Distance between cities(table)

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ChessWriter: Senior | Next Rank: 100 Posts; Posts: 54; Joined: Sun Jan 01, 2012 4:24 am

by ChessWriter » Tue Feb 28, 2012 10:24 am

116. Each "¢ in the mileage table above represents an entry
indicating the distance between a pair of the ï¬�ve
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 [spoiler](B)435[/spoiler]

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Source: Beat The GMAT — Problem Solving | Tue Feb 28, 2012 10:24 am

ChessWriter: Senior | Next Rank: 100 Posts; Posts: 54; Joined: Sun Jan 01, 2012 4:24 am

by ChessWriter » 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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ChessWriter: Senior | Next Rank: 100 Posts; Posts: 54; Joined: Sun Jan 01, 2012 4:24 am

by ChessWriter » 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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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » 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

Brent Hanneson - Creator of GMATPrepNow.com

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Feb 28, 2012 11:23 am

ChessWriter wrote:

116. Each "¢ in the mileage table above represents an entry
indicating the distance between a pair of the ï¬�ve
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 [spoiler](B)435[/spoiler]

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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ChessWriter: Senior | Next Rank: 100 Posts; Posts: 54; Joined: Sun Jan 01, 2012 4:24 am

by ChessWriter » Tue Feb 28, 2012 11:53 am

Thanks Brent and Mitch. That was very helpful..........

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santhosh_katkurwar: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Fri Aug 11, 2017 1:40 am

by santhosh_katkurwar » 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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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » 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: https://www.destination-nz.com/wp-conten ... island.jpg

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

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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