Eureka Airlines operates 5 daylong flight sequences serving City A, City B, City C, City D, and City E. The...
GMAT Two Part Analysis : (TPA) Questions
Eureka Airlines operates 5 daylong flight sequences serving City A, City B, City C, City D, and City E. The following list shows the order of the cities served by each flight sequence.
- A-B-C-A (3 flights)
- A-D-B-A (3 flights)
- A-E-D-A (3 flights)
- A-E-B-D-A (4 flights)
- A-C-B-D-A (4 flights)
Due to the number of available flight crews, Eureka can operate only 3 flight sequences in a given day. Exactly 10 flights are flown each day. No sequence is used for more than 2 consecutive days, and no sequence is idle for 2 consecutive days.
For City A, select the number of Eureka flights that will arrive in City A over the span of any 3-day period. For City D, select the number of Eureka flights that will arrive in City D over the span of any 3-day period. Make only two selections, one in each column.
Phase 1: Owning the Dataset
Understanding Flight Sequences
Let's map out each sequence and count the flights:
| Sequence | Route | Total Flights | Arrivals at A | Arrivals at D |
|---|---|---|---|---|
| 1 | A→B→C→A | 3 | 1 (from C) | 0 |
| 2 | A→D→B→A | 3 | 1 (from B) | 1 (from A) |
| 3 | A→E→D→A | 3 | 1 (from D) | 1 (from E) |
| 4 | A→E→B→D→A | 4 | 1 (from D) | 1 (from B) |
| 5 | A→C→B→D→A | 4 | 1 (from D) | 1 (from B) |
Key insight: Each sequence has exactly 1 arrival at City A (the final flight returns to A).
Daily Flight Requirements
We need exactly 10 flights per day with 3 sequences. Let's check combinations:
- \(3 + 3 + 3 = 9\) ❌
- \(3 + 3 + 4 = 10\) ✓
- \(3 + 4 + 4 = 11\) ❌
- \(4 + 4 + 4 = 12\) ❌
So each day MUST have: Two 3-flight sequences + One 4-flight sequence
This means daily we need:
- 2 sequences from {1, 2, 3}
- 1 sequence from {4, 5}
Phase 2: Understanding the Constraints
Usage Constraints Over 3 Days
The constraints state:
- No sequence used for more than 2 consecutive days
- No sequence idle for 2 consecutive days
This means over any 3-day period:
- Each sequence must be used at least once (otherwise it's idle for 2+ days)
- Each sequence can be used at most twice (otherwise it's used for 3 consecutive days)
With 5 sequences and 3 sequences/day × 3 days = 9 total uses:
- We need exactly 9 sequence uses
- With 5 sequences each used 1-2 times
- This means: 4 sequences used twice, 1 sequence used once
Forced Distribution
Given our daily requirements:
- Sequences {1, 2, 3} collectively used 6 times over 3 days (2 per day)
- Sequences {4, 5} collectively used 3 times over 3 days (1 per day)
With the at-most-twice constraint:
- All three of {1, 2, 3} must be used exactly twice (\(2×3 = 6\))
- One of {4, 5} used twice, other used once (\(2+1 = 3\))
Phase 3: Finding the Answer
Arrivals at City A
Each sequence brings exactly 1 arrival to A.
Over 3 days: 9 sequence uses = 9 arrivals at City A
Arrivals at City D
Let's calculate based on our forced distribution:
- Sequence 1: 0 arrivals × 2 uses = 0
- Sequence 2: 1 arrival × 2 uses = 2
- Sequence 3: 1 arrival × 2 uses = 2
- Sequences 4&5: 1 arrival each × 3 total uses = 3
Total arrivals at D: \(0 + 2 + 2 + 3 = 7\) arrivals at City D
Phase 4: Solution
Based on our analysis, over any 3-day period:
- City A: 9 arrivals
- City D: 7 arrivals
These numbers are fixed due to the constraints forcing a specific distribution of sequence usage.