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In the table above, what is the least number of table entries that are
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24 Jan 2014, 02:51
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68% (01:11) correct 33% (01:14) wrong based on 979 sessions
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Re: In the table above, what is the least number of table entries that are
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31 Jul 2014, 23:26
Just count the colored boxes below the diagonal = 5 + 4 + 3 + 2 + 1 = 15 Answer = A
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Re: In the table above, what is the least number of table entries that are
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20 Jul 2014, 01:59
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 !
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Re: In the table above, what is the least number of table entries that are
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01 Apr 2015, 02:09
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionAttachment: 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 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you! Answer 15 A
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Re: In the table above, what is the least number of table entries that are
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12 Oct 2015, 07:39
Total possible pairs =6*6=36 Pairs with same letters=6 Pairs with same combinations (AB=BA etc) \(\frac{366}{2}\)=15 Answer=36615=15
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Re: In the table above, what is the least number of table entries that are
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12 Oct 2015, 09:50
5+4+3+2+1=15 AB,AC,AD,AE,AF 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
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04 Apr 2017, 15:20
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 Answer A
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Re: In the table above, what is the least number of table entries that are
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29 Sep 2018, 04:59
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 Answer: Option C
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Re: In the table above, what is the least number of table entries that are
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