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End the PS Party With this and Go Sit Your GMAT

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End the PS Party With this and Go Sit Your GMAT

by dtweah » Tue May 12, 2009 6:24 am
After all twenty participants in a figure skating tournament skated, each of the 9 judges ordered
the participants from place 1 (the best) to place 20 (the worst). It turned out that for each
participant, the places assigned by different judges were not more than 3 apart. The sum of the
places for each participant was calculated and the sums were ordered: c1 ≤ c2 ≤ ... ≤ c20. What
is the largest possible value of c1?

a. 18
b. 19
c. 21
d. 22
e. 24
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by avenus » Tue May 12, 2009 2:24 pm
I can get a combination that yields 24, although I would need to think a bit more to prove that it is the maximum in a compelling way. I'm not positive. Since 24 is the greatest of the answers provided, I'll go for E.
In the matrix below, the nth column represents the number of judges that awarded a mark n

c1 2 2 2 3
c2 2 2 2 3
c3 2 2 2 3
c4 3 3 3 2

and then you organise the other ones as you like, for example:
Cij = i with i=5,6,...,20 j=1,2,...,9
Cij is the mark ith skater received from judge j
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by joegmat680 » Fri May 15, 2009 5:29 am
the answer should be C 21
think about it, C1 has to be have the number 1 so the greatest it can become is 4. So the way to give the most amount to C1 is to make C1 to C4 as equal as possible. That is to give each one (C1, C2, C3 C4) two votes of 1 , two votes of 2, two votes of 3, and two votes of 4. and then in order to make C1 the smallest, the ninth vote should be C1 1 C2 2 C3 3 C4 4
It should look like this:

C1 C2 C3 C4
4 3 2 1
4 3 2 1
3 2 1 4
3 2 1 4
2 1 4 3
2 1 4 3
1 4 3 2
1 4 3 2

so far the sums are equal. so for the final vote it would be:

C1 C2 C3 C4
1 2 3 4

Add up C1's votes and you have 21 :D
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Re: End the PS Party With this and Go Sit Your GMAT

by dtweah » Fri May 15, 2009 6:28 am
dtweah wrote:After all twenty participants in a figure skating tournament skated, each of the 9 judges ordered
the participants from place 1 (the best) to place 20 (the worst). It turned out that for each
participant, the places assigned by different judges were not more than 3 apart. The sum of the
places for each participant was calculated and the sums were ordered: c1 ≤ c2 ≤ ... ≤ c20. What
is the largest possible value of c1?

a. 18
b. 19
c. 21
d. 22
e. 24
OA

If all judges gave 1st place to the same participant, then c1 = 9. If
exactly 2 people got 1st places, one of them got at least 5 ones. His
score is at most 5×1+4×4 (the worst he can get from other judges).
So c1 ≤ 5+16 = 21. If 3 people got 1st places, the sum of their scores is
at most 9×1+9×4+9×3 = 72. So c1 ≤ 24. If 4 people got 1st places,
then the sum of their scores is at least 9 × 1 + 9 × 4 + 9 × 3 + 9 × 2 =
90, and c1 ≤ 22. Five or more 1st places is not possible. Example
for c1 = 24: Each of the top 3 gets the scores 1,1,1,3,3,3,4,4,4 (not
in the same order). The next gets 2,2,2,2,2,5,5,5,5 and the next gets
5,5,5,5,5,2,2,2,2. The rest get arbitrary grades with difference ≤ 3.
This gives c1 = 24. The answer is e.
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by joegmat680 » Fri May 15, 2009 9:07 am
Posted: Fri May 15, 2009 7:28 am Post subject: Re: End the PS Party With this and Go Sit Your GMAT
dtweah wrote:
After all twenty participants in a figure skating tournament skated, each of the 9 judges ordered
the participants from place 1 (the best) to place 20 (the worst). It turned out that for each
participant, the places assigned by different judges were not more than 3 apart. The sum of the
places for each participant was calculated and the sums were ordered: c1 ≤ c2 ≤ ... ≤ c20. What
is the largest possible value of c1?

a. 18
b. 19
c. 21
d. 22
e. 24


OA

If all judges gave 1st place to the same participant, then c1 = 9. If
exactly 2 people got 1st places, one of them got at least 5 ones. His
score is at most 5×1+4×4 (the worst he can get from other judges).
So c1 ≤ 5+16 = 21. If 3 people got 1st places, the sum of their scores is
at most 9×1+9×4+9×3 = 72. So c1 ≤ 24. If 4 people got 1st places,
then the sum of their scores is at least 9 × 1 + 9 × 4 + 9 × 3 + 9 × 2 =
90, and c1 ≤ 22. Five or more 1st places is not possible. Example
for c1 = 24: Each of the top 3 gets the scores 1,1,1,3,3,3,4,4,4 (not
in the same order). The next gets 2,2,2,2,2,5,5,5,5 and the next gets
5,5,5,5,5,2,2,2,2. The rest get arbitrary grades with difference ≤ 3.
This gives c1 = 24. The answer is e.

I'm not sure you're calculating correctly. As you mentioned, if there are four first place winners the sum of their scores should be 90. If C1 is the least from the individual sums, then the individual sums should look like this :
C1- 21
C2- 22
C3- 23
c4- 24
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by avenus » Sat May 16, 2009 8:17 am
Cool OE. It wouldn't hurt if they elaborated a bit more, though...
I think where it reads:
If 4 people got 1st places, then the sum of their scores is at least 9 × 1 + 9 × 4 + 9 × 3 + 9 × 2 = 90, and c1 ≤ 22.
it should read is exactly. If there are exactly 4 people with 1st places, the sum of their scores must be 90.

@dtweah
This is an interesting question, so are many of the ones you post. Where do you get them?

@joegmat680
I'm not sure you're calculating correctly. As you mentioned, if there are four first place winners the sum of their scores should be 90. If C1 is the least from the individual sums, then the individual sums should look like this :
C1- 21
C2- 22
C3- 23
c4- 24
OE provides you with a valid example that yields c1 = 24, so there's not much point in arguing...

.
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