Question

Which of the following can possibly be the sequence of the number of broken pencils in Boxes 7-16?

• A. 5,7,7,7,9,11,15,20,20,20
• B. 7,7,9,9,11,13,15,19,20,20
• C. 7,7,7,7,11,15,15,19,20,20
• D. 5,6,6,6,11,15,15,20,20,20
• E. 6,7,9, 11,15,19,20,20,20,29
• A. No box numbered 1-6 has more broken pencils than any box numbered 17-20.
• B. Three among the boxes numbered 17 to 20 have 29, 31 and 33 broken pencils in some order.
• C. Four among the boxes numbered 7 to 16 have less than 10 broken pencils.
• D. Exactly three boxes have 20% broken pencils.
• J. A box with the highest percentage of broken pencils has 100 pencils.
• A. Box no. 17 contains more defective pencils than any box from among boxes no. 1-1d.
• B. Boxes no. 17-20 contain a total of 108 defective pencils.
• C. Boxes no. 7-16 contains a total of 124 defective pencils.
• D. Boxes no. 11-16 contain a total of 101 defective pencils.
• O. Boxes no. 7-16 contain a total of 133 defective pencils.

Answer 1

The Correct Option is C

To determine the possible sequence of the number of broken pencils in Boxes 7-16, we need to understand the constraints and available data:

• Boxes 7-16 contain 100 pencils each.
• No box has less than 5% or more than 20% broken pencils.

First, calculate the permissible range for broken pencils in these boxes:

• 5% of 100 pencils = 0.05 × 100 = 5 pencils (minimum).
• 20% of 100 pencils = 0.20 × 100 = 20 pencils (maximum).

Thus, each box numbered 7 through 16 can have between 5 and 20 broken pencils. Now, review the given options to find one that adheres to these constraints:

• Option 1: 5,7,7,7,9,11,15,20,20,20
• Option 2: 7,7,9,9,11,13,15,19,20,20
• Option 3: 7,7,7,7,11,15,15,19,20,20
• Option 4: 5,6,6,6,11,15,15,20,20,20
• Option 5: 6,7,9,11,15,19,20,20,20,29

Analysis shows:

• Options 1 and 4 start with fewer than 7 for multiple boxes, which can be valid.
• Option 5 has a value (29) that exceeds the maximum of 20; hence it is immediately disqualified.

The sequences in Options 1, 2, and 4 need careful examination, but the correct sequence adhering strictly to given percentages and avoiding any overlaps with disqualified limits is: 7,7,7,7,11,15,15,19,20,20 (Option 3). This option maintains all boxes between 5 and 20 broken pencils, making it valid under the given constraints.

Answer 2

To determine which option cannot be conclusively inferred, we need to analyze each one based on the information provided.

Information Summary:

• Boxes 1-6 have 50 pencils each.
• Boxes 7-16 have 100 pencils each.
• Boxes 17-20 have 200 pencils each.
• No box has less than 5% or more than 20% broken pencils.

Box Number Range	Number of Pencils
1-6	50
7-16	100
17-20	200

Options Analysis:

1. No box numbered 1-6 has more broken pencils than any box numbered 17-20: Since boxes numbered 17-20 have 200 pencils, even the minimum allowed broken percentage (5%) results in 10 broken pencils. For boxes 1-6, the maximum allowed (20%) would be 10 broken pencils. Therefore, boxes 1-6 cannot have more broken pencils than boxes 17-20.
2. Three among the boxes numbered 17 to 20 have 29, 31 and 33 broken pencils in some order: It is not necessary to determine conclusively through information given, but the statement doesn't contain contradictions and cannot be outright rejected due to lack of evidence, as it could be a possible range.
3. Four among the boxes numbered 7 to 16 have less than 10 broken pencils: Without specific numbers for boxes 7 to 16, it's feasible considering they have a tolerance for 5% which is possible to achieve with 100 pencils, but not confirmed with data.
4. Exactly three boxes have 20% broken pencils: For 20% broken pencils, boxes 1-6 yield 10, 7-16 yield 20, and 17-20 yield 40. Conclusive determination of three such boxes requires more specific figures.
5. A box with the highest percentage of broken pencils has 100 pencils: The maximum percentage of broken pencils is given (20%). Therefore, boxes 7-16 can be assumed to have as such, but without specific data, this remains speculative.

The option that cannot be conclusively inferred is: Exactly three boxes have 20% broken pencils because, without specific box data showing percentages, concluding exactly three such boxes is speculative.

Answer 3

The problem requires determining which piece of additional information is insufficient to uniquely determine the number of defective pencils in boxes numbered 17-20. Let's analyze each option:

• Box no. 17 contains more defective pencils than any box from among boxes no. 1-16.
  This information specifies that Box 17 has a higher number of defective pencils than all preceding boxes, helping to limit the range of possibilities.
• Boxes no. 17-20 contain a total of 108 defective pencils.
  Knowing the total defective pencils in these boxes establishes a constraint on the distribution of defects, narrowing down possible combinations.
• Boxes no. 7-16 contain a total of 124 defective pencils.
  This provides constraints on boxes 7-16 but does not directly impact the unique identification of defective content in boxes 17-20.
• Boxes no. 11-16 contain a total of 101 defective pencils.
  Specifying the sum for a different subset of boxes might help cross-verify other constraints but does not directly aid in distinguishing boxes 17-20.
• Boxes no. 7-16 contain a total of 133 defective pencils.
  This option conflicts with the previous given sum for boxes 7-16 (124 pencils). As it stands, this piece of information cannot be paired or validated with other data to directly infer unique configurations for boxes 17-20.

Given the above analysis, it's clear that option Boxes no. 7-16 contain a total of 133 defective pencils (the fifth option) is less useful for deriving the unique configuration for boxes 17-20. This is because its validity is questionable compared to related data, and it doesn't contribute additional insights into the distribution pattern in boxes 17-20.