A certain company conducted phone interviews with each of 6 candidates–Candidates U through Z–for the purpose of determining who should...

OWNING THE DATASET

Let's start by understanding what we're working with in this candidate evaluation table:

Candidate	Criterion A	Criterion B	Criterion C	Criterion D	Criterion E
Z	4.8	4.4	4.6	4.3	4.3
Y	4.6	4.6	4.4	4.4	5.0
U	4.6	4.8	4.5	4.2	4.2
W	4.4	4.0	4.2	4.0	3.9
X	4.2	3.8	4.0	3.8	4.0
V	3.8	3.6	3.8	3.6	3.7

Key Insights:

• We have 6 candidates (Z, Y, U, W, X, V) evaluated on 5 criteria (A-E)
• All scores are on a scale that appears to range from approximately 3.6 to 5.0
• No candidate has the highest score across all criteria
• Looking at the range of scores, we can see some natural groupings forming

Question: For candidates to be invited, they must outperform all non-invited candidates on ALL five criteria. Determine which of the following statements is true.

Rather than checking every candidate combination individually (which would be time-consuming), let's use sorting to quickly identify patterns.

ANALYZING STATEMENT 3

Statement 3 Translation:
Original: "Exactly 3 candidates can be invited."
What we're looking for:

• A group of exactly 3 candidates where every member outperforms every non-member on all criteria
• This means \(\mathrm{MIN(invited\,group)} > \mathrm{MAX(non\text{-}invited\,group)}\) for each criterion

In other words: Can we find 3 candidates who completely dominate the other 3 candidates across all criteria?

Let's approach this strategically by sorting the data for each criterion to see if any clear groupings emerge:

Sorting by Criterion A:
\(\mathrm{Z}\,(4.8) > \mathrm{Y}\,(4.6) = \mathrm{U}\,(4.6) > \mathrm{W}\,(4.4) > \mathrm{X}\,(4.2) > \mathrm{V}\,(3.8)\)

Sorting by Criterion B:
\(\mathrm{U}\,(4.8) > \mathrm{Y}\,(4.6) > \mathrm{Z}\,(4.4) > \mathrm{W}\,(4.0) > \mathrm{X}\,(3.8) > \mathrm{V}\,(3.6)\)

Sorting by Criterion C:
\(\mathrm{Z}\,(4.6) > \mathrm{U}\,(4.5) > \mathrm{Y}\,(4.4) > \mathrm{W}\,(4.2) > \mathrm{X}\,(4.0) > \mathrm{V}\,(3.8)\)

Sorting by Criterion D:
\(\mathrm{Y}\,(4.4) > \mathrm{Z}\,(4.3) > \mathrm{U}\,(4.2) > \mathrm{W}\,(4.0) > \mathrm{X}\,(3.8) > \mathrm{V}\,(3.6)\)

Sorting by Criterion E:
\(\mathrm{Y}\,(5.0) > \mathrm{Z}\,(4.3) > \mathrm{U}\,(4.2) > \mathrm{X}\,(4.0) > \mathrm{W}\,(3.9) > \mathrm{V}\,(3.7)\)

Looking at these sorted lists, we can immediately notice a pattern: Z, Y, and U consistently rank in the top three positions across all criteria, while W, X, and V consistently rank in the bottom three positions.

Now, instead of checking all possible combinations, we can apply the boundary principle:

• For {Z,Y,U} to qualify as the invited group, the lowest score among them must exceed the highest score among {W,X,V} for each criterion.

Let's check the boundary values:

• Criterion A: \(\mathrm{MIN(Z,Y,U)} = 4.6 > \mathrm{MAX(W,X,V)} = 4.4\) ✓
• Criterion B: \(\mathrm{MIN(Z,Y,U)} = 4.4 > \mathrm{MAX(W,X,V)} = 4.0\) ✓
• Criterion C: \(\mathrm{MIN(Z,Y,U)} = 4.4 > \mathrm{MAX(W,X,V)} = 4.2\) ✓
• Criterion D: \(\mathrm{MIN(Z,Y,U)} = 4.2 > \mathrm{MAX(W,X,V)} = 4.0\) ✓
• Criterion E: \(\mathrm{MIN(Z,Y,U)} = 4.2 > \mathrm{MAX(W,X,V)} = 4.0\) ✓

Since the lowest score in {Z,Y,U} exceeds the highest score in {W,X,V} for all five criteria, exactly 3 candidates (Z, Y, and U) can be invited.

Teaching Callout: Notice how sorting revealed a natural grouping that would have taken much longer to discover through individual comparisons. By focusing on the boundary values (lowest of top group vs. highest of bottom group), we efficiently verified all requirements without having to check each candidate pair individually.

Statement 3 is Yes.

ANALYZING STATEMENT 1

Statement 1 Translation:
Original: "Exactly 1 candidate can be invited."
What we're looking for:

• A single candidate who outperforms all other candidates on all criteria
• This means the candidate must have the highest score on every criterion

In other words: Does any candidate rank #1 across all five criteria?

We've already sorted by each criterion above, so we can quickly check if any candidate consistently appears in the top position:

• Criterion A: Z ranks highest (4.8)
• Criterion B: U ranks highest (4.8)
• Criterion C: Z ranks highest (4.6)
• Criterion D: Y ranks highest (4.4)
• Criterion E: Y ranks highest (5.0)

We can immediately see that no single candidate ranks highest across all five criteria. Z tops the list in criteria A and C, U tops criterion B, and Y tops criteria D and E.

For a single candidate to be invited, they would need to outperform all others on all criteria, which is not the case here.

Teaching Callout: The sorting we did earlier saved us time here. Instead of individually comparing each candidate against all others (which would require many comparisons), we simply looked at who ranked first in each criterion and saw that no single candidate dominated across all five.

Statement 1 is No.

ANALYZING STATEMENT 2

Statement 2 Translation:
Original: "Exactly 2 candidates can be invited."
What we're looking for:

• A pair of candidates where both outperform all non-invited candidates on all criteria
• This means \(\mathrm{MIN(invited\,pair)} > \mathrm{MAX(non\text{-}invited\,group)}\) for each criterion

In other words: Can we find 2 candidates who completely dominate the other 4 candidates across all criteria?

Since we've already identified Z, Y, and U as our top performers, let's check if any pair from this group can dominate all others:

For a pair to work, both candidates must exceed all others in all criteria. Let's check the most promising pairs:

1. Z+Y:
   • Criterion B: Z has 4.4, but U has 4.8 → Fail
2. Z+U:
   • Criterion E: U has 4.2, but Y has 5.0 → Fail
3. Y+U:
   • Criterion A: Both have 4.6, but Z has 4.8 → Fail

None of the possible pairs work because in each case, at least one member of the non-invited group outperforms at least one member of the invited pair on at least one criterion.

Teaching Callout: After our initial sorting and analysis of Statement 3, we had strong insights about the data structure. We knew the top three candidates (Z, Y, U) had clear strengths, so we only needed to check if any pair within this group could dominate all others. This targeted approach saved us from checking all 15 possible pairs of candidates.

Statement 2 is No.

FINAL ANSWER COMPILATION

Statement 1: No - No single candidate ranks highest across all criteria.
Statement 2: No - No pair of candidates outperforms all others across all criteria.
Statement 3: Yes - The trio of Z, Y, and U outperforms W, X, and V across all criteria.

Therefore, the answer is Statement 3 only.

LEARNING SUMMARY

Skills We Used

• Pattern Recognition through Sorting: Sorting by each criterion instantly revealed natural groupings that would have taken much longer to discover through calculation.
• Boundary Analysis: Instead of checking every candidate against every other candidate, we focused on the critical boundary values between groups.
• Leveraging Insights Across Statements: Our analysis of Statement 3 provided valuable insights that made evaluating Statements 1 and 2 much faster.

Strategic Insights

• In table analysis questions, sorting is almost always the fastest first step. It transforms a complex dataset into visually scannable patterns.
• The "boundary principle" (MIN of one group > MAX of another group) provides a single check that replaces dozens of individual comparisons.
• When evaluating multiple statements, start with the one that gives you the most information about the dataset structure.

Common Mistakes We Avoided

• We didn't manually compare each candidate against every other candidate (which would require 15 pair comparisons for 6 candidates).
• We didn't check each criterion separately for every possible grouping (which would have been 5 criteria × multiple groupings).
• We recognized that after sorting, many answers become visually obvious without additional calculation.

Remember, the fastest approach to table analysis questions is often to transform the data first (through sorting or other reorganization) before beginning detailed analysis. This reveals patterns that make the solution path much clearer and more efficient.