In a certain learning experiment, each participant had three trials and was assigned, for each trial, a score of either...

1. Translate the problem requirements: Anne's final score = \(\mathrm{(first\ trial\ score) + 2 \times (second\ trial\ score) + 3 \times (third\ trial\ score)}\). Anne got scores of 1 and -1 in her first two trials in some order, and her third trial score can be any of \(\{-2, -1, 0, 1, 2\}\). We need to find which final score is impossible.
2. Identify the possible scenarios for first two trials: Since Anne received 1 and -1 for her first two trials in some order, there are exactly two cases to consider based on which score came first versus second.
3. Calculate all possible final scores for each scenario: For each of the two cases from the previous step, systematically compute the final score for all five possible third trial scores.
4. Determine the impossible final score: Compare the complete list of possible final scores against the answer choices to identify which one cannot occur.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we know in everyday language first. Anne took three trials, and each trial gave her a score between -2 and 2. But here's the key: the trials don't all count equally toward her final score. The first trial counts once, the second trial counts double (multiplied by 2), and the third trial counts triple (multiplied by 3).

We know Anne got a score of 1 on one trial and -1 on another trial in her first two attempts. We don't know which order they came in, and we don't know what she got on her third trial yet.

So Anne's final score formula is: \(\mathrm{Final\ Score = (first\ trial\ score) + 2 \times (second\ trial\ score) + 3 \times (third\ trial\ score)}\)

Process Skill: TRANSLATE - Converting the problem's weighted scoring system into a clear mathematical relationship

2. Identify the possible scenarios for first two trials

Since Anne received 1 and -1 for her first two trials in some order, let's think about the two possibilities:

Scenario A: Anne got 1 on her first trial and -1 on her second trial
Scenario B: Anne got -1 on her first trial and 1 on her second trial

These are the only two ways to arrange the scores 1 and -1 across the first two trials. Now we need to see what final scores are possible in each scenario when we consider all possible third trial scores.

Process Skill: CONSIDER ALL CASES - Systematically identifying all possible arrangements of the known scores

3. Calculate all possible final scores for each scenario

Let's work through each scenario systematically. Remember, the third trial score can be any of: -2, -1, 0, 1, or 2.

Scenario A: First trial = 1, Second trial = -1
\(\mathrm{Final\ Score = 1 + 2 \times (-1) + 3 \times (third\ trial)}\)
\(\mathrm{Final\ Score = 1 - 2 + 3 \times (third\ trial)}\)
\(\mathrm{Final\ Score = -1 + 3 \times (third\ trial)}\)

Now let's try each possible third trial score:

• If third trial = -2: \(\mathrm{Final\ Score = -1 + 3 \times (-2) = -1 - 6 = -7}\)
• If third trial = -1: \(\mathrm{Final\ Score = -1 + 3 \times (-1) = -1 - 3 = -4}\)
• If third trial = 0: \(\mathrm{Final\ Score = -1 + 3 \times (0) = -1 + 0 = -1}\)
• If third trial = 1: \(\mathrm{Final\ Score = -1 + 3 \times (1) = -1 + 3 = 2}\)
• If third trial = 2: \(\mathrm{Final\ Score = -1 + 3 \times (2) = -1 + 6 = 5}\)

Scenario B: First trial = -1, Second trial = 1
\(\mathrm{Final\ Score = -1 + 2 \times (1) + 3 \times (third\ trial)}\)
\(\mathrm{Final\ Score = -1 + 2 + 3 \times (third\ trial)}\)
\(\mathrm{Final\ Score = 1 + 3 \times (third\ trial)}\)

Now let's try each possible third trial score:

• If third trial = -2: \(\mathrm{Final\ Score = 1 + 3 \times (-2) = 1 - 6 = -5}\)
• If third trial = -1: \(\mathrm{Final\ Score = 1 + 3 \times (-1) = 1 - 3 = -2}\)
• If third trial = 0: \(\mathrm{Final\ Score = 1 + 3 \times (0) = 1 + 0 = 1}\)
• If third trial = 1: \(\mathrm{Final\ Score = 1 + 3 \times (1) = 1 + 3 = 4}\)
• If third trial = 2: \(\mathrm{Final\ Score = 1 + 3 \times (2) = 1 + 6 = 7}\)

So all possible final scores for Anne are: \(\{-7, -5, -4, -2, -1, 1, 2, 4, 5, 7\}\)

4. Determine the impossible final score

Now let's check each answer choice against our list of possible final scores:

• Choice A: -4 ✓ (possible in Scenario A with third trial = -1)
• Choice B: -2 ✓ (possible in Scenario B with third trial = -1)
• Choice C: 1 ✓ (possible in Scenario B with third trial = 0)
• Choice D: 5 ✓ (possible in Scenario A with third trial = 2)
• Choice E: 6 ✗ (NOT in our list of possible scores)

Looking at our complete list \(\{-7, -5, -4, -2, -1, 1, 2, 4, 5, 7\}\), we can see that 6 is not among the possible final scores.

Final Answer

The answer is E. 6 could NOT be Anne's final score. We systematically calculated all possible final scores by considering both ways to arrange her first two trial scores (1 and -1) and all five possible third trial scores (-2, -1, 0, 1, 2). The final score of 6 never appears in any valid combination.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the weighted scoring system
Students often miss that the trials have different weights in the final score calculation. They might think all three trials count equally (1+1+1) instead of recognizing the 1:2:3 weighting pattern. This leads them to use an incorrect formula like \(\mathrm{Final\ Score = trial1 + trial2 + trial3}\) instead of \(\mathrm{Final\ Score = trial1 + 2 \times trial2 + 3 \times trial3}\).

2. Overlooking the constraint about trial order
Students may assume Anne got 1 on the first trial and -1 on the second trial without considering that the problem states these scores occurred "not necessarily in that order." This causes them to miss one entire scenario, leading to an incomplete list of possible final scores.

3. Forgetting the range of possible third trial scores
Students might focus so much on the given information about the first two trials that they forget the third trial can be any score from -2 to 2. Some may assume the third trial must also be 1 or -1, significantly limiting their calculation of possible outcomes.

Errors while executing the approach

1. Arithmetic errors in weighted calculations
When calculating \(\mathrm{Final\ Score = first + 2 \times second + 3 \times third}\), students commonly make sign errors, especially with negative numbers. For example, calculating \(\mathrm{1 + 2 \times (-1) + 3 \times 2}\) as \(\mathrm{1 - 2 + 6 = 5}\) instead of correctly getting \(\mathrm{1 - 2 + 6 = 5}\), or making errors like \(\mathrm{1 + 2 \times (-1) = 1 + (-2) = -1}\) instead of \(\mathrm{1 - 2 = -1}\).

2. Incomplete enumeration of scenarios
Students may correctly identify both scenarios (1,-1) and (-1,1) for the first two trials but then fail to systematically calculate all five possible third trial scores for each scenario. They might calculate only a few cases and miss others, leading to an incomplete list of possible final scores.

3. Mixing up scenarios during calculation
When working through multiple scenarios simultaneously, students often confuse which trial scores belong to which scenario. They might start calculating Scenario A (first=1, second=-1) but accidentally use the formula from Scenario B (first=-1, second=1), leading to incorrect final scores.

Errors while selecting the answer

1. Checking answer choices against an incomplete list
If students missed some possible final scores during their calculations, they might incorrectly conclude that a valid score is impossible. For example, if they only calculated one scenario, they might think -2 is impossible when it actually occurs in Scenario B with third trial = -1.

2. Misreading the question requirement
Students may forget that the question asks for what could NOT be Anne's final score and instead select an answer that IS possible. This is a classic trap where students find a score that can be achieved and mistakenly choose it, rather than looking for the score that cannot be achieved.