There are x teams and each team plays each other once

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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

The OA is B.

Please, can anyone assist me with this PS question? I'm confused. Thanks!

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by GMATGuruNY » Mon Jun 04, 2018 1:49 pm
BTGmoderatorLU 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
Each game is played by a PAIR OF TEAMS.
Since every team must play every other team exactly once, the total number of games is to equal to the number of possible pairs that can be formed from the x teams.
We can PLUG IN THE ANSWERS, which represent the value of x.
When the correct answer is plugged in:
(number of pairs that can be formed from x-1 teams)/(number of pairs that can be formed from x teams) = 3/4.

B: x = 8 teams
From 8 teams, the number of pairs that can be formed = 8C2 = (8*7)/(2*1) = 28.
If the number of teams decreases by 1, the number of pairs that can be formed from the remaining 7 teams = (7*6)/(2*1) = 21.
21/28 = 3/4.
Success!

The correct answer is B.
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by Scott@TargetTestPrep » Wed Jun 06, 2018 4:02 pm
BTGmoderatorLU 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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