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PS - 329 (OG 10)

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nick_sun
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PS - 329 (OG 10) [#permalink] New post   05 Apr 2007, 04:50
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A
B
C
D
E

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(N/A)

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Guys, please consider the problem attached below. The official explanation is a little bit confusing.



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Re: PS - 329 (OG 10) [#permalink] New post   05 Apr 2007, 04:58
Answer is A (15).

Total cities = 6

Total entries = 6*6 = 36

Entries in same cities (A and A, B and B and so) = 6
Entries left = 36-6 = 30

Every entry repeated twice (e.g. A and B = B and A) = 30/2 = 15

So the answer is 15.
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Re: PS - 329 (OG 10) [#permalink] New post   05 Apr 2007, 05:01
Thinking logically will allow you to find the answer.........since the answer asks for the minimum b/tw cities you need only account for the mileage once. Start at the top left corner, one down, and then follow diagonally down the line.....all blocks below will count, the number should add up to 15

not sure this is coherent.....
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Re: PS - 329 (OG 10) [#permalink] New post   05 Apr 2007, 05:36
Isn't the way you solve this very similar to the way you would solve the problem where 11 teams play one game against each other, to get the total number of games played?

The number of points would have to be:

the 6th city against 5 others, the 5th city against 4 others (since the one against 6th is already counted), the 4th city against 3 others etc

= 5+4+3+2+1 = 15

I am probably not being very clear here, but the first thing that striked me abt this problem was that it was like the other problem.
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Re: PS - 329 (OG 10) [#permalink] New post   05 Apr 2007, 08:18
Thank you, guys! Your replies are very comprehensive :P :panel
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Re: PS - 329 (OG 10) [#permalink] New post   05 Apr 2007, 17:46
Least number of combinations will be -
5+4+3+2+1 = 15
Answer is 'A'



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Re: PS - 329 (OG 10) [#permalink]
05 Apr 2007, 17:46
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