December 11, 2018 December 11, 2018 09:00 PM EST 10:00 PM EST Strategies and techniques for approaching featured GMAT topics. December 11 at 9 PM EST. December 13, 2018 December 13, 2018 08:00 AM PST 09:00 AM PST What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 16 Feb 2010
Posts: 177

In the Mundane Goblet competition, 6 teams compete in a
[#permalink]
Show Tags
Updated on: 09 Oct 2012, 03:00
Question Stats:
55% (01:11) correct 45% (01:32) wrong based on 431 sessions
HideShow timer Statistics
In the Mundane Goblet competition, 6 teams compete in a “round robin” format: that is, each team plays every other team exactly once. A team gets 3 points for a win, 1 point for a tie (a draw), and 0 points for a loss. What is the difference between the maximum total points and the minimum total points that can be gained by all teams (added together) in the Mundane Goblet competition? (A) 15 (B) 30 (C) 45 (D) 60 (E) 75
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by zisis on 08 Jul 2010, 13:00.
Last edited by Bunuel on 09 Oct 2012, 03:00, edited 2 times in total.
Edited the question.




Math Expert
Joined: 02 Sep 2009
Posts: 51100

Re: MBAmission: The Quest for 700
[#permalink]
Show Tags
08 Jul 2010, 15:06
zisis wrote: In the Mundane Goblet competition, 6 teams compete in a “round robin” format: that is, each team plays every other team exactly once. A team gets 3 points for a win, 1 point for a tie (a draw), and 0 points for a loss. What is the difference between the maximum total points and the minimum total points that can be gained by all teams (added together) in the Mundane Goblet competition?
(A) 15
(B) 30
(C) 45
(D) 60
(E) 75 There will be \(C^2_6=15\) games needed so that each team to play every other team exactly once (\(C^2_6=15\) is the # of ways we can pick two different teams to play each other). Now, in one game max points (3 points) will be obtained if one team wins and another looses and min points (1+1=2 points) will be obtained if there will be a tie. Hence, maximum points that can be gained by all teams will be 15 games * 3 points=45 and the minimum points that can be gained by all teams will be 15 games * 2 points=30, difference is 4530=15. Answer: A. Hope it helps.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Manager
Joined: 16 Feb 2010
Posts: 177

Re: MBAmission: The Quest for 700
[#permalink]
Show Tags
08 Jul 2010, 13:02
No OA yet: my method:
max points: A B C D E F 15 12 9 6 3 0 = 45POINTS
min points A B C D E F 5 5 5 5 5 5 = 30
4530= 15 points ====> (A)



Senior Manager
Status: Yeah well whatever.
Joined: 18 Sep 2009
Posts: 309
Location: United States
GMAT 1: 660 Q42 V39 GMAT 2: 730 Q48 V42
GPA: 3.49
WE: Analyst (Insurance)

Re: MBAmission: The Quest for 700
[#permalink]
Show Tags
09 Jul 2010, 07:02
Maybe this is a dumb question but I have to ask. Is \(C^2_6\) the same thing as 6!/(2!*4!)? That’s the only way I could get the 15 by which you multiplied 3 and 2 and then subtracted 30 from 45.
_________________
He that is in me > he that is in the world.  source 1 John 4:4



Senior Manager
Affiliations: Volunteer Operation Smile India, Creative Head of College IEEE branch (200910), Chief Editor College Magazine (2009), Finance Head College Magazine (2008)
Joined: 25 Jul 2010
Posts: 372
Location: India
WE2: Entrepreneur (Ecommerce  The Laptop Skin Vault)
Concentration: Marketing, Entrepreneurship
WE: Marketing (Other)

Re: MBAmission: The Quest for 700
[#permalink]
Show Tags
06 Aug 2010, 11:14
Its just 3(total teams  1)  0 Coz max pts are if 1 team won all the matches & min pts are 0
In This case 3(61)  0 = 15  0 = 15
HTH



Manager
Joined: 15 Jan 2011
Posts: 101

logical combinatorics
[#permalink]
Show Tags
21 Aug 2011, 08:00
"In the Mundane Goblet competition, 6 teams compete in a "round robin" format: that is, each team plays every other team exactly once. A team gets 3 points for a win, 1 point for a tie (a draw), and 0 points for a loss. What is the difference between the maximum total points and the minimum total points that can be gained by all teams (added together) in the Mundane Goblet competition? *15 *30 *45 *60 *75 the decision:6!/(2!*4!)=15 games Maximum=3*15=45 Minimum=2*15=30, so 4530=15 the correct answer is A" Could somebody explain me please why the minimum score i had to calculate as 2*15, instead of 0*15??? I solved it as 3*150*15=45...or i just missed that there are no any earned scores when the team loses the game?



Manager
Joined: 19 Jul 2011
Posts: 91
Concentration: Finance, Economics
GPA: 3.9

Re: logical combinatorics
[#permalink]
Show Tags
21 Aug 2011, 08:11
When a team wins one team gets 3 points and the other 0 meaning 3 points are earned between the two teams in the match. When the teams draw both teams get a point each meaning 2 points are earned between the teams in the match (hence 2*15).
_________________
Show Thanks to fellow members with Kudos its shows your appreciation and its free



Manager
Joined: 20 Aug 2011
Posts: 128

Re: logical combinatorics
[#permalink]
Show Tags
21 Aug 2011, 08:27
good point nammers the points are split between the two teams, and the question says together so it's 2 i.e. a tie
_________________
Hit kudos if my post helps you. You may send me a PM if you have any doubts about my solution or GMAT problems in general.



Director
Joined: 01 Feb 2011
Posts: 659

Re: logical combinatorics
[#permalink]
Show Tags
21 Aug 2011, 17:43
the question talks about total points of all teams  meaning they are not just interested in any specific team. every match is played between two teams  two possible scenarios  one team wins and other loses => total points in that match = 3+0 = 3 its tie  each team gets point => total points in that match = 1+1 = 2 so maximum possible points in any match = 3 and minimum =2 altogether we have 15 games, all 15 can fall in one of those scenarios above. That's why maximum possible points = 15*3 = 45 minimum possible points = 15*2 = 30 difference is 15. Galiya wrote: "In the Mundane Goblet competition, 6 teams compete in a "round robin" format: that is, each team plays every other team exactly once. A team gets 3 points for a win, 1 point for a tie (a draw), and 0 points for a loss. What is the difference between the maximum total points and the minimum total points that can be gained by all teams (added together) in the Mundane Goblet competition? *15 *30 *45 *60 *75 the decision:6!/(2!*4!)=15 games Maximum=3*15=45 Minimum=2*15=30, so 4530=15 the correct answer is A" Could somebody explain me please why the minimum score i had to calculate as 2*15, instead of 0*15??? I solved it as 3*150*15=45...or i just missed that there are no any earned scores when the team loses the game?



Senior Manager
Joined: 06 Aug 2011
Posts: 339

Re: Mundane Goblet competition
[#permalink]
Show Tags
08 Oct 2012, 08:18
Why we r not taking minumun value 0?? Bunuell..what actually i was doing..i multiplied 3*15=45.. and 15*0 =0...4515=30.. what i think because if we take 0?? then one team loss is another team win..so it will take same 45 points ... my concept behind it is rite?? Thanks
_________________
Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !



Math Expert
Joined: 02 Sep 2009
Posts: 51100

Re: Mundane Goblet competition
[#permalink]
Show Tags
09 Oct 2012, 03:03



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8659
Location: Pune, India

Re: Mundane Goblet competition
[#permalink]
Show Tags
09 Oct 2012, 21:27
sanjoo wrote: Why we r not taking minumun value 0?? Bunuell..what actually i was doing..i multiplied 3*15=45.. and 15*0 =0...4515=30.. what i think because if we take 0?? then one team loss is another team win..so it will take same 45 points ... my concept behind it is rite?? Thanks To add to what Bunuel said above, we are interested in the total points obtained by all the teams together i.e. the total points earned in all the matches. In any one match, if one team loses and gets 0, the other team wins and gets 3. So, in that match, a total of 3 points are generated. On the other hand, if the two teams draw, they both get a point each and in that game, a total of 2 points are generated. No game can generate less than 2 points. So the minimum points generated by a match are 2 and the maximum are 3. Hence we do 3*15  2*15.
_________________
[b]Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Senior Manager
Joined: 06 Aug 2011
Posts: 339

Re: In the Mundane Goblet competition, 6 teams compete in a
[#permalink]
Show Tags
10 Oct 2012, 02:20
Thanks alot Karishma and Bunuel U both rocked ..
_________________
Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !



Intern
Joined: 04 Jan 2013
Posts: 2

Re: MBAmission: The Quest for 700
[#permalink]
Show Tags
14 Jan 2013, 12:51
[quote]
There will be \(C^2_6=15\) games needed so that each team to play every other team exactly once (\(C^2_6=15\) is the # of ways we can pick two different teams to play each other).
[end quote]
Pardon me if the answer to my question has been posted already, but I do not understand the method some of you used to get 15 games:
Bunuel I am not familiar at all with the notation you used ( \(C^2_6=15\) )
Furthermore, some others here used 6!/(2!4!). Although I can see that it yields the right answer, I fail to understand the reasoning in how and why they chose those particular numbers. ie: 2! and 4! in the denominator. I guess I need to be enlightened about why factorials are a profitable tool here.
In my case, I used n(n1)/2 which seemed to work okay. Is one method preferable to the other?
Hope my question was clear.



Intern
Joined: 30 Apr 2011
Posts: 9

Re: MBAmission: The Quest for 700
[#permalink]
Show Tags
14 Jan 2013, 17:40
IanElliott wrote: Quote: There will be \(C^2_6=15\) games needed so that each team to play every other team exactly once (\(C^2_6=15\) is the # of ways we can pick two different teams to play each other).
[end quote]
Pardon me if the answer to my question has been posted already, but I do not understand the method some of you used to get 15 games:
Bunuel I am not familiar at all with the notation you used ( \(C^2_6=15\) )
Furthermore, some others here used 6!/(2!4!). Although I can see that it yields the right answer, I fail to understand the reasoning in how and why they chose those particular numbers. ie: 2! and 4! in the denominator. I guess I need to be enlightened about why factorials are a profitable tool here.
In my case, I used n(n1)/2 which seemed to work okay. Is one method preferable to the other?
Hope my question was clear.
\(C^2_6=15\) is a quick notation to say: a combination (order doesn't matter) of 2 items in a pool of 6 items In order words: how many possibilities to pick 2 different teams among 6 teams More generally \(C^r_n\) is a combination of r items in a pool of n items Which leads to the general formula: \(n!/(nr)!r!\) \(6!/(2!4!)\)



Senior Manager
Joined: 27 Jun 2012
Posts: 374
Concentration: Strategy, Finance

Re: MBAmission: The Quest for 700
[#permalink]
Show Tags
14 Jan 2013, 18:42
IanElliott wrote: Quote: There will be \(C^2_6=15\) games needed so that each team to play every other team exactly once (\(C^2_6=15\) is the # of ways we can pick two different teams to play each other).
[end quote]
Pardon me if the answer to my question has been posted already, but I do not understand the method some of you used to get 15 games:
Bunuel I am not familiar at all with the notation you used ( \(C^2_6=15\) )
Furthermore, some others here used 6!/(2!4!). Although I can see that it yields the right answer, I fail to understand the reasoning in how and why they chose those particular numbers. ie: 2! and 4! in the denominator. I guess I need to be enlightened about why factorials are a profitable tool here.
In my case, I used n(n1)/2 which seemed to work okay. Is one method preferable to the other?
Hope my question was clear.
IanElliott, check below link about Combinatorics topic and practice problems. You may find it useful to understand above concepts. mathcombinatorics87345.html
_________________
Thanks, Prashant Ponde
Tough 700+ Level RCs: Passage1  Passage2  Passage3  Passage4  Passage5  Passage6  Passage7 Reading Comprehension notes: Click here VOTE GMAT Practice Tests: Vote Here PowerScore CR Bible  Official Guide 13 Questions Set Mapped: Click here Finance your Student loan through SoFi and get $100 referral bonus : Click here



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8659
Location: Pune, India

Re: MBAmission: The Quest for 700
[#permalink]
Show Tags
14 Jan 2013, 19:44
IanElliott wrote: Quote: There will be \(C^2_6=15\) games needed so that each team to play every other team exactly once (\(C^2_6=15\) is the # of ways we can pick two different teams to play each other).
[end quote]
Pardon me if the answer to my question has been posted already, but I do not understand the method some of you used to get 15 games:
Bunuel I am not familiar at all with the notation you used ( \(C^2_6=15\) )
Furthermore, some others here used 6!/(2!4!). Although I can see that it yields the right answer, I fail to understand the reasoning in how and why they chose those particular numbers. ie: 2! and 4! in the denominator. I guess I need to be enlightened about why factorials are a profitable tool here.
In my case, I used n(n1)/2 which seemed to work okay. Is one method preferable to the other?
Hope my question was clear.
There are various ways to get a particular answer. People choose a particular method depending on how they think. Your thought would have been something like this: There are 6 teams. The first team plays with every other team i.e. with 5 teams and goes away. Now, the second plays with the rest of the 4 teams and goes away. Now the third team plays with the rest of the 3 teams and so on... So total there are 5 + 4+ 3 + 2 + 1 and you used n(n1)/2 here. Another way to think about it is this: There are total 6 teams. Every pair of two teams played one match. Out of 6 teams, we will try to make as many distinct pairs of 2 teams each as possible. For this, we use 6C2 = 6!/2!*4! There is absolutely nothing wrong with either one of the methods. But you must be able to think in nCr terms too because it comes in very handy in many questions. nCr = choose r items from n For a background on combinations, check this post: http://www.veritasprep.com/blog/2011/11 ... binations/
_________________
[b]Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 16 Nov 2012
Posts: 25
Location: United States
Concentration: Operations, Social Entrepreneurship
GMAT Date: 08272013
GPA: 3.46
WE: Project Management (Other)

Re: MBAmission: The Quest for 700
[#permalink]
Show Tags
19 Jan 2013, 08:20
IanElliott wrote: Quote: There will be \(C^2_6=15\) games needed so that each team to play every other team exactly once (\(C^2_6=15\) is the # of ways we can pick two different teams to play each other).
[end quote]
Pardon me if the answer to my question has been posted already, but I do not understand the method some of you used to get 15 games:
Bunuel I am not familiar at all with the notation you used ( \(C^2_6=15\) )
Furthermore, some others here used 6!/(2!4!). Although I can see that it yields the right answer, I fail to understand the reasoning in how and why they chose those particular numbers. ie: 2! and 4! in the denominator. I guess I need to be enlightened about why factorials are a profitable tool here.
In my case, I used n(n1)/2 which seemed to work okay. Is one method preferable to the other?
Hope my question was clear.
Hi ,the method is called permutations and combinations.here we want to find the total number of games played by all teams together.Here we have to choose 2 teams from 6 teams to play game i.e here we are using combinations so the formula to find out the total combination how each team can play with other exactly ones the formula is (n C r) nrepresents total teams ,C  combination and r  represents minimum no.of teams required to play a game .In this problem n=6 and r= 2 n C r = n! /( (nr)! * r !) = (1*2*...*n)/((1*2.....(nr))*(1*2.....r)) .here 6!/(4! * 2!) = 15. I hope you understood
_________________
......................................................................................... Please give me kudos if my posts help.



Intern
Joined: 16 Nov 2012
Posts: 25
Location: United States
Concentration: Operations, Social Entrepreneurship
GMAT Date: 08272013
GPA: 3.46
WE: Project Management (Other)

Re: In the Mundane Goblet competition, 6 teams compete in a
[#permalink]
Show Tags
19 Jan 2013, 08:38
total no of games played = 15 max points gained by all teams together = 15* 3 = 45 minimum no.of points gained by all teams together is = 15*2 =30 difference between max and min points = 4530=15 The procedure to calculate min number of points: here we require the minimum number of points can be gained by all teams together is 2 points i.e if the game is tied each time will get 1 points making min points each game is 2 points so if all the games are tie ,the minimum points of all games is = 15* 2=30
_________________
......................................................................................... Please give me kudos if my posts help.



Manager
Joined: 14 Nov 2011
Posts: 123
Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.61
WE: Consulting (Manufacturing)

Re: MBAmission: The Quest for 700
[#permalink]
Show Tags
26 May 2013, 03:16
VeritasPrepKarishma wrote: IanElliott wrote: Quote: There will be \(C^2_6=15\) games needed so that each team to play every other team exactly once (\(C^2_6=15\) is the # of ways we can pick two different teams to play each other).
[end quote]
Pardon me if the answer to my question has been posted already, but I do not understand the method some of you used to get 15 games:
Bunuel I am not familiar at all with the notation you used ( \(C^2_6=15\) )
Furthermore, some others here used 6!/(2!4!). Although I can see that it yields the right answer, I fail to understand the reasoning in how and why they chose those particular numbers. ie: 2! and 4! in the denominator. I guess I need to be enlightened about why factorials are a profitable tool here.
In my case, I used n(n1)/2 which seemed to work okay. Is one method preferable to the other?
Hope my question was clear.
There are various ways to get a particular answer. People choose a particular method depending on how they think. Your thought would have been something like this: There are 6 teams. The first team plays with every other team i.e. with 5 teams and goes away. Now, the second plays with the rest of the 4 teams and goes away. Now the third team plays with the rest of the 3 teams and so on... So total there are 5 + 4+ 3 + 2 + 1 and you used n(n1)/2 here. Another way to think about it is this: There are total 6 teams. Every pair of two teams played one match. Out of 6 teams, we will try to make as many distinct pairs of 2 teams each as possible. For this, we use 6C2 = 6!/2!*4! There is absolutely nothing wrong with either one of the methods. But you must be able to think in nCr terms too because it comes in very handy in many questions. nCr = choose r items from n For a background on combinations, check this post: http://www.veritasprep.com/blog/2011/11 ... binations/Hi Karishma, Can you please explain why did we add = 5+4+3+2+1 here?




Re: MBAmission: The Quest for 700 &nbs
[#permalink]
26 May 2013, 03:16



Go to page
1 2
Next
[ 26 posts ]



