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Each • in the mileage table above represents an entry indica [#permalink]
02 Jun 2010, 09:32
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00:00
A
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Difficulty:
45% (medium)
Question Stats:
67% (02:21) correct
33% (01:46) wrong based on 781 sessions
Attachment:
Table.png [ 23.23 KiB | Viewed 13700 times ]
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?
Re: PS Q: OG-12 #116 [#permalink]
02 Jun 2010, 10:42
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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
Re: PS Q: OG-12 #116 [#permalink]
02 Jun 2010, 10:54
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Expert's post
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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?
Re: PS Q: OG-12 #116 [#permalink]
03 Jan 2013, 11:54
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?
Re: Each • in the mileage table above represents an entry indica [#permalink]
10 Nov 2013, 22:42
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.
Re: Each • in the mileage table above represents an entry indica [#permalink]
16 Feb 2014, 01:47
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. _________________
learn the rules of the game, then play better than anyone else.
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 "\(C^n_k = \frac{n!}{k!(n-k)!}\)"
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.
Re: PS Q: OG-12 #116 [#permalink]
10 May 2014, 04:06
Expert's post
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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? \(C^2_5=10\).
Re: PS Q: OG-12 #116 [#permalink]
15 May 2014, 15:02
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? \(C^2_5=10\).
Re: Each • in the mileage table above represents an entry indica [#permalink]
13 Oct 2014, 02:08
1
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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.
Re: Each • in the mileage table above represents an entry indica [#permalink]
08 Nov 2014, 10:10
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?
Re: Each • in the mileage table above represents an entry indica [#permalink]
14 Apr 2015, 07:28
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?
Each • in the mileage table above represents an entry indica [#permalink]
15 Apr 2015, 09:32
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
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