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

quantumliner: thanks, but how did you get this equation: = (x-1)(1+x-1)/2 = x(x-1)/2.

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

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.

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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I still dont understand, is there another method to approach this question?

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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