To determine the maximum number of perfect-10 scores possible in this test, let's break down the information provided:

• Mean: 8
• Median: 8
• Mode: 7
• Number of scores (I): We are not explicitly given the number of scores, so we'll denote it as $I$.

Step 1: Calculate the Total Sum of Scores

Given that the mean of the scores is 8, the total sum of all the scores is: $\text{Total sum of scores} = \text{Mean} \times I = 8I$

Step 2: Consider the Median

Since the median is 8, this means that the middle score (or the average of the two middle scores) in the ordered list of scores is 8. This implies that at least half of the scores are 8 or above.

Step 3: Consider the Mode

The mode is 7, meaning that 7 is the most frequently occurring score. However, the problem asks for the maximum number of perfect scores (which are 10s). Thus, we need to minimize the number of 7s while keeping it as the mode to maximize the number of 10s.

Step 4: Maximize the Number of 10s

To maximize the number of 10s, we want to distribute the remaining scores in such a way that the conditions for the mode and median are still met. Let's outline the possibilities:

1. Case 1: Minimizing the number of scores less than 8 and maintaining the mode of 7: If the mode is 7, there must be at least 2 scores of 7. If there were only 1 score of 7, it could not be the mode.

2. Case 2: Rest of the Scores: Assume we have $x$ perfect 10s. The rest of the scores must satisfy the mode of 7, and the median and mean conditions.

Let’s consider $I = 11$ (as there are 11 scores in the quiz):

$\text{Total sum of scores} = 8 \times 11 = 88$

Let $x$ be the number of perfect 10s, and $y$ the number of scores of 7. To satisfy the mode, $y \geq 2$. Then the remaining scores (which are not 10s or 7s) should be distributed as necessary to meet the mean and median conditions.

Given: $10x + 7y + z = 88$ where $z$ is the sum of the remaining scores.

• To maximize $x$, set $y = 2$ (minimum mode).
• Set the median score of 8. Assume the simplest distribution, where other non-7, non-10 scores can be 8.

For example:

• Let $x = 6$ (6 perfect 10s),
• $y = 2$ (2 scores of 7),
• The remaining 3 scores can be 8 to maintain the median.

Now check: $10(6) + 7(2) + 8(3) = 60 + 14 + 24 = 98 \quad \text{(not equal to 88)}$ So, we reduce the number of 10s.

Let $x = 5$: $10(5) + 7(2) + 8(4) = 50 + 14 + 32 = 96 \quad \text{(still too large)}$ Let $x = 4$: $10(4) + 7(2) + 8(5) = 40 + 14 + 40 = 94 \quad \text{(still too large)}$ Let $x = 3$: $10(3) + 7(2) + 8(6) = 30 + 14 + 48 = 92 \quad \text{(still too large)}$ Let $x = 2$: $10(2) + 7(2) + 8(7) = 20 + 14 + 56 = 90 \quad \text{(still too large)}$ Finally, $x = 1$: $10(1) + 7(3) + 8(7) = 10 + 21 + 56 = 87 \quad \text{(still too large)}$ Hence the largest value that satisfies $88$ is: $x = 1, \text{so the max is } 4 perfect scores.$