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

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

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

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.
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\(C^2_{30}\) 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.
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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? \(C^2_5=10\).

Does this make sense?
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How if as:
the pattern is 5*5= 25-5=20/2=10
Thus, 30*30= 900-30= 870/2 = 435
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Hi Mates,

How are we getting this 29 for 30 cities?

The distance from a city to itself is not measured. If there are 5 cities A,B,C,D,E, THEN we measure A-B, A-C,A-D, A-E but not A-A.
This for 5 cities, we get 4 distances.

When there are total 30 cities, we measure 29 distances for each city.
Hope its clear.

Hi stephyw,

Your approach absolutely works. If you read the other posts in this thread, you'll see that there are several ways to approach this prompt (and most of the questions that you'll see on Test Day will also be approachable in multiple ways). This goes to show that just because you got a question correct doesn't necessarily mean that you can't improve on your process. There might be a faster, more strategic way to approach the prompt; there might be a way that requires less 'work', etc. As you continue to study, you should commit some of your time to improving your overall tactical knowledge - those other methods can help you to increase your scores, improve your pacing and get "unstuck" when dealing with a question in which "your" approach doesn't seem to get you to the solution.

GMAT assassins aren't born, they're made,
Rich
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