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# In the table above, what is the least number of table entries that are

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Math Expert
Joined: 02 Sep 2009
Posts: 59075
In the table above, what is the least number of table entries that are  [#permalink]

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24 Jan 2014, 03:51
2
34
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Difficulty:

35% (medium)

Question Stats:

65% (01:11) correct 35% (01:17) wrong based on 963 sessions

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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

Problem Solving
Question: 56
Category: Arithmetic Interpretation of tables
Page: 69
Difficulty: 600

The Official Guide For GMAT® Quantitative Review, 2ND Edition

Attachment:

Untitled.png [ 49.83 KiB | Viewed 16836 times ]

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

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24 Jan 2014, 03:52
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5
SOLUTION

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

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

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01 Aug 2014, 00:26
8
2
Just count the colored boxes below the diagonal

= 5 + 4 + 3 + 2 + 1

= 15

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Schools: WHU MBA"20 (A$) GMAT 1: 580 Q46 V24 GPA: 3.88 WE: Information Technology (Consulting) Re: In the table above, what is the least number of table entries that are [#permalink] ### Show Tags 20 Jul 2014, 02:59 5 1 For A horizontal: 5 entries (eliminate A to A) For B horizontal: 4 entries (A to B = B to A / eliminate B to B ) For C horizontal: 3 entries (we had already A to C & B to C previously/ eliminate C to C) For D horizontal: 2 entries (we had already A to D & B to D & C to D previously/ eliminate D to D) For E horizontal: 1 entry (we had already A to E & B to E & C to E & D to E previously/ eliminate E to E) For F horizontal: 0 entries (we had already A to F & B to F & C to F & D to F & E to F previously/ eliminate F to F) Hope it helps ! _________________ When you’re up, your friends know who you are. When you’re down, you know who your friends are. Share some Kudos, if my posts help you. Thank you ! 800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50 GMAT PREP 670 MGMAT CAT 630 KAPLAN CAT 660 Senior Manager Joined: 10 Mar 2013 Posts: 461 Location: Germany Concentration: Finance, Entrepreneurship Schools: WHU MBA"20 (A$)
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Re: In the table above, what is the least number of table entries that are  [#permalink]

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12 Oct 2015, 08:39
5
Total possible pairs =6*6=36
Pairs with same letters=6
Pairs with same combinations (AB=BA etc) $$\frac{36-6}{2}$$=15
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Re: In the table above, what is the least number of table entries that are  [#permalink]

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01 Apr 2015, 03:09
3
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

Attachment:
The attachment Untitled.png is no longer available
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

Problem Solving
Question: 56
Category: Arithmetic Interpretation of tables
Page: 69
Difficulty: 600

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

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04 Apr 2017, 16:20
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Bunuel wrote:

Attachment:
Untitled.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

This problem can be best solved using combinations.

This problem is similar to one in which 6 sports teams are playing in a tournament in which every team plays with each other team exactly once. No team plays itself, obviously, and the order of each pairing doesn't matter. [For example, if Team A plays Team B, the pairing of (Team A vs. Team B) is identical to (Team B vs. Team A)]. We would calculate 6C2, or the number of combinations of 6 items taken 2 at a time:

6C2 =6!/2! (6 - 2)! = (6 x 5)/2! = 15

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

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12 Oct 2015, 10:50
5+4+3+2+1=15
BC,BD,BE,BF
CD,CE,CF
DE.DF
EF
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Re: In the table above, what is the least number of table entries that are  [#permalink]

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29 Sep 2018, 05:59
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

One entry must come between each pair of two cities

Two cities out of six cities (ABCDEF) can be chosen in 6C2 = 15 ways

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

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03 Oct 2019, 04:59
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Re: In the table above, what is the least number of table entries that are   [#permalink] 03 Oct 2019, 04:59
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