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There are x teams and each team plays each other once. If there was on
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Updated on: 28 Jun 2017, 08:02
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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
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Originally posted by ManSab on 28 Jun 2017, 07:29.
Last edited by Bunuel on 28 Jun 2017, 08:02, edited 1 time in total.
Renamed the topic and edited the question.



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Re: Combination Problem
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28 Jun 2017, 08:10
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+ ...+ (X1)
= (x1)(1+x1)/2 = x(x1)/2
If there are 'X1' teams, then the total number of games played by each team once is given by = 1+2+3+4+5+6+7+ ...+ (X2)
= (x2)(1+x2)/2 = (x2)(x1)/2
As per given information:
(x2)(x1)/2 = 3/4 * x(x1)/2
==> (x2)(x1)=3/4 * x(x1)
==> (x2) = 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
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28 Jun 2017, 08:22
quantumliner: thanks, but how did you get this equation: = (x1)(1+x1)/2 = x(x1)/2.



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Re: There are x teams and each team plays each other once. If there was on
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28 Jun 2017, 08:37
If you notice  1+2+3+4+5+6+7+ ...+ (X1) 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
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28 Jun 2017, 08:42
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 "\({x1}\)" teams and for a game we need 2 teams So number of Games with "\(x\)" teams = \(C^x_2\) and number of Games with "\({x1}\)" teams = \(C^{x1}_2\) as per the question \(\frac{3}{4}\)*\(C^x_2\) = \(C^{x1}_2\). Solving this we will get \(\frac{3}{4}*x*(x1)\) = \((x1)*(x2)\), or \(3x = 4x8\) 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
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30 Jun 2017, 08:24
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 (x1)C2. Thus, we can create the following equation: (x1)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
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16 Jul 2017, 17:34
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
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17 Jul 2017, 08:59
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
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06 Jun 2018, 15:52
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 (x1)C2. Thus, we can create the following equation: (x1)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
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07 Jun 2018, 11:53
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 (x1)(x2)/x(x1)=3/4 ➡(x2)/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
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03 Aug 2018, 11:32
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
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03 Aug 2018, 22:30
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, (x1)C2= (3/4) * (xC2) > 3/4 of originally played games Now Apply Combinations. (x1)(x2) / 2 = (3/4) * (x)(x1)/2 x2 = 3/4 * (x) 4x8=3x x=8



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Re: There are x teams and each team plays each other once. If there was on
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03 Aug 2018, 23:20
I didn't understand. Pls can anyone give any easier solution? Why x2 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
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04 Aug 2018, 01:59
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*(x1).
So, y=x*(x1). 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 x1 in our above Eq.A and replace y with 3y/4. Effectively, 3y/4=(x1) *(x11) which is 3y/4=(x1)*(x2). 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
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04 Aug 2018, 02:18
01520102735 wrote: I didn't understand. Pls can anyone give any easier solution? Why x2 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 (m1) teams > we have (m1) games The 2nd team plays with (m2) 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 (m2) games ... So when there are m teams, each team plays each other once, the number of games = (m1) + (m2) + ... +2+1 = \(\frac{[(m1)+1]*(m1)}{2}\) = \(\frac{m*(m1)}{2}\) In this question: when there are x teams, the number of games = \(\frac{x*(x1)}{2}\) when there are (x1) teams, the number of games = \(\frac{(x1)*[(x1)1]}{2}\) = \(\frac{(x1)*(x2)}{2}\) Now we have \(\frac{(x1)*(x2)}{2}\) = \(\frac{3}{4}\) * \(\frac{x*(x1)}{2}\) 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
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05 Aug 2018, 23:54
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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