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In the table above what is the least number of table entries

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In the table above what is the least number of table entries [#permalink] New post 15 Mar 2012, 19:59
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In the table above, what is the least number of table entries that are needed to show the mileage between each city and each of the other five cities?

A. 15
B. 21
C. 25
D. 30
E. 36

Sorry for the messy picture.. :oops:
[Reveal] Spoiler: OA

Last edited by Bunuel on 16 Mar 2012, 04:00, edited 1 time in total.
Edited the question and the image
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Re: PT #13 PS2 Q 13 [#permalink] New post 15 Mar 2012, 21:28
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eybrj2 wrote:
In the table above, what is the least number of table entries that are needed to show the mileage between each city and each of the other five cities?

a) 15

b) 21

c) 25

d) 30

e) 36



Sorry for the messy picture.. :oops:


Total number of entries 6*6(6rows*6columns) =36
Now 6 entries are representing mileage with the city itself so subtract that => 36-6
Minimum entries required = half the Total = 30/2 = 15
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Re: PT #13 PS2 Q 13 [#permalink] New post 16 Mar 2012, 03:58
eybrj2 wrote:
In the table above, what is the least number of table entries that are needed to show the mileage between each city and each of the other five cities?

A. 15
B. 21
C. 25
D. 30
E. 36

Sorry for the messy picture.. :oops:


The least number of table entries will be if we use only one entry for each pair of the cities. How many entries would the table then have? Or how many different pairs can be selected out of 6 cities?

C^2_{6}=15

Answer: A.

Similar question to practice: each-dot-in-the-mileage-table-above-represents-an-entry-95162.html

Hope it helps.
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Re: PT #13 PS2 Q 13   [#permalink] 16 Mar 2012, 03:58
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