If in a chess tournament each contestant plays once against each of the other and in all 45 games are played, then the number of participants, is

Options

A. 9
B. 10
C. 15
D. none of these

Answer

To find the number of participants in a chess tournament where each contestant plays once against each of the other contestants, and a total of 45 games are played, we can follow these steps:

### Step 1: Understanding the Problem
In a chess tournament with \( n \) participants, each participant plays against every other participant exactly once. The total number of games played can be represented mathematically.

### Step 2: Formula for Total Games
The total number of games played in a round-robin tournament (where each participant plays against every other participant) is given by the formula:
\[ \text{Total Games} = \frac{n(n-1)}{2} \]
This formula arises because each of the \( n \) participants plays \( n-1 \) games, but since each game is counted twice (once for each participant), we divide by 2.

### Step 3: Setting Up the Equation
According to the problem, the total number of games played is 45. Therefore, we can set up the equation:
\[ \frac{n(n-1)}{2} = 45 \]

### Step 4: Solving the Equation
To eliminate the fraction, multiply both sides of the equation by 2:
\[ n(n-1) = 90 \]
Now we need to rearrange this into a standard quadratic equation:
\[ n^2 - n - 90 = 0 \]

### Step 5: Factoring the Quadratic Equation
Next, we will factor the quadratic equation:
\[ n^2 - n - 90 = (n - 10)(n + 9) = 0 \]
Setting each factor to zero gives us the possible solutions:
\[ n - 10 = 0 \quad \Rightarrow \quad n = 10 \]
\[ n + 9 = 0 \quad \Rightarrow \quad n = -9 \]
Since the number of participants cannot be negative, we discard \( n = -9 \).

### Step 6: Conclusion
Thus, the number of participants in the chess tournament is:
\[ \boxed{10} \]