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In a certain learning experiment, each participant had three trials and was assigned, for each trial, a score of either \(-2, -1, 0, 1, \text{ or } 2\). The participant's final score consisted of the sum of the first trial score, \(2\) times the second trial score, and \(3\) times the third trial score. If Anne received scores of \(1 \text{ and } -1\) for her first two trials, not necessarily in that order, which of the following could NOT be her final score?
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
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
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:
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:
So all possible final scores for Anne are: \(\{-7, -5, -4, -2, -1, 1, 2, 4, 5, 7\}\)
Now let's check each answer choice against our list of possible final 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.
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.
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.
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.
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.