Question

On a True/False test, Amy answered the first three questions wrong but answered the rest of the questions correctly. On the same test, Scott answered exactly two questions wrong. He answered the last question wrong, the 26 questions before it correctly, and the question before that wrong. If Amy and Scott were both incorrect on the same question exactly once, what is the greatest possible percent of the total number of questions that Amy could have answered correctly?

Answer 1

Let $n$ represent the total number of questions on the test. Amy's performance can be summarized as follows: she answered the first three questions incorrectly and all remaining questions correctly. Therefore, her number of correct answers can be expressed as:

$$\text{Amy's Correct Answers}=n-3$$

Next, we analyze Scott's performance. We know he answered two questions incorrectly: the last question (question $n$) and the question before it (question $n-1$). Since he answered $26$ questions before the last two correctly, it means he answered correctly questions $1$ through $n-2$.

Now we can summarize Scott’s correct answers:

$$\text{Scott's Correct Answers}=n-2-2=n-4$$

Given that Amy and Scott were incorrect on the same question exactly once, and knowing Amy answered the first three questions incorrectly, the only question they can both have answered incorrectly is one of these three questions: 1, 2, or 3.

Next, we evaluate scenarios based on the conditions described:

1. If Amy and Scott both answered question $1$ incorrectly, then Amy answered questions $2$ and $3$ also incorrectly (and all other questions correctly). Scott must have gotten question $2$ or $3$ correct, because he has exactly two wrong answers (both of which are $n-1$ and $n$). This scenario leads to a contradiction since Scott cannot have two incorrect answers among the first three.

2. If they both answered question $2$ incorrectly, then Amy answered questions $1$ and $3$ incorrectly (leaving her other answers correct). Scott answered answers $1$ correctly but must also have answered $3$ correctly since his two incorrect answers are $n-1$ and $n$, which again leads to contradictions about the number of incorrect answers.

3. The viable option is that they both answered question $3$ incorrectly. Then, Amy has answered questions $1$ and $2$ correctly (and all afterward), giving her correct answers as:

$$\text{Amy's Correct Answers}=n-3+2=n-1$$

Now, Scott's answer situation leads him to have questions $1,2$ correct and $n-1,n$ incorrect, leading to:

$$\text{Scott's Correct Answers}=n-4$$

Both have now been shown to have actual unanswered overlap. Therefore, we can analyze the maximal $n$ ensuring the constraints hold.

To maximize the proportion of correct answers Amy has, we examine her score to avoid contradictions between the answer conditions. Thus her final correct total is:

$$\text{Correct percentage for Amy}=\frac{n-1}{n}=1-\frac{1}{n}$$

To find the maximum percentage of correct answers, we seek to minimize the total number of questions, $n$. Since $n\ge 3$, let’s analyze $n=3$:

$$\text{Correct percentage}=1-\frac{1}{3}=\frac{2}{3}$$

In conclusion, the greatest possible fraction translates mathematically into:

$$\text{Extra left or missing right}$$

The result here shows:

$$\text{Extra open brace or missing close brace}$$

This appeals to significant refining of the question structure, maintaining sound thresholds for Amy invoking the distributed statements in encapsulated solutions across correct engagements among question queries on examination.