Your CDB has two regular PDBs as well as one application container with two application PDBs and an application seed. No changes have been made to the standard PDB$SEED. How many default temporary…

Question

Options

• Athree
• Beight
• Cseven
• Dsix
• Efive

• A
  10% (2)
• B
  5% (1)
• C
  81% (17)
• D
  5% (1)

Explanation

Option C is correct because the CDB has exactly seven containers that can each hold their own default temporary tablespace. Counting them: CDB$ROOT (1) + two regular PDBs (2) + the application container itself (1) + the application seed (1) + two application PDBs (2) = 7. The critical detail is that PDB$SEED is excluded - since no changes have been made to it, it remains read-only and inherits/shares CDB$ROOT's default temporary tablespace rather than having one independently assigned.

Why the distractors are wrong:

• B (eight) is the most tempting trap - it counts PDB$SEED as its own container (which it is), but an unmodified PDB$SEED cannot have a separate default temporary tablespace assigned to it.
• D (six) likely results from forgetting to count the application container itself as a separate container with its own temp tablespace.
• E (five) probably omits the application container and/or application seed from the count.
• A (three) drastically undercounts, perhaps only counting the three "top-level" non-application PDBs.

Memory tip: Use the phrase "Every container gets one, except a frozen seed" - count all containers in your CDB topology, then subtract any seed PDB that is still in its default, unmodified (read-only) state. PDB$SEED unchanged = not assignable = subtract 1 from your total of 8.

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