A batsman's runs just before the last match of the season, adds up to 750. In his last two innings, he scores only 6 runs, and his average drops by 2. Find his final average of the season.

Answer

To solve the problem, we need to find the final average of the batsman after his last two innings. Here’s a step-by-step solution:

### Step 1: Determine the number of innings before the last match
Let’s denote the number of innings before the last match as \( n \). The total runs scored before the last match is given as 750.

### Step 2: Calculate the average before the last match
The average runs scored before the last match can be calculated using the formula:
\[ \text{Average} = \frac{\text{Total Runs}}{\text{Number of Innings}} = \frac{750}{n} \]

### Step 3: Calculate the total runs after the last two innings
In the last two innings, the batsman scores 6 runs. Therefore, the total runs after the last two innings will be:
\[ \text{Total Runs after last two innings} = 750 + 6 = 756 \]

### Step 4: Determine the new number of innings
After the last two innings, the total number of innings becomes:
\[ \text{Total Innings} = n + 2 \]

### Step 5: Calculate the new average
The new average after the last match can be calculated as:
\[ \text{New Average} = \frac{\text{Total Runs after last two innings}}{\text{Total Innings}} = \frac{756}{n + 2} \]

### Step 6: Set up the equation based on the average drop
According to the problem, the new average drops by 2 runs compared to the previous average:
\[ \frac{756}{n + 2} = \frac{750}{n} - 2 \]

### Step 7: Solve the equation
To solve the equation, we first eliminate the fractions by multiplying both sides by \( n(n + 2) \):
\[ 756n = 750(n + 2) - 2n(n + 2) \]
Expanding both sides gives:
\[ 756n = 750n + 1500 - 2n^2 - 4n \]
Rearranging the equation results in:
\[ 2n^2 - 2n + 1500 = 0 \]
Dividing the entire equation by 2 simplifies it to:
\[ n^2 - n + 750 = 0 \]

### Step 8: Use the quadratic formula
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -1, c = 750 \):
\[ n = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 750}}{2 \cdot 1} \]
\[ n = \frac{1 \pm \sqrt{1 - 3000}}{2} \]
Since the discriminant is negative, we need to check our calculations.

### Step 9: Re-evaluate the equation
Let’s go back to the equation:
\[ 756n = 750n + 1500 - 2n^2 - 4n \]
Rearranging gives:
\[ 2n^2 - 4n - 1500 = 0 \]
Dividing by 2:
\[ n^2 - 2n - 750 = 0 \]
Using the quadratic formula again:
\[ n = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-750)}}{2 \cdot 1} \]
\[ n = \frac{2 \pm \sqrt{4 + 3000}}{2} \]
\[ n = \frac{2 \pm \sqrt{3004}}{2} \]
Calculating the square root:
\[ n = \frac{2 \pm 54.77}{2} \]
Taking the positive root:
\[ n \approx 28.39 \]
Since \( n \) must be a whole number, we round down to 28.

### Step 10: Calculate the final average
Now substituting \( n = 28 \) back into the average formula:
\[ \text{Previous Average} = \frac{750}{28} \approx 26.79 \]
The new average after the last two innings:
\[ \text{New Average} = 26.79 - 2 = 24.79 \]

Thus, the final average of the season is approximately **24.79**.