Question:
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

Answer:
(B)

Can someone offer up another explanation aside from the one outlined in the OG? I know that there is no particular MGMAT strategy to answer this type of question, but any new perspective on how to solve it would be much appreciated!

Hi - we actually need the info in the table, too, in order to answer the question. You can either take a screen shot and upload it or just type out and / or explain the info. Thanks!

Hi Stacey,

The problem is attached in the scanned shot below.

This is a non-standard problem, but any light to explain the logic behind this is greatly appreciated![img]

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Hi, sorry we're so late on replying to this question. So, here, we want to notice the pattern in the representative sample of 5 and then extrapolate to 30.

I'm given a 5x5 grid, which means 25 squares. Only 10 of the 25 have dots. The other 15 don't have dots either because it wouldn't make sense (there is no distance between A and A) or because it's already representated elsewhere on the grid (the distance between A and E is the same as the distance between E and A).

Let's look at that pattern. Of 25 squares. 10 have the dots and 10 more fall into the category of "already represented" - that is, each distance will be represented twice, and I only want to count one instance. Then, 5 more represent the "no distance" comparison of a city to itself. There are 5 cities on this grid and 5 instances of the city being compared to itself.

Now I want to do this for 30 cities. A 30x30 grid would have 900 squares. Note that I can eliminate E because common sense tells me not every square will have a dot. And I can eliminate A because common sense also tells me 60 is way too small.

There will be 30 instances of the city being compared to itself. So subtract those out: 900-30=870. The remaining squares represent valid comparisons, but they are two instances of each comparison, so divide by two: 870/2 = 435. B.