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OG12: Mileage table

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by nhai2003 » Sun Sep 13, 2009 2:11 am
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

Can someone shed some light on this problem?
Thanks
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by bpgen » Thu Apr 01, 2010 8:33 pm
take total combination of 2 cities out of 30 cities, i.e 30C2=>30*29/2=>435
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by eaakbari » Thu Apr 01, 2010 9:04 pm
IMO D
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by eaakbari » Thu Apr 01, 2010 9:13 pm
My method may be a little convoluted, so have patience

In the table since only entries on one side of the diagonal will matter (as the distances are to be stated once).
If you notice it forms an A.P witch first term as 1 (from city one) and common difference one and last term 30 (from city 30)

Sum = n(a + l)/2
= 15(31)
=465


Hence D
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