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Distance between cities(table)

This topic has 3 expert replies and 4 member replies
ChessWriter Senior | Next Rank: 100 Posts Default Avatar
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Distance between cities(table)

Post Tue Feb 28, 2012 10:24 am
ttp://postimage.org/image/cm91b5b0f/" target="_blank">


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

This is a question from the Official guide 12, Question number 116, Page number 223

The Official answer is (B)435

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Post Wed Sep 27, 2017 2:23 am
Hi Brent,

Can you please explain this in detail. what exactly is the question asking and we need to find?

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GMAT/MBA Expert

Post Wed Sep 27, 2017 6:47 am
santhosh_katkurwar wrote:
Hi Brent,

Can you please explain this in detail. what exactly is the question asking and we need to find?
This is a mileage table that helps you find the driving distance between any two cities. Just find the row representing one city and the column representing the other city, and the place where the row and column intersect has a box indicating the distance between the two cities.

Here's an example: http://www.destination-nz.com/wp-content/uploads/2013/12/distance-charts_north-island.jpg

The question asks us to determine the number of entries (boxes) needed to create a mileage table with 30 cities.

Cheers,
Brent

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ChessWriter Senior | Next Rank: 100 Posts Default Avatar
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Post Tue Feb 28, 2012 11:53 am
Thanks Brent and Mitch. That was very helpful..........

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Post Tue Feb 28, 2012 11:23 am
ChessWriter wrote:
ttp://postimage.org/image/cm91b5b0f/" target="_blank">


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

This is a question from the Official guide 12, Question number 116, Page number 223

The Official answer is (B)435
Every PAIR of cities is represented by a dot.
Thus, the total number of dots is equal to the total number of pairs that can be formed from the 30 cities.
The number of combinations of 2 that can formed from 30 choices = (30*29)/(2*1) = 435.

The correct answer is B.

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Post Tue Feb 28, 2012 10:41 am
ChessWriter wrote:
I have posted the official guide explanation below. But, My method for solving the question is as follows.

Since for 5 cities we get only 4 entries per row - The total number of entries is 1+2+3+4 = 10

Therefore, for 30 cities we would have to add numbers 1+2+3+4+5......till 30. The formula for this is n(n+1)/2.

When applied to this problem, it becomes 30(30+1)/2 = 465

The official guide gives a different figure of 435 as the answer.

So can anyone tell me what is the problem with my method of solving it.
Your solution is almost perfect.
For 5 cities, the answer is 1+2+3+4=10
For 30 cities, the answer is 1+2+3+4+5....+28+29

In your solution, you are adding 1+2+3+4+5....+29+30

Cheers,
Brent

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ChessWriter Senior | Next Rank: 100 Posts Default Avatar
Joined
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Posted:
54 messages
Post Tue Feb 28, 2012 10:40 am
This is the official explanation given in the Official guide 12th edition:

In a table with 30 cities, there are 30(30) = 900
boxes for entries. However, since a city does not
need to have any entry for a distance to and from
itself, 30 entries are not needed on the diagonal
through the table. Th us, the necessary number of
entries is reduced to 900 - 30 = 870 entries. Th en,
it is given that each pair of cities only needs one
table entry, not two as the table allows; therefore,
the table only needs to have 870/2 = 435 entries.
The correct answer is B.

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ChessWriter Senior | Next Rank: 100 Posts Default Avatar
Joined
01 Jan 2012
Posted:
54 messages
Post Tue Feb 28, 2012 10:38 am
I have posted the official guide explanation below. But, My method for solving the question is as follows.

Since for 5 cities we get only 4 entries per row - The total number of entries is 1+2+3+4 = 10

Therefore, for 30 cities we would have to add numbers 1+2+3+4+5......till 30. The formula for this is n(n+1)/2.

When applied to this problem, it becomes 30(30+1)/2 = 465

The official guide gives a different figure of 435 as the answer.

So can anyone tell me what is the problem with my method of solving it.

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