GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Oct 2019, 07:02

### 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

# Several teams are competing in a basketball tournament, and each team

Author Message
TAGS:

### Hide Tags

GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4009
Several teams are competing in a basketball tournament, and each team  [#permalink]

### Show Tags

05 Oct 2018, 09:32
1
Top Contributor
5
00:00

Difficulty:

65% (hard)

Question Stats:

31% (02:37) correct 69% (02:09) wrong based on 16 sessions

### HideShow timer Statistics

Several teams are competing in a basketball tournament, and each team plays every other team once. Each game has exactly 1 winner and 1 loser (no ties). If 4 teams lost exactly 5 games, 5 teams won exactly 3 games, and each of the remaining teams won all of its games, what is the total number of games played during the tournament?

A) 36
B) 45
C) 55
D) 66
E) 78

*kudos for all correct solutions

_________________
Test confidently with gmatprepnow.com
Intern
Joined: 19 Nov 2017
Posts: 18
Re: Several teams are competing in a basketball tournament, and each team  [#permalink]

### Show Tags

05 Oct 2018, 10:31
1
if the total number of games = 55 then no of team = n*(n-1)/2 ie., N = 11

If total team is 11 : 4 teams lost 5 games ie., they won 5=10-5 games since(win = Total - Loss) . so total wins = 4*5 = 20 wins
similarly 5 teams won 3 so total wins = 15 wins and rest of the team (11-5-4) = 2 won all games (10 games) =2*10 = 20 wins
Total = 20+20+15 = 55
Math Expert
Joined: 02 Aug 2009
Posts: 7984
Re: Several teams are competing in a basketball tournament, and each team  [#permalink]

### Show Tags

05 Oct 2018, 10:51
1
GMATPrepNow wrote:
Several teams are competing in a basketball tournament, and each team plays every other team once. Each game has exactly 1 winner and 1 loser (no ties). If 4 teams lost exactly 5 games, 5 teams won exactly 3 games, and each of the remaining teams won all of its games, what is the total number of games played during the tournament?

A) 36
B) 45
C) 55
D) 66
E) 78

*kudos for all correct solutions

Hi..
Brent, shifted it to PS
_________________
Math Expert
Joined: 02 Aug 2009
Posts: 7984
Re: Several teams are competing in a basketball tournament, and each team  [#permalink]

### Show Tags

05 Oct 2018, 11:16
2
GMATPrepNow wrote:
Several teams are competing in a basketball tournament, and each team plays every other team once. Each game has exactly 1 winner and 1 loser (no ties). If 4 teams lost exactly 5 games, 5 teams won exactly 3 games, and each of the remaining teams won all of its games, what is the total number of games played during the tournament?

A) 36
B) 45
C) 55
D) 66
E) 78

*kudos for all correct solutions

Another way..
Number of matches lost = number of matches won

Number of matches lost
1) 4 teams lost 5 matches so 4*5=20
2) 5 teams won 3 so lost (n-1-(3))=n-4, thus 5*(n-4)
3) all remaining teams won all matches. So 0
Total matches lost = 20+5*(n-4)=20+5n-20=5n

Number of matches won
1) 4 teams lost 5 matches so they won (n-1-(5))=n-6, thus 4*(n-6)
2) 5 teams won 3 so 5*3=15
3) all remaining teams won all matches. Each team plays n-1 matches a there are n teams. Now remaining teams will be n-(4+5)=n-9.. so (n-9)(n-1)
Total matches won = 4*(n-6)+15+(n-9)(n-1)=4n-24+15+n^2-10n+9=n^2-6n

Thus won=lost....5n=n^2-6n......n^2-11n=0....n(n-11)=0
So n 0 or 11, but there are more than 9 teams so n=11

Matches =11C2=11*10/2=55

C
_________________
GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Several teams are competing in a basketball tournament, and each team  [#permalink]

### Show Tags

05 Oct 2018, 14:04
1
I have changed the question stem slightly, because in the original wording it would be impossible to each of two teams win all its games (one played the other)!

Quote:
Several teams are competing in a basketball tournament, and each team plays every other team once. Each game has exactly 1 winner and 1 loser (no ties). If each of 5 teams lost exactly 5 games, each of 4 teams won exactly 3 games, and each of the remaining teams lost exactly 1 game, what is the total number of games played during the tournament?

A) 36
B) 45
C) 55
D) 66
E) 78

Very nice problem, Brent. Congrats! (kudos!)

Let´s suppose there are (N+1) teams. Hence each one will play the N other teams, so that

$$? = {{\left( {N + 1} \right) \cdot N} \over 2}$$

Each of 5 teams lost 5 games and won (N-5) games.
Each of 4 teams lost (N-3) games and won 3 games.
Each of the remaining (N+1-9) teams lost 1 game and won N-1 games. (*)

(*) If there were at least one SINGLE team who lost exactly 5 games AND won exactly 3 games, we would have N=8 (hence 9 teams), impossible.
(No team would have lost exactly one game, contradicting the question stem.)

There were no ties, therefore the number of games lost must equal the number of games won, hence:

$$5 \cdot 5 + 4\left( {N - 3} \right) + \left[ {\left( {N + 1} \right) - 9} \right] \cdot 1 = 5\left( {N - 5} \right) + 4 \cdot 3 + \left[ {\left( {N + 1} \right) - 9} \right] \cdot \left( {N - 1} \right)$$

$${N^2} - 9N - 10 = 0\,\,\,\,\mathop \Rightarrow \limits^{{\rm{sum}} = 9\,\,,\,\,\,{\rm{product}} = - 10} \,\,\,\,\,N = 10\,\,\,\,{\rm{or}}\,\,\,\,N = - 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,N = 10$$

$$? = {{\left( {N + 1} \right) \cdot N} \over 2} = 55$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Several teams are competing in a basketball tournament, and each team   [#permalink] 05 Oct 2018, 14:04
Display posts from previous: Sort by