Distance between cities(table)
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116. Each "¢ in the mileage table above represents an entry
indicating the distance between a pair of the �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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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.
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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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.
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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Your solution is almost perfect.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.
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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Every PAIR of cities is represented by a dot.ChessWriter wrote:
116. Each "¢ in the mileage table above represents an entry
indicating the distance between a pair of the �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]
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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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
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Hi Brent,
Can you please explain this in detail. what exactly is the question asking and we need to find?
Can you please explain this in detail. what exactly is the question asking and we need to find?
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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.santhosh_katkurwar wrote:Hi Brent,
Can you please explain this in detail. what exactly is the question asking and we need to find?
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