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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]

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15 Mar 2012, 19:59
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Table.png [ 3.59 KiB | Viewed 11942 times ]
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
[Reveal] Spoiler: OA

Last edited by Bunuel on 29 Jul 2014, 09:02, edited 2 times in total.
Edited the question and the image

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Re: In the table above what is the least number of table entries [#permalink]

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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..

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: In the table above what is the least number of table entries [#permalink]

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16 Mar 2012, 03:58
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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..

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?

$$C^2_{6}=15$$

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

Hope it helps.
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Re: In the table above what is the least number of table entries [#permalink]

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21 Oct 2015, 23:19
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Re: In the table above what is the least number of table entries [#permalink]

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29 Jul 2014, 07:30
Hi Bunuel!

I understand how we have arrived at 15. Here, we assume that the distance from a city to another city is the same even when the origin and destination is flipped.

But, there is a possibility to travel from City A to City B in 5 Kilometers and from City B to City A in 10 kilometers (since the route is a one way or something). The question merely asks what the least number of table entries must be and not the least number of table entries in the shortest possible route (which could remove the possible assumption that there are no one ways). So, shouldn't the answer be 30?

Bunuel wrote:
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..

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?

$$C^2_{6}=15$$

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

Hope it helps.

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Cheers!!

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Re: In the table above what is the least number of table entries [#permalink]

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29 Jul 2014, 09:04
joseph0alexander wrote:
Hi Bunuel!

I understand how we have arrived at 15. Here, we assume that the distance from a city to another city is the same even when the origin and destination is flipped.

But, there is a possibility to travel from City A to City B in 5 Kilometers and from City B to City A in 10 kilometers (since the route is a one way or something). The question merely asks what the least number of table entries must be and not the least number of table entries in the shortest possible route (which could remove the possible assumption that there are no one ways). So, shouldn't the answer be 30?

Bunuel wrote:
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..

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?

$$C^2_{6}=15$$

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

Hope it helps.

You are over-thinking. If the distance from A to B is 5 miles, then the distance from B to A is also 5 miles.
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Re: In the table above what is the least number of table entries [#permalink]

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02 Feb 2017, 17:28
Another alternative is if you cant think about conceptually mark each box. But keep in mind not to double mark the city on the horizontal axis with the city on the vertical axis....

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Re: In the table above what is the least number of table entries [#permalink]

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10 Jun 2017, 06:47
Hi -- quick question

--- i see a table is represented to signify between A and B

--- why isnt a table entry represented by signify B and A

i understand that A to B is 5 miles, so will B to 5, be 5 miles

but even if the numbers the same (5 miles), why should there be no table entry to represent B to A ...

Is it because the question asks for "least" number of table entries, i.e. the reader is expected to believe if A - B is 5 miles, so will B - A, be 5 miles ?

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Re: In the table above what is the least number of table entries [#permalink]

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27 Jul 2017, 20:34
eybrj2 wrote:
Attachment:
Table.png
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

we can also do this question with brutal force.

Since there is no mileage between a city and itself and since the mileage for
each pair of cities needs to be entered only once. This gives entries as-
1+2+3+4+5 = 15

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Re: In the table above what is the least number of table entries   [#permalink] 27 Jul 2017, 20:34
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