In order to compensate for a diffecult midterm exam. Danialle's mathematics teacher adjusted each of the 25 students' midterm exam scores by replacing it by one half of the original score increased by 50. If the mean of the revised set of midterm scores is 82, what was the mean of the original set of scores?

Answer

To find the mean of the original set of scores, we can follow these steps:

### Step 1: Understand the adjustment made to the scores
The teacher adjusted each student's score by taking half of the original score and then adding 50. This means if \( x \) is the original score, the adjusted score \( y \) can be expressed as:
\[ y = \frac{1}{2}x + 50 \]

### Step 2: Set up the equation for the mean of the adjusted scores
Let \( x_1, x_2, \ldots, x_{25} \) be the original scores of the 25 students. The mean of the original scores \( x_m \) is given by:
\[ x_m = \frac{x_1 + x_2 + \ldots + x_{25}}{25} \]
The mean of the adjusted scores \( y_m \) is given as 82. Therefore, we can express the mean of the adjusted scores as:
\[ y_m = \frac{y_1 + y_2 + \ldots + y_{25}}{25} \]

### Step 3: Substitute the adjusted scores in terms of original scores
Using the adjustment formula, we can rewrite the mean of the adjusted scores:
\[ y_m = \frac{\left(\frac{1}{2}x_1 + 50\right) + \left(\frac{1}{2}x_2 + 50\right) + \ldots + \left(\frac{1}{2}x_{25} + 50\right)}{25} \]
This simplifies to:
\[ y_m = \frac{\frac{1}{2}(x_1 + x_2 + \ldots + x_{25}) + 25 \times 50}{25} \]

### Step 4: Simplify the equation
Now, we can factor out the terms:
\[ y_m = \frac{1}{2} \cdot \frac{x_1 + x_2 + \ldots + x_{25}}{25} + \frac{25 \times 50}{25} \]
This further simplifies to:
\[ y_m = \frac{1}{2}x_m + 50 \]

### Step 5: Set up the equation with the known mean of adjusted scores
We know that the mean of the adjusted scores \( y_m \) is 82:
\[ 82 = \frac{1}{2}x_m + 50 \]

### Step 6: Solve for the original mean \( x_m \)
To isolate \( x_m \), we can rearrange the equation:
\[ 82 - 50 = \frac{1}{2}x_m \]
\[ 32 = \frac{1}{2}x_m \]
Multiplying both sides by 2 gives:
\[ x_m = 64 \]

### Conclusion
The mean of the original set of scores is \( \boxed{64} \).

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