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

Question

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 · Answer

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

ManSab · Answer

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

quantumliner · Answer

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.

niks18 · Answer

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

ScottTargetTestPrep · 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 = ¾ * xC2
 
(x - 1)(x - 2)/2 = ¾ 
 
x - 2 = ¾(x)
 
4(x - 2) = 3x
 
4x - 8 = 3x
 
x = 8 
 
Answer: B

mvictor · Answer

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.

ManSab · Answer

Hi ScottTargetTestPrep, Could you please breakdown how we get this expression-> (x - 1)(x - 2)/2 = ¾  ?

Thanks

ScottTargetTestPrep · 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 - 2 =3x/4

4(x - 2) = 3x

4x - 8 = 3x

x = 8

Answer: B

gracie · Answer

(x-1)(x-2)/x(x-1)=3/4
➡(x-2)/x=3/4
x=8
B

Sandy56 · Answer

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

elavendan1 · Answer

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    . 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

01520102735 · Answer

I didn't understand. Pls can anyone give any easier solution? Why x-2 is coming everything time?

Posted from my mobile device

kunalkankariya · Answer

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.

Izzyjolly · Answer

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)}{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)* }{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.

Seryozha · Answer

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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