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There are x teams and each team plays each other once. If there was on

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There are x teams and each team plays each other once. If there was on  [#permalink]

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New post Updated on: 28 Jun 2017, 09:02
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There are x teams and each team plays each other once. If there was one less team, there would be 3/4 as many games. What is the value of x?

A) 9
B) 8
C) 7
D) 6
E) 5

Originally posted by ManSab on 28 Jun 2017, 08:29.
Last edited by Bunuel on 28 Jun 2017, 09:02, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Combination Problem  [#permalink]

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New post 28 Jun 2017, 09:10
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If there are 'X' teams, then the total number of games played by each team once is given by = 1+2+3+4+5+6+7+ ...+ (X-1)

= (x-1)(1+x-1)/2 = x(x-1)/2

If there are 'X-1' teams, then the total number of games played by each team once is given by = 1+2+3+4+5+6+7+ ...+ (X-2)

= (x-2)(1+x-2)/2 = (x-2)(x-1)/2

As per given information:


(x-2)(x-1)/2 = 3/4 * x(x-1)/2


==> (x-2)(x-1)=3/4 * x(x-1)

==> (x-2) = 3/4 * x

==> 4x - 8 = 3x

==> x = 8

Answer is B
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Re: There are x teams and each team plays each other once. If there was on  [#permalink]

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New post 28 Jun 2017, 09:22
quantumliner: thanks, but how did you get this equation: = (x-1)(1+x-1)/2 = x(x-1)/2.
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Re: There are x teams and each team plays each other once. If there was on  [#permalink]

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New post 28 Jun 2017, 09:37
If you notice - 1+2+3+4+5+6+7+ ...+ (X-1) is a sequence in Arithmetic Progression

So you can use the formula to find the sum of all terms, which gives you the total number of games played.
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Re: There are x teams and each team plays each other once. If there was on  [#permalink]

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New post 28 Jun 2017, 09:42
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ManSab wrote:
There are x teams and each team plays each other once. If there was one less team, there would be 3/4 as many games. What is the value of x?

A) 9
B) 8
C) 7
D) 6
E) 5


we have "\(x\)" and "\({x-1}\)" teams and for a game we need 2 teams

So number of Games with "\(x\)" teams = \(C^x_2\) and
number of Games with "\({x-1}\)" teams = \(C^{x-1}_2\)
as per the question \(\frac{3}{4}\)*\(C^x_2\) = \(C^{x-1}_2\). Solving this we will get
\(\frac{3}{4}*x*(x-1)\) = \((x-1)*(x-2)\), or \(3x = 4x-8\)
therefore \(x = 8\)

Option B
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Re: There are x teams and each team plays each other once. If there was on  [#permalink]

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New post 30 Jun 2017, 09:24
ManSab wrote:
There are x teams and each team plays each other once. If there was one less team, there would be 3/4 as many games. What is the value of x?

A) 9
B) 8
C) 7
D) 6
E) 5


The number of games played if there are x teams is xC2 and the number of games played if there were one less team would be (x-1)C2. Thus, we can create the following equation:

(x-1)C2 = ¾ * xC2

(x - 1)(x - 2)/2 = ¾[x(x - 1)/2]

x - 2 = ¾(x)

4(x - 2) = 3x

4x - 8 = 3x

x = 8

Answer: B
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Re: There are x teams and each team plays each other once. If there was on  [#permalink]

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New post 16 Jul 2017, 18:34
akh...forgot the statistics, and did the "long" way, by drawing a table...
if we had 7 teams, we would have had 21 games, but we see that this number is not divisible by 4.
if we had 8 teams, we would have had 28 games, and this number is divisible by 4.
so 8 must be the answer.
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Re: There are x teams and each team plays each other once. If there was on  [#permalink]

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New post 17 Jul 2017, 09:59
Hi ScottTargetTestPrep, Could you please breakdown how we get this expression-> (x - 1)(x - 2)/2 = ¾[x(x - 1)/2] ?

Thanks
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Re: There are x teams and each team plays each other once. If there was on  [#permalink]

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New post 06 Jun 2018, 16:52
ManSab wrote:
There are x teams and each team plays each other once. If there was one less team, there would be 3/4 as many games. What is the value of x?

A) 9
B) 8
C) 7
D) 6
E) 5


The number of games played if there are x teams is xC2 and the number of games played if there were one less team would be (x-1)C2. Thus, we can create the following equation:

(x-1)C2 = 3/4 * xC2

(x - 1)(x - 2)/2 = 3/4[x(x - 1)/2]

x - 2 =3x/4

4(x - 2) = 3x

4x - 8 = 3x

x = 8

Answer: B
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There are x teams and each team plays each other once. If there was on  [#permalink]

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New post 07 Jun 2018, 12:53
ManSab wrote:
There are x teams and each team plays each other once. If there was one less team, there would be 3/4 as many games. What is the value of x?

A) 9
B) 8
C) 7
D) 6
E) 5


(x-1)(x-2)/x(x-1)=3/4
➡(x-2)/x=3/4
x=8
B
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Re: There are x teams and each team plays each other once. If there was on  [#permalink]

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New post 03 Aug 2018, 12:32
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I still dont understand, is there another method to approach this question?
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Re: There are x teams and each team plays each other once. If there was on  [#permalink]

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New post 03 Aug 2018, 23:30
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Sandy56 wrote:
I still dont understand, is there another method to approach this question?


I think the question is not properly worded. It should be
"There are x teams and each team plays each other once. If there was one less team, there would be 3/4 as many games [as originally played ]. What is the value of x?

Now, let's say we have x teams. Number of matches played among these teams would be xC2. Condition given is if there's one less team, then 3/4 games as originally played.
This boils down to,
(x-1)C2= (3/4) * (xC2) --> 3/4 of originally played games

Now Apply Combinations.
(x-1)(x-2) / 2 = (3/4) * (x)(x-1)/2
x-2 = 3/4 * (x)
4x-8=3x
x=8
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Re: There are x teams and each team plays each other once. If there was on  [#permalink]

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New post 04 Aug 2018, 00:20
I didn't understand. Pls can anyone give any easier solution? Why x-2 is coming everything time?

Posted from my mobile device
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Re: There are x teams and each team plays each other once. If there was on  [#permalink]

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New post 04 Aug 2018, 02:59
Assume there are three teams in total A B and C. A plays 2 matches against B &C, B plays 2 matches against C&A, C plays 2 matches against A&B. Total of 6 matches.

With 3 teams in total the number of matches are 3 * 2 =6. Similarly, extend the example to 4 teams ( A,B,C&D) and you will find that they play total of 12 matches which is 4*3. It is safe to say that, for any 'x' teams the total number of matches, lets call it 'Y' will be x*(x-1).

So, y=x*(x-1). Lets call this Eq. A.

Based on the condition given, you reduce x by 1 and the total number of matches reduce to 3/4th which is 3y/4.

Substitue x with x-1 in our above Eq.A and replace y with 3y/4. Effectively, 3y/4=(x-1) *(x-1-1) which is 3y/4=(x-1)*(x-2). Lets call this Eq.B.

Equate both Eq A and Eq B. You get, the value of x as 8.
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Re: There are x teams and each team plays each other once. If there was on  [#permalink]

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New post 04 Aug 2018, 03:18
01520102735 wrote:
I didn't understand. Pls can anyone give any easier solution? Why x-2 is coming everything time?

Posted from my mobile device


which part didn't you understand?

Let's say we have 3 teams A, B, C. Each team plays each other once so we have 3 games: AB, AC, BC
In case of 4 teams A, B, C, D, we have 6 games : AB, AC, AD, BC, BC, CD
You can see the pattern here. When we have m teams:
The 1st team plays with other (m-1) teams -> we have (m-1) games
The 2nd team plays with (m-2) teams (as the 2nd team can't play game with itself and the game between 1st team and 2nd team is already counted in number of games that 1st team plays) --> we have (m-2) games
...
So when there are m teams, each team plays each other once, the number of games = (m-1) + (m-2) + ... +2+1 = \(\frac{[(m-1)+1]*(m-1)}{2}\) = \(\frac{m*(m-1)}{2}\)
In this question:
when there are x teams, the number of games = \(\frac{x*(x-1)}{2}\)
when there are (x-1) teams, the number of games = \(\frac{(x-1)*[(x-1)-1]}{2}\) = \(\frac{(x-1)*(x-2)}{2}\)

Now we have \(\frac{(x-1)*(x-2)}{2}\) = \(\frac{3}{4}\) * \(\frac{x*(x-1)}{2}\)
Solve this equation we have the result x = 8.

Answer: B.

Hope it helps.
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Re: There are x teams and each team plays each other once. If there was on  [#permalink]

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New post 06 Aug 2018, 00:54
It is a simple logic:

Order does not matter!
Let's assume there are 8 teams: a b c d e f g h.
ab,ac,ad,ae,af,ag,ah (7 games) then, bc,bd,be,bf,bg,bh (6games), so the logic stands as: 7+6+5+4+3+2+1=28 games in total.
Now, when one team is absent, let's assume a is absent: bc,bd,be,bf,bg,bh (6 games) then 5 games then 4. So, 6+5+4+3+2+1=21 games without one team.
Now, 21/28 = 3/4
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Re: There are x teams and each team plays each other once. If there was on &nbs [#permalink] 06 Aug 2018, 00:54
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