Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Each • in the mileage table above represents an entry indica [#permalink]
02 Jun 2010, 09:32

21

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

67% (02:22) correct
33% (01:48) wrong based on 691 sessions

Attachment:

Table.png [ 23.23 KiB | Viewed 11703 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

9

This post received KUDOS

4

This post was BOOKMARKED

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

7

This post received KUDOS

Expert's post

6

This post was BOOKMARKED

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

1

This post was BOOKMARKED

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

This post received KUDOS

1

This post was BOOKMARKED

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

On September 6, 2015, I started my MBA journey at London Business School. I took some pictures on my way from the airport to school, and uploaded them on...

When I was growing up, I read a story about a piccolo player. A master orchestra conductor came to town and he decided to practice with the largest orchestra...

Amy Cuddy, Harvard Business School professor, at TED Not all leadership looks the same; there is no prescribed formula for what makes a good leader. Rudi Gassner believed that...

We are thrilled to welcome the Class of 2017 to campus today, and data from the incoming class of students indicates that Kellogg’s community is about to reach a...