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Register now and save up to $200 Available with Beat the GMAT members only code • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • Get 300+ Practice Questions 25 Video lessons and 6 Webinars for FREE Available with Beat the GMAT members only code ## OG 13th Q124 tagged by: Brent@GMATPrepNow This topic has 5 expert replies and 4 member replies anurag_7 Junior | Next Rank: 30 Posts Joined 09 Oct 2013 Posted: 20 messages #### OG 13th Q124 Tue Aug 05, 2014 9:25 pm Elapsed Time: 00:00 • Lap #[LAPCOUNT] ([LAPTIME]) OG 13th Q124- Each • in the mileage table above represents an entry indicating the distance between a pair of the five cities. If the table were extended to represent the distances between all pairs of 30 cities and each distance were to be represented by only one entry, how many entries would the table then have? (A) 60 (B) 435 (0 450 (D) 465 (E) 900 What is wrong in my method???? 2 cities can be shown via one entry. For 3 cities we need two entries and so on. So for 30 cities we need 29 entries. Thus 1+2+...........+28+29 = 450. But the answer is wrong... Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums! ### GMAT/MBA Expert Rich.C@EMPOWERgmat.com Elite Legendary Member Joined 23 Jun 2013 Posted: 8714 messages Followed by: 460 members Thanked: 2742 times GMAT Score: 800 Tue Aug 05, 2014 9:32 pm Hi anurag_7, Your calculation involves a minor math mistake: 1+2+3+….29 does NOT total 450. I'm going to explain why below, but you should retry your calculation first to see if you can find the mistake: 1 to 29, inclusive is 29 terms… The average of those terms is 15…. 29 x 15 = 435 Also, you'll notice that the first column in that chart is completely empty, so… FEWER than HALF of the 900 squares would have a dot in them Final Answer: B GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at Rich.C@empowergmat.com Thanked by: anurag_7 anurag_7 Junior | Next Rank: 30 Posts Joined 09 Oct 2013 Posted: 20 messages Tue Aug 05, 2014 9:40 pm Thanks a lot rich...I took 30 terms instead of 29... ### GMAT/MBA Expert GMATGuruNY GMAT Instructor Joined 25 May 2010 Posted: 13372 messages Followed by: 1780 members Thanked: 12897 times GMAT Score: 790 Wed Aug 06, 2014 2:42 am anurag_7 wrote: OG 13th Q124- Each • in the mileage table above represents an entry indicating the distance between a pair of the five cities. If the table were extended to represent the distances between all pairs of 30 cities and each distance were to be represented by only one entry, how many entries would the table then have? (A) 60 (B) 435 (C) 450 (D) 465 (E) 900 Each PAIR OF CITIES requires a dot in the table. From 30 cities, the number of pairs that can be formed = 30C2 = (30*29)/(2*1) = 435. The correct answer is B. _________________ Mitch Hunt GMAT Private Tutor GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "Thank" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Thanked by: anurag_7 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. ### GMAT/MBA Expert Brent@GMATPrepNow GMAT Instructor Joined 08 Dec 2008 Posted: 10764 messages Followed by: 1213 members Thanked: 5153 times GMAT Score: 770 Wed Aug 06, 2014 8:11 am anurag_7 wrote: OG 13th Q124- Each • in the mileage table above represents an entry indicating the distance between a pair of the five cities. If the table were extended to represent the distances between all pairs of 30 cities and each distance were to be represented by only one entry, how many entries would the table then have? (A) 60 (B) 435 (0 450 (D) 465 (E) 900 What is wrong in my method???? 2 cities can be shown via one entry. For 3 cities we need two entries and so on. So for 30 cities we need 29 entries. Thus 1+2+...........+28+29 = 450. But the answer is wrong... Another way to find the sum 1+2+...........+28+29 is to apply the following formula: The sum of the integers from 1 to n inclusive = (n)(n+1)/2 So, 1+2+...........+28+29 = (29)(29+1)/2 = (29)(30)/2 = (29)(15) = 435 Cheers, Brent _________________ Brent Hanneson – Founder of GMATPrepNow.com Use our video course along with Check out the online reviews of our course Come see all of our free resources Thanked by: anurag_7 GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months! GMATinsight Legendary Member Joined 10 May 2014 Posted: 998 messages Followed by: 21 members Thanked: 203 times Wed Aug 06, 2014 10:46 am anurag_7 wrote: OG 13th Q124- Each • in the mileage table above represents an entry indicating the distance between a pair of the five cities. If the table were extended to represent the distances between all pairs of 30 cities and each distance were to be represented by only one entry, how many entries would the table then have? (A) 60 (B) 435 (0 450 (D) 465 (E) 900 What is wrong in my method???? 2 cities can be shown via one entry. For 3 cities we need two entries and so on. So for 30 cities we need 29 entries. Thus 1+2+...........+28+29 = 450. But the answer is wrong... To make an entry you need one Start point City and one destination City and two Cities out of 30 Cities can be chosen in 30C2 ways 30C2 = 30!/(2!x28!) = 435 Answer: Option B _________________ Prosper!!! Bhoopendra Singh & Sushma Jha "GMATinsight" Contact Us Testimonials To register for One-on-One FREE ONLINE DEMO Class Call/e-mail e-mail: info@GMATinsight.com Mobile: +91-9999687183 / +91-9891333772 Get in touch for SKYPE-Based Interactive Private Tutoring One-On-One Classes fee - US$40 per hour &
for FULL COURSE (38 LIVE Sessions)-US$1000 "Please click on 'Thank' if you like my post/response." Classroom Centres Address: GMATinsight Dwarka, New Delhi-110075 and Shivalik New Delhi Thanked by: anurag_7 GMATinsight Legendary Member Joined 10 May 2014 Posted: 998 messages Followed by: 21 members Thanked: 203 times Wed Aug 06, 2014 10:52 am ALTERNATE For a Table with 2x2 Matrix, number of entries = 1 = 2x1/2 For a Table with 3x3 Matrix, number of entries = 3 = 3x2/2 For a Table with 4x4 Matrix, number of entries = 6 = 4x3/2 For a Table with 5x5 Matrix, number of entries = 10 = 5x4/2 ..... ..... For a Table with 30x30 Matrix, number of entries = 30x29/2 = 435 Answer: Option B _________________ Prosper!!! Bhoopendra Singh & Sushma Jha "GMATinsight" Contact Us Testimonials To register for One-on-One FREE ONLINE DEMO Class Call/e-mail e-mail: info@GMATinsight.com Mobile: +91-9999687183 / +91-9891333772 Get in touch for SKYPE-Based Interactive Private Tutoring One-On-One Classes fee - US$40 per hour &
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melguy Master | Next Rank: 500 Posts
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Fri Oct 28, 2016 8:53 pm
Hello All

I am still a bit lost with this Q even after reading all the explanations. If I may ask

- What is this question even asking?
- Which topic of GMAT is this question covering (P&C?)
- How did we get to '30 cities'. We can pair A -> B , A -> C, A -> D, A -> E and the same for rest 4 cities so there should be a total of 4 x 4 = 16 cities?

Thanks
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GMATGuruNY GMAT Instructor
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Sat Oct 29, 2016 2:44 am
melguy wrote:
Hello All

I am still a bit lost with this Q even after reading all the explanations. If I may ask

- What is this question even asking?
- Which topic of GMAT is this question covering (P&C?)
- How did we get to '30 cities'. We can pair A -> B , A -> C, A -> D, A -> E and the same for rest 4 cities so there should be a total of 4 x 4 = 16 cities?

Thanks
Prompt:
Each • in the mileage table above represents an entry
indicating the distance between a pair of the five
cities. If the table were extended to represent the
distances between all pairs of 30 cities
and each
distance were to be represented by only one entry,
how many entries would the table then have?

The portion in blue indicates that there are 30 cities in total.

In the mileage chart, every distance between two cities must be represented by a dot:
The distance between A AND B --> dot.
The distance between A AND C --> dot.
The distance between A AND D --> dot.
And so on.

Since every pair of cities requires a dot, the total number of dots is equal to the total number of PAIRS that can be formed from the 30 cities.
From 30 cities, the number of combinations of 2 that can be formed = 30C2 = (30*29)/(2*1) = 435.
Thus, the mileage chart requires a total of 435 dots.

This problem asks us to count COMBINATIONS: the total number of pairs that can be formed from 30 options.

_________________
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GMATGuruNY@gmail.com
If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
Available for tutoring in NYC and long-distance.

Thanked by: melguy
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Scott@TargetTestPrep GMAT Instructor
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Tue Nov 01, 2016 5:17 am
anurag_7 wrote:
OG 13th Q124- Each • in the mileage table above represents an entry
indicating the distance between a pair of the five
cities. If the table were extended to represent the
distances between all pairs of 30 cities and each
distance were to be represented by only one entry,
how many entries would the table then have?
(A) 60
(B) 435
(0 450
(D) 465
(E) 900
This problem can best be solved using combinations.

This problem is similar to one in which 30 sports teams are playing in a tournament where every team plays every 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 30C2, or the number of combinations of 30 items taken 2 at a time.

We can solve this problem in the same way:

30C2 = 30! / [2! x (30 - 2)!]

(30 x 29 x 28!) / [2! x 28!]

(30 x 29)/2!

(30 x 29) / 2

15 x 29 = 435

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