That means there's only one possible solution to this question.

Final Answer: D

GMAT assassins aren't born, they're made,
Rich
Hi,

Thanks for the explanation.

This is not a gmat practice question, however, I found it somewhere and I thought this would help build my concept on averages which is why I decided to post it here. Even though it made sense that there should only be one possible arrangement, I wasn't very convinced. Your explanation makes sense and given me an idea on how to approach questions based on averages in general.
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by Matt@VeritasPrep » Thu Jul 07, 2016 3:49 pm
Seems like it could be a GMAT question to me: it has a very clever solution that doesn't require much technical ability, but DOES require some pretty nifty analytical reasoning.
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by ceilidh.erickson » Sat Jul 09, 2016 9:40 am
Matt@VeritasPrep wrote:Seems like it could be a GMAT question to me: it has a very clever solution that doesn't require much technical ability, but DOES require some pretty nifty analytical reasoning.
Well... except that this question has notation that would NEVER be found on the GMAT. GMAT combinatoric problems will never use abbreviations like 55C3.
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by danielle07 » Thu Aug 31, 2017 3:44 am
The answer would be C. 1
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by Matt@VeritasPrep » Thu Aug 31, 2017 4:18 pm
ceilidh.erickson wrote:
Matt@VeritasPrep wrote:Seems like it could be a GMAT question to me: it has a very clever solution that doesn't require much technical ability, but DOES require some pretty nifty analytical reasoning.
Well... except that this question has notation that would NEVER be found on the GMAT. GMAT combinatoric problems will never use abbreviations like 55C3.
True, but ...

1) I didn't say it was an official GMAT problem, only that it was in scope;
2) No poster would ever misabbreviate, mistype, or otherwise distort an answer choice on this forum :)
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by ceilidh.erickson » Thu Aug 31, 2017 5:23 pm
Matt@VeritasPrep wrote:
ceilidh.erickson wrote:
Matt@VeritasPrep wrote:Seems like it could be a GMAT question to me: it has a very clever solution that doesn't require much technical ability, but DOES require some pretty nifty analytical reasoning.
Well... except that this question has notation that would NEVER be found on the GMAT. GMAT combinatoric problems will never use abbreviations like 55C3.
True, but ...

1) I didn't say it was an official GMAT problem, only that it was in scope;
2) No poster would ever misabbreviate, mistype, or otherwise distort an answer choice on this forum :)
Ha! Too true.
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