Each • in the mileage table above represents an entry indica

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Each • in the mileage table above represents an entry indica [#permalink]

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

[Reveal] Spoiler: OA

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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]

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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Re: PS Q: OG-12 #116 [#permalink]

02 Jun 2010, 11:54

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.
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Re: PS Q: OG-12 #116 [#permalink]

30 Jun 2010, 02:16

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]

30 Jun 2010, 07:45

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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Question about Cities' Distance [#permalink]

20 Mar 2012, 01:13

A • • • •
City B • • •
City C • •
City D •
City E

116. 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

I solved this question before, but I forgot how to do so and when I tried solving it again it took me a very long time to do so.

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Re: Question about Cities' Distance [#permalink]

20 Mar 2012, 01:25

dzodzo85 wrote:

A • • • •
City B • • •
City C • •
City D •
City E

116. 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

I solved this question before, but I forgot how to do so and when I tried solving it again it took me a very long time to do so.

Merging similar topics. Please ask if anything remains unclear.
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Re: PS Q: OG-12 #116 [#permalink]

03 Jan 2013, 12:54

Bunuel wrote:

snkrhed wrote:

Attachment:

img.jpg

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]

03 Jan 2013, 13:37

Combinations formula.

30 C 2. Answer 435.

Re: Each • in the mileage table above represents an entry indica [#permalink] 03 Jan 2013, 13:37

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Each • in the mileage table above represents an entry indica

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Each • in the mileage table above represents an entry indica

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