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Each • in the mileage table above represents an entry indica [#permalink]
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 02 Jun 2010, 10:32
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Each • in the mileage table above represents an entry indicating the distance between a pair of the five cities. If the table were extended to represent the distances between all pairs of 30 cities and each distance were to be represented by only one entry, how many entries would the table then have? (A) 60 (B) 435 (C) 450 (D) 465 (E) 900
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Last edited by Bunuel on 12 Dec 2012, 10:09, edited 3 times in total.
Renamed the topic and edited the question.
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Re: PS Q: OG-12 #116 [#permalink]
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02 Jun 2010, 11:42
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snkrhed wrote: So there's a chart that looks a lot like this:
-E D C B A
A * * * *
B * * *
C * *
D *
E
Each * in the mileage table above represents an entry indicating the distance between all pairs of 30 cities and each distance were to be represented by only one entry, how many entries would the table then have?
(A) 60
(B) 435
(C) 450
(D) 465
(E) 900 City B, the second city has 1 point City C the third city has 2 points City D, the fourth city has 3 points What's the pattern? Number of cities minus 1 so the 30th city is going to have 29 points Then it becomes a matter of adding the consecutive integers from 1 to 29 The sum is the average * number of terms average = 15 number of terms = 29 29*15 = 435
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Each • in the mileage table above represents an entry indica [#permalink]
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02 Jun 2010, 11:54
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Re: PS Q: OG-12 #116 [#permalink]
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30 Jun 2010, 02:16
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Because the GMAT is multiple choice.
I first found the pattern mentioned each additional city adds (total cities - 1) to the table.
So I figure with 30 cities the last 3 cities added 29+28+27 to our total number of entries so (A) is ridiculous
And I knew the whole table would be 30x30 with 900 entries. Since I know I will not have entries for each box I ruled out (E) 900
Looking at the remaining choices I thought
(B) 435 = less than half the table is filled
(C) 450 = exactly half the table is filled
(D) 465 = more than half the table is filled
The table given shows a 5x5 table with only 10 entries. 10< .5(25)
So I chose (B).
This method works for these answer choices, but if the choices were 430, 435, 440 I would be screwed right?
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Re: PS Q: OG-12 #116 [#permalink]
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30 Jun 2010, 07:45
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This can be done in n * (n - 1) / 2 ways.
Hence -> 30 * 29 / 2 = 435 ways.
Correct answer choice is B. Thank You.
Thanks,
Akhil M.Parekh
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Re: PS Q: OG-12 #116 [#permalink]
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03 Jan 2013, 12:54
Bunuel wrote: snkrhed wrote: Attachment: img.jpg Each dot in the mileage table above represents an entry indicating the distance between a pair of the five cities. If the table were extended to represent the distances of 30 cities and each distance were to be represented by only one entry, how many entries would the table then have? (A) 60 (B) 435 (C) 450 (D) 465 (E) 900 We are there told that there should be one entry for each pair. How many entries would the table then have? Or how many different pairs can 30 cities give? \(C^2_{30}=435\) Answer: B. Hello, Bunuel Which formula did you use here ? Thanks,
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Re: Each • in the mileage table above represents an entry indica [#permalink]
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03 Jan 2013, 13:37
Combinations formula.
30 C 2. Answer 435.
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Re: Each • in the mileage table above represents an entry indica [#permalink]
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10 Nov 2013, 23:42
How would this question be solved using a consecutive integer format? Can you find the average on the consecutive integers and then multiply by the number of terms? I ask because this question is listed as a consecutive integer question in the MGAMT quant guide.
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Re: Each • in the mileage table above represents an entry indica [#permalink]
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stevennu wrote: How would this question be solved using a consecutive integer format? Can you find the average on the consecutive integers and then multiply by the number of terms? I ask because this question is listed as a consecutive integer question in the MGAMT quant guide. If second entry =1 third entry = 2 30th entry = 29 etc Thus S(n)=n/2(2a+(n-1)d) where a=1, d=1, n=29 plug in and you get the answer.
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Re: PS Q: OG-12 #116 [#permalink]
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13 Apr 2014, 12:30
Bunuel wrote: snkrhed wrote: Attachment: img.jpg Each dot in the mileage table above represents an entry indicating the distance between a pair of the five cities. If the table were extended to represent the distances of 30 cities and each distance were to be represented by only one entry, how many entries would the table then have? (A) 60 (B) 435 (C) 450 (D) 465 (E) 900 We are there told that there should be one entry for each pair. How many entries would the table then have? Or how many different pairs can 30 cities give? \(C^2_{30}=435\) Answer: B. Hi Bunuel, Can you please elaborate on how this formula works? Thanks! EDIT: I did it via the table method but i've seen your formula pop up quite often and I'm failing miserably at it. That might explain the horrible score in NP.  I understand what formula to use but i'm having a hard time connecting the formula to the problem "\(C^n_k = \frac{n!}{k!(n-k)!}\)"
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Re: PS Q: OG-12 #116 [#permalink]
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14 Apr 2014, 01:04
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russ9 wrote: Bunuel wrote: snkrhed wrote: Attachment: img.jpg Each dot in the mileage table above represents an entry indicating the distance between a pair of the five cities. If the table were extended to represent the distances of 30 cities and each distance were to be represented by only one entry, how many entries would the table then have? (A) 60 (B) 435 (C) 450 (D) 465 (E) 900 We are there told that there should be one entry for each pair. How many entries would the table then have? Or how many different pairs can 30 cities give? \(C^2_{30}=435\) Answer: B. Hi Bunuel, Can you please elaborate on how this formula works? Thanks! EDIT: I did it via the table method but i've seen your formula pop up quite often and I'm failing miserably at it. That might explain the horrible score in NP. I understand what formula to use but i'm having a hard time connecting the formula to the problem "\(C^n_k = \frac{n!}{k!(n-k)!}\)" \(C^2_{30}\) is choosing 2 out of 30. There are 30 cities and each pair of cities need an entry, hence 30 cites need \(C^2_{30}\) entries. Hope it's clear.
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Re: PS Q: OG-12 #116 [#permalink]
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09 May 2014, 15:22
Bunuel wrote:
\(C^2_30\) is choosing 2 out of 30. There are 30 cities and each pair of cities need an entry, hence 30 cites need \(C^2_30\) entries.
Hope it's clear.
Hi Bunuel, Unfortunately, still not clear. Why are we choosing 2 out of 30?
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Re: PS Q: OG-12 #116 [#permalink]
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10 May 2014, 05:06
russ9 wrote: Bunuel wrote:
\(C^2_30\) is choosing 2 out of 30. There are 30 cities and each pair of cities need an entry, hence 30 cites need \(C^2_30\) entries.
Hope it's clear.
Hi Bunuel, Unfortunately, still not clear. Why are we choosing 2 out of 30? Consider the table given in the original post:  A and B have 1 entry; A and C have 1 entry; A and D have 1 entry; A and E have 1 entry; B and C have 1 entry; B and D have 1 entry; B and E have 1 entry; C and D have 1 entry; C and E have 1 entry; D and E have 1 entry. So, each pair of letters from {A, B, C, D, E} has 1 entry, total of 10 entries. How many pairs can we have? \(C^2_5=10\). Does this make sense?
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Re: PS Q: OG-12 #116 [#permalink]
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15 May 2014, 16:02
Bunuel wrote: russ9 wrote: Bunuel wrote:
\(C^2_30\) is choosing 2 out of 30. There are 30 cities and each pair of cities need an entry, hence 30 cites need \(C^2_30\) entries.
Hope it's clear.
Hi Bunuel, Unfortunately, still not clear. Why are we choosing 2 out of 30? Consider the table given in the original post: A and B have 1 entry; A and C have 1 entry; A and D have 1 entry; A and E have 1 entry; B and C have 1 entry; B and D have 1 entry; B and E have 1 entry; C and D have 1 entry; C and E have 1 entry; D and E have 1 entry. So, each pair of letters from {A, B, C, D, E} has 1 entry, total of 10 entries. How many pairs can we have? \(C^2_5=10\). Does this make sense? Yes, now it does. Thanks!
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Re: Each • in the mileage table above represents an entry indica [#permalink]
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Combination formula is no doubt easiest and fastest. But other method is
Imagine it was an excel spreadsheet. Remove Cells A1, B2, C3, D4 etc, basically a diagonal across. Total will be 30 such cells.
So now we have 900 - 30 = 870.
On both sides of the diagonal distance (between cities) is shown twice.
Therefore divide 870 into half.
Answer 435.
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Re: Each • in the mileage table above represents an entry indica [#permalink]
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08 Nov 2014, 11:10
zubinator wrote: Combination formula is no doubt easiest and fastest. But other method is
Imagine it was an excel spreadsheet. Remove Cells A1, B2, C3, D4 etc, basically a diagonal across. Total will be 30 such cells.
So now we have 900 - 30 = 870.
On both sides of the diagonal distance (between cities) is shown twice.
Therefore divide 870 into half.
Answer 435. Hi Bunuel, I solved this problem as permutation 5P2= 870 then By symmetry we need only half of the table so total # of dots=870/2 = 435.........Answer B. So is there any problem to solve it like using permutation?
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Re: Each • in the mileage table above represents an entry indica [#permalink]
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Re: Each • in the mileage table above represents an entry indica [#permalink]
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14 Apr 2015, 08:28
Mo2men wrote: zubinator wrote: Combination formula is no doubt easiest and fastest. But other method is
Imagine it was an excel spreadsheet. Remove Cells A1, B2, C3, D4 etc, basically a diagonal across. Total will be 30 such cells.
So now we have 900 - 30 = 870.
On both sides of the diagonal distance (between cities) is shown twice.
Therefore divide 870 into half.
Answer 435. Hi Bunuel, I solved this problem as permutation 5P2= 870 then By symmetry we need only half of the table so total # of dots=870/2 = 435.........Answer B. So is there any problem to solve it like using permutation? Why shall I go for 5 rather 30?
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Each • in the mileage table above represents an entry indica [#permalink]
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15 Apr 2015, 10:32
The distance from each city to its own is not represented in the table.
So in the case of 5 cities, each city can have a distance w.r.t another 4 cities. ( A-B,A-C,A-D,A-E; BUT NOT A-A)
Hence these 5 cities can have 5*4 = 20 distances.
However we are representing each distance (to and fro) only once instead of twice. e.g A-B is same as B-A. Hence divide 20/2 = 10 dots
Similarly, in case of 30 cities, total distances will be 30*29 = 870
But we want to represent each distance only once instead of twice, so 870/2= 435 dots
Hope its clear!
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Re: Each • in the mileage table above represents an entry indica [#permalink]
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18 Apr 2015, 09:14
Hi Mates,
How are we getting this 29 for 30 cities?
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Re: Each • in the mileage table above represents an entry indica
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