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!
There are x teams and each team plays each other once
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Each game is played by a PAIR OF TEAMS.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
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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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: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
(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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