- Translate the problem requirements: We have 10 children contributing 28 total pieces of clothing with a range of 2 (difference between highest and lowest contributions). We need to determine which values (0, 5, or 9) could represent how many children contributed exactly 3 pieces each.
- Establish the contribution constraints: Since the range is 2, if the minimum contribution is 'min', then the maximum is \(\mathrm{min + 2}\). All contributions must fall within this 2-number span.
- Test each scenario systematically: For each given option (0, 5, or 9 children contributing 3 pieces), determine what values of 'min' would make the total equal 28, then verify if 3 falls within the allowable range.
- Validate feasibility: Check that each valid scenario can actually be achieved with 10 children contributing integer amounts within the established range.
Execution of Strategic Approach
1. Translate the problem requirements
Let's break down what we know in simple terms:
- We have 10 children total
- They contributed 28 pieces of clothing combined
- The range is 2, meaning the difference between the highest and lowest individual contributions is exactly 2
- We need to find which of these could be true: 0 children contributed 3 pieces, OR 5 children contributed 3 pieces, OR 9 children contributed 3 pieces
Process Skill: TRANSLATE - Converting the word problem into clear mathematical constraints
2. Establish the contribution constraints
Since the range is 2, if the smallest contribution anyone made was 'min' pieces, then the largest contribution was \(\mathrm{min + 2}\) pieces. This means every single child contributed either 'min', \(\mathrm{min + 1}\), or \(\mathrm{min + 2}\) pieces - no other amounts are possible.
For example, if the minimum contribution was 2 pieces, then all contributions must be 2, 3, or 4 pieces only.
Let's say:
- x = number of children who contributed 'min' pieces
- y = number of children who contributed \(\mathrm{min + 1}\) pieces
- z = number of children who contributed \(\mathrm{min + 2}\) pieces
We know: \(\mathrm{x + y + z = 10}\) (total children)
And: \(\mathrm{x(min) + y(min + 1) + z(min + 2) = 28}\) (total pieces)
Process Skill: APPLY CONSTRAINTS - Setting up the mathematical relationships
3. Test each scenario systematically
Let's test each option to see if having that many children contribute exactly 3 pieces is possible.
Testing Option I: 0 children contributed 3 pieces
If no one contributed 3 pieces, then 3 is not one of our three allowed values (min, min+1, min+2).
This means our allowed contributions could be {1,2,3} but no one chose 3, or {2,3,4} but no one chose 3, etc.
Let's try min = 2, so allowed values are {2,3,4} but everyone chose either 2 or 4:
- If a children contributed 2 pieces and b children contributed 4 pieces:
- \(\mathrm{a + b = 10}\) and \(\mathrm{2a + 4b = 28}\)
- From the second equation: \(\mathrm{2a + 4b = 28}\), so \(\mathrm{a + 2b = 14}\)
- Solving: \(\mathrm{a = 10 - b}\) and \(\mathrm{(10 - b) + 2b = 14}\), so \(\mathrm{b = 4}\) and \(\mathrm{a = 6}\)
- Check: 6 children gave 2 pieces, 4 children gave 4 pieces = \(\mathrm{6(2) + 4(4) = 12 + 16 = 28}\) ✓
So 0 children contributing 3 pieces IS possible.
Testing Option II: 5 children contributed 3 pieces
If 5 children contributed 3 pieces, then 3 must be one of our allowed values.
This means either min = 1 (so values are {1,2,3}), min = 2 (values {2,3,4}), or min = 3 (values {3,4,5}).
Let's try min = 2, so values are {2,3,4}:
- 5 children contributed 3 pieces = 15 pieces total from these children
- Remaining 5 children contributed 28 - 15 = 13 pieces
- These 5 children can only contribute 2 or 4 pieces each
- If c children contributed 2 and d children contributed 4: \(\mathrm{c + d = 5}\) and \(\mathrm{2c + 4d = 13}\)
- From \(\mathrm{2c + 4d = 13}\), we get \(\mathrm{c + 2d = 6.5}\)
- Since c and d must be whole numbers, this is impossible!
Let's try min = 1, so values are {1,2,3}:
- 5 children contributed 3 pieces = 15 pieces
- Remaining 5 children contributed 13 pieces using only values 1 and 2
- If c contributed 1 and d contributed 2: \(\mathrm{c + d = 5}\) and \(\mathrm{c + 2d = 13}\)
- Solving: \(\mathrm{d = 8}\), but we only have 5 remaining children, so impossible!
So 5 children contributing 3 pieces is NOT possible.
Testing Option III: 9 children contributed 3 pieces
If 9 children contributed 3 pieces, then 3 is one of our allowed values, and only 1 child contributed something else.
- 9 children contributed 3 pieces = 27 pieces
- 1 remaining child contributed 28 - 27 = 1 piece
- Our allowed values must include both 1 and 3
- This works if min = 1, giving us allowed values {1,2,3}
- Check: 1 child gave 1 piece, 0 children gave 2 pieces, 9 children gave 3 pieces = \(\mathrm{1(1) + 0(2) + 9(3) = 28}\) ✓
So 9 children contributing 3 pieces IS possible.
Process Skill: CONSIDER ALL CASES - Systematically checking each option
4. Validate feasibility
Let's double-check our valid scenarios:
Option I (0 children with 3 pieces): 6 children gave 2 pieces, 4 children gave 4 pieces
- Total children: \(\mathrm{6 + 4 = 10}\) ✓
- Total pieces: \(\mathrm{6(2) + 4(4) = 28}\) ✓
- Range: \(\mathrm{4 - 2 = 2}\) ✓
Option III (9 children with 3 pieces): 1 child gave 1 piece, 9 children gave 3 pieces
- Total children: \(\mathrm{1 + 9 = 10}\) ✓
- Total pieces: \(\mathrm{1(1) + 9(3) = 28}\) ✓
- Range: \(\mathrm{3 - 1 = 2}\) ✓
Both scenarios work perfectly!
5. Final Answer
Options I and III are possible, but Option II is not possible.
Therefore, the answer is D) I and III only.
Common Faltering Points
Errors while devising the approach
- Misunderstanding the range constraint: Students often think that if the range is 2, then only two different values are possible (minimum and maximum), forgetting that the range allows for three possible consecutive values (min, min+1, min+2). This leads to setting up incorrect equations with only two variables instead of three.
- Overlooking the "could be" nature of the question: Students may try to find a single definitive answer instead of recognizing that they need to test each option separately to see which scenarios are mathematically possible given the constraints.
- Incorrectly assuming 3 must always be included: Students might assume that since the question asks about children contributing 3 pieces, the value 3 must always be one of the allowed contribution amounts, missing that Option I specifically tests when 0 children contribute 3 pieces.
Errors while executing the approach
- Arithmetic errors in solving systems of equations: When testing each option, students often make calculation mistakes when solving the linear equations, such as incorrectly solving \(\mathrm{a + 2b = 14}\) and \(\mathrm{a + b = 10}\) simultaneously, or making errors when checking if solutions yield integer values.
- Failing to check all possible minimum values: For each option, students may only test one possible minimum value (like min = 2) and conclude the option is impossible, without systematically checking other possible minimums (like min = 1 or min = 3) that might work.
- Incorrectly handling fractional solutions: When students get non-integer solutions like \(\mathrm{c + 2d = 6.5}\), they might round or continue with fractional children instead of recognizing this immediately indicates the scenario is impossible.
Errors while selecting the answer
- Misreading the Roman numeral format: Students correctly identify that Options I and III work but Option II doesn't, then mistakenly select answer choice C (I and II only) instead of the correct D (I and III only) due to confusion with the Roman numerals.
- Incomplete verification: Students might find that one or two options work but fail to complete the verification of all three options, leading them to select a partial answer like A (I only) when multiple options are actually possible.
