Miguel's online banking password is nine characters long and includes at least one character of each of five types: digits,...
GMAT Two Part Analysis : (TPA) Questions
Miguel's online banking password is nine characters long and includes at least one character of each of five types: digits, punctuation marks, uppercase (capital) letters, lowercase (noncapital) letters, and certain other characters. The password begins or ends with one of these other characters, of which it contains exactly one. No consecutive characters in the password are of the same one of these types. (For example, no capital letter is preceded or followed by another capital letter.) The password contains more lowercase letters than characters of any one of the other four types. It also contains fewer punctuation marks than digits, lowercase letters, or uppercase letters. The first three characters of the password are M, ?, and G.
In the table, select a type of character that the fifth character in the password must be and a type of character that the seventh character must be. Make only two selections, one in each column.
Solution: Password Character Type Analysis
Creating Visual Representation
Password Structure (9 positions):
Position: 1 2 3 4 5 6 7 8 9 Known: M ? G _ _ _ _ _ _ Type: U P U ? ? ? ? ? ?
Types: D (Digit), P (Punctuation), U (Uppercase), L (Lowercase), O (Other)
Drawing Immediate Inferences
Key Constraints:
- 'Other' character placement: Exactly one, must be at position 1 or 9
- Position 1 is M (Uppercase), so Other must be at position 9
- No consecutive same types: Creates alternating pattern requirement
- Frequency constraints:
- Lowercase > any other single type
- Punctuation < Digits, Lowercase, Uppercase
Seeking the Critical Insight
Determining Character Counts:
- We have 2 Uppercase letters already (positions 1 and 3)
- For Lowercase to be most frequent, it must be at least 3
- This limits other types to maximum 2 each
- Since Punctuation < others and we have 1, Punctuation = 1
- Total must be 9: L=3, U=2, D=2, P=1, O=1
Processing the Solution
Mapping Remaining Positions:
After position 3, we need to place:
- 3 Lowercase letters
- 2 Digits
- 0 more Uppercase or Punctuation
- 1 Other (at position 9)
Alternating Pattern Analysis:
Position 3: U (last Uppercase) Position 4: Must be L or D (cannot repeat U) Position 5: Must alternate from position 4 Position 6: Continue alternating Position 7: Continue alternating Position 8: Must be L or D Position 9: O (fixed)
Testing Configuration:
With 3 L's and 2 D's to place in positions 4-8:
- Starting with L at position 4: L-D-L-D-L ✓ (uses exactly 3 L's and 2 D's)
- Starting with D at position 4: D-L-D-L-? (would need 3rd D, but only have 2)
Therefore, positions 4-8 must be: L-D-L-D-L
Final Solution Synthesis
Complete Password Pattern:
Position: 1 2 3 4 5 6 7 8 9 Type: U P U L D L D L O
Answer:
- Fifth character must be: Digit
- Seventh character must be: Digit
Key Insights:
- The 'Other' character constraint immediately fixes position 9
- Character count requirements force a unique distribution
- Alternating constraint with fixed counts determines exact sequence
- Only one valid configuration exists given all constraints
Exam Strategy:
When facing TPA questions with alternating patterns and count constraints, first determine the fixed positions and character distributions. This often reveals a unique solution without testing multiple scenarios.