Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 04 May 2015, 12:46

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If each participant of a chess tournament plays exactly one

Author Message
TAGS:
Intern
Joined: 07 Nov 2012
Posts: 13
Followers: 0

Kudos [?]: 12 [1] , given: 10

If each participant of a chess tournament plays exactly one [#permalink]  10 Nov 2012, 13:06
1
KUDOS
3
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

68% (02:25) correct 32% (01:49) wrong based on 131 sessions
If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants.

A. 15
B. 16
C. 17
D. 18
E. 19
[Reveal] Spoiler: OA
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5453
Location: Pune, India
Followers: 1334

Kudos [?]: 6782 [3] , given: 177

Re: If each participant of a chess tournament plays exactly one [#permalink]  10 Nov 2012, 18:00
3
KUDOS
Expert's post
1
This post was
BOOKMARKED
derekgmat wrote:
If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants.

A. 15
B. 16
C. 17
D. 18
E. 19

If the number of participants is 3 (say A, B, C) the number of games played will be 2 (A plays against B and C) + 1 (B plays against C) = 3
Using the same logic, if the number of participants is n, the number of games played will be (n-1) + (n - 2) + (n - 3) + ... 3 + 2 + 1

Given that this sum = 153 = 1 + 2 + 3 + ... ( n - 1)

Sum of first m positive integers is given by m(m+1)/2. So sum of first (n-1) positive integers is (n-1)*n/2

153 = (n-1)*n/2
(n-1)*n = 306
17*18 = 306 (We know that 15^2 = 225 so the two consecutive numbers must be greater than 15. Also, 20^2 = 400 so the two numbers must be less than 20. The pair of numbers in between 15 and 20 whose product ends with 6 is 17 and 18)

So n = 18

_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Veritas Prep GMAT course is coming to India. Enroll in our weeklong Immersion Course that starts March 29!

Veritas Prep Reviews

Math Expert
Joined: 02 Sep 2009
Posts: 27215
Followers: 4227

Kudos [?]: 41004 [3] , given: 5654

Re: If each participant of a chess tournament plays exactly one [#permalink]  11 Nov 2012, 04:40
3
KUDOS
Expert's post
1
This post was
BOOKMARKED
derekgmat wrote:
If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants.

A. 15
B. 16
C. 17
D. 18
E. 19

We are basically told that we can choose 153 groups of two players out of n players, thus $$C^2_n=153$$ --> $$\frac{n!}{(n-2)!2!}=153$$ --> $$\frac{(n-1)n}{2}=153$$ --> $$(n-1)n=306$$ --> $$n=18$$.

Similar questions to practice:
how-many-diagonals-does-a-polygon-with-21-sides-have-if-one-101540.html
if-10-persons-meet-at-a-reunion-and-each-person-shakes-hands-110622.html
how-many-different-handshakes-are-possible-if-six-girls-129992.html
15-chess-players-take-part-in-a-tournament-every-player-55939.html
there-are-5-chess-amateurs-playing-in-villa-s-chess-club-127235.html

Hope it helps.
_________________
Intern
Joined: 07 Nov 2012
Posts: 13
Followers: 0

Kudos [?]: 12 [2] , given: 10

Re: If each participant of a chess tournament plays exactly one [#permalink]  11 Nov 2012, 00:29
2
KUDOS
Thanks Karishma

I must admit I am struggling to work through your solution. I looked at the question again and answered it with the approach of working through each of the choices long-hand.

If A, then 15 players means they each will play 14 games (so 14*15), divide the result by two for double counted games:
14*15 = 210
201/2 = 105, not 153, next

If B, 15*16 = 240
240/2 = 120, not 153, next

If C, 16*17 = 274
274/2 = 137, not 153, next

If D, 17*18 = 306

Am I missing any important concepts by answering the question with this long-hand method as opposed to your more structured approach?

Kind regards

Derek
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5453
Location: Pune, India
Followers: 1334

Kudos [?]: 6782 [2] , given: 177

Re: If each participant of a chess tournament plays exactly one [#permalink]  11 Nov 2012, 18:40
2
KUDOS
Expert's post
derekgmat wrote:
Thanks Karishma

I must admit I am struggling to work through your solution. I looked at the question again and answered it with the approach of working through each of the choices long-hand.

If A, then 15 players means they each will play 14 games (so 14*15), divide the result by two for double counted games:
14*15 = 210
201/2 = 105, not 153, next

If B, 15*16 = 240
240/2 = 120, not 153, next

If C, 16*17 = 274
274/2 = 137, not 153, next

If D, 17*18 = 306

Am I missing any important concepts by answering the question with this long-hand method as opposed to your more structured approach?

Kind regards

Derek

Your method is absolutely fine. Your logic of multiplying n players by (n-1) games and dividing by 2 is great. The issue with your approach is that you need to calculate for every option. If the numbers were a little larger, you would end up doing calculations a number of times.
For each option, you are doing this calculation (n-1)*n/2
(A) 14*15/2
(B) 15*16/2
etc
You are trying to find the option that will give you the result 153. I have done the same thing. I get (n-1)*n/2 = 153.
Instead of trying all options, you should multiply 153 by 2 to get 306 and then see which two numbers will end in 6. That way you will save a lot of calculations.

14*15 - No
15*16 - No
16*17 - No
17*18 - Yes
18*19 - No

As for the approach used by me to arrive at n*(n-1)/2:
Think of it this way - you make all participants stand in a straight line. The first one comes up and play a game with everyone else i.e. (n-1) games and goes away. The next one comes up and plays a game with all remaining people i.e. (n-2) people and goes away too. This goes on till last two people are left and one comes up, play a game against the other and they both go away. Hence, they end up playing
(n-1) + (n - 2) + .... 3 + 2 + 1 games (total number of games)

You must learn that sum of first m positive integers is given by m(m+1)/2 (very useful to know this)
We need to sum first (n-1) numbers so their sum will be (n-1)n/2

Bunuel has used the combinatorics approach to arrive at n(n-1)/2. There are n people and you want to select as many distinct two people teams as you can (since each person can play against the other only once)
In how many ways can you do that? nC2 ways
nC2 = n(n-1)/2

So the calculation involved is the same in every case. The method of arriving at the equation is different.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Veritas Prep GMAT course is coming to India. Enroll in our weeklong Immersion Course that starts March 29!

Veritas Prep Reviews

Current Student
Status: Done with formalities.. and back..
Joined: 15 Sep 2012
Posts: 648
Location: India
Concentration: Strategy, General Management
Schools: Olin - Wash U - Class of 2015
WE: Information Technology (Computer Software)
Followers: 38

Kudos [?]: 401 [1] , given: 23

Re: If each participant of a chess tournament plays exactly one [#permalink]  11 Nov 2012, 04:07
1
KUDOS
derekgmat wrote:
Thanks Karishma

I must admit I am struggling to work through your solution. I looked at the question again and answered it with the approach of working through each of the choices long-hand.

If A, then 15 players means they each will play 14 games (so 14*15), divide the result by two for double counted games:
14*15 = 210
201/2 = 105, not 153, next

If B, 15*16 = 240
240/2 = 120, not 153, next

If C, 16*17 = 274
274/2 = 137, not 153, next

If D, 17*18 = 306

Am I missing any important concepts by answering the question with this long-hand method as opposed to your more structured approach?

Kind regards

Derek

both are actually doing the same thing. You approach is a logical one while Karishma's is a structured one. But eventually both are doing same thing:
n*(n-1)/2 = 153 ( or 17*18/2 =153)
_________________

Lets Kudos!!!
Black Friday Debrief

Senior Manager
Joined: 13 Aug 2012
Posts: 464
Concentration: Marketing, Finance
GMAT 1: Q V0
GPA: 3.23
Followers: 17

Kudos [?]: 257 [1] , given: 11

Re: If each participant of a chess tournament plays exactly one [#permalink]  28 Dec 2012, 05:32
1
KUDOS
derekgmat wrote:
If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants.

A. 15
B. 16
C. 17
D. 18
E. 19

$$\frac{P!}{2!(P-2)!}=153$$
$$\frac{P * (P-1) * (P-2)!}{(P-2)!}=2*9*17$$
$$P*(P-1) = 18*17$$
$$P = 18$$

_________________

Impossible is nothing to God.

Intern
Joined: 05 Jun 2012
Posts: 25
WE: Marketing (Retail)
Followers: 0

Kudos [?]: 10 [1] , given: 0

Re: If each participant of a chess tournament plays exactly one [#permalink]  29 Dec 2012, 02:59
1
KUDOS
derekgmat wrote:
If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants.

A. 15
B. 16
C. 17
D. 18
E. 19

Num of games - 153, The total number of players that can be chosen as a pair would be nC2.

Now look through the options & start form the middle number as the options are in ascending order - 17C2 = 17*16/2 = 6 as last digit (17 *8) ignore
Next number 18c2 = 18*17/2 = 3 as last digit this is the answer

Choice D
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 4767
Followers: 296

Kudos [?]: 52 [0], given: 0

Re: If each participant of a chess tournament plays exactly one [#permalink]  04 Jul 2014, 20:46
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Senior Manager
Joined: 23 Jan 2013
Posts: 284
Schools: Cambridge'16
Followers: 2

Kudos [?]: 43 [0], given: 31

Re: If each participant of a chess tournament plays exactly one [#permalink]  01 Oct 2014, 03:32
x!/2!*(x-2)!=153, so

Backsolving looks better:

get C=17
17!/2!*15!=17*16/2=136

go D=18
18!/2!*16!=18*17/2=153 (correct)

D
Re: If each participant of a chess tournament plays exactly one   [#permalink] 01 Oct 2014, 03:32
Similar topics Replies Last post
Similar
Topics:
Two teams, X and Y played against each other in a tournament 1 18 Feb 2012, 00:54
6 There are 5 chess amateurs playing in Villa's chess club 10 07 Feb 2012, 20:36
2 GMAT Study Tip: Play Chess 3 13 Feb 2010, 17:05
There are 10 people to play in the tournament in which a 4 22 Jul 2006, 08:37
In a basketball tournament, each of 4 players must play each 1 23 Feb 2006, 03:21
Display posts from previous: Sort by