Let's denote the unreadable score as $x$.

Step 1: Define the problem variables.

• There are 10 students in total.
• The mean score for all 10 students is 81.
• The mean score for the 9 readable scores is 85.

Step 2: Use the mean formula for the entire group.

The mean score formula is:

$\text{Mean} = \frac{\text{Sum of the scores}}{\text{Number of students}}$

For all 10 students, we have:

$81 = \frac{S}{10}$

where $S$ is the total sum of the scores for all 10 students. Solving for $S$:

$S = 81 \times 10 = 810$

So, the total sum of the 10 scores is 810.

Step 3: Calculate the sum of the 9 readable scores.

For the 9 readable scores, the mean is 85, so:

$85 = \frac{S_9}{9}$

where $S_9$ is the sum of the 9 readable scores. Solving for $S_9$:

$S_9 = 85 \times 9 = 765$

So, the sum of the 9 readable scores is 765.

Step 4: Solve for the unreadable score.

Since the total sum of the 10 scores is 810, and the sum of the 9 readable scores is 765, the unreadable score $x$ is:

$x = S - S_9 = 810 - 765 = 45$

Thus, the unreadable score is $\boxed{45}$.

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Would you like more details or have any questions?

Here are 5 questions that expand on this problem:

1. How would the calculation change if the total number of students was different?
2. What happens to the mean if the unreadable score is higher or lower than 45?
3. Can you derive the formula for calculating a missing score given the mean and other scores?
4. How does the sum of the scores relate to the mean of a dataset?
5. How would you calculate the median score in this situation?

Tip: Always double-check calculations when working with averages and totals, as small errors can lead to incorrect conclusions.