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DBMS/Phase 2 · Relational Model & Constraints

Checkpoint CP-2 · Item 35 of 119

Phase 2 Checkpoint (15 questions)

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15 questions · 24 marks · 40 minutes. Negative marking on MCQ only: −1/3 mark for a wrong 1-mark MCQ, −2/3 for a wrong 2-mark MCQ. No negative marking for MSQ or NAT; MSQ has no partial credit.

Q1 · NAT · 1 mark(s) — Degree

A relation has attributes (a, b, c, d) and 50 tuples. What is its degree?

Numerical answer: ________

Q2 · MCQ · 1 mark(s) — Properties

Which is true in the relational model?

  1. (A) Every value is atomic or NULL
  2. (B) A relation may contain duplicate tuples
  3. (C) Attributes must have numeric domains
  4. (D) Tuples are ordered by primary key

Q3 · NAT · 2 mark(s) — NULL

Column x: 3, NULL, 3, 8, NULL, 1. How many rows satisfy NOT (x = 3)? (NOT of UNKNOWN is UNKNOWN.)

Numerical answer: ________

Q4 · NAT · 2 mark(s) — Superkeys

R(A, B, C, D, E, F) has candidate keys {A, B} and {A, C}. How many superkeys?

Numerical answer: ________

Q5 · MCQ · 2 mark(s) — Keys

Instance of R(A, B, C): (1, a, x), (2, a, y), (3, b, x), (4, b, y). Which can be a candidate key based on this instance?

  1. (A) {B}
  2. (B) {C}
  3. (C) {B, C}
  4. (D) {A, B}

Q6 · MSQ · 2 mark(s) — Foreign keys

Which are allowed for a foreign key FK in R₁ referencing R₂'s primary key?

  1. (A) An FK value missing from R₂
  2. (B) The same FK value in many tuples of R₁
  3. (C) FK referencing R₁ itself
  4. (D) FK value NULL (FK not part of R₁'s PK)

Q7 · MCQ · 1 mark(s) — Constraints

Updating the cgpa of a student (non-key, non-FK) can violate:

  1. (A) domain constraint
  2. (B) key constraint
  3. (C) entity integrity
  4. (D) referential integrity

Q8 · MSQ · 2 mark(s) — Constraints

Which operations can never violate referential integrity?

  1. (A) Inserting into the referencing relation
  2. (B) Inserting into the referenced relation
  3. (C) Deleting from the referenced relation
  4. (D) Deleting from the referencing relation

Q9 · NAT · 2 mark(s) — Cascade

R(A, B) with B → R.A ON DELETE CASCADE: (1, NULL), (2, 1), (3, 1), (4, 2), (5, 4), (6, 3), (7, NULL), (8, 7). Delete A = 2. How many tuples remain?

Numerical answer: ________

Q10 · MCQ · 2 mark(s) — Set null

Same R but ON DELETE SET NULL. Delete A = 1. Which tuples change?

  1. (A) (2, 1) and (3, 1) become (2, NULL) and (3, NULL)
  2. (B) (2, 1), (3, 1), (4, 2), (6, 3) all change
  3. (C) all tuples are deleted
  4. (D) nothing; the delete is rejected

Q11 · MCQ · 1 mark(s) — Mapping

A composite attribute is mapped to:

  1. (A) a foreign key
  2. (B) a separate table
  3. (C) its simple components as columns
  4. (D) a multivalued column

Q12 · NAT · 1 mark(s) — Minimum tables

E1 —M:N— E2, where E1 has 1 multivalued attribute. Minimum tables?

Numerical answer: ________

Q13 · NAT · 2 mark(s) — Minimum tables

E1 —1:N— E2 —1:1 (both total)— E3, E2 is a weak entity owned by E1 through the 1:N relationship. Minimum tables?

Numerical answer: ________

Q14 · MCQ · 2 mark(s) — Ternary

A ternary relationship R(E1, E2, E3) with all sides many and no attributes needs:

  1. (A) one table whose PK is the three keys together
  2. (B) no table; FKs are added to the entity tables
  3. (C) one table whose PK is E1's key
  4. (D) three tables, one per pair

Q15 · MCQ · 1 mark(s) — Specialisation

A total, disjoint specialisation of ACCOUNT into SAVINGS and CURRENT can be mapped without NULLs and without a superclass table using:

  1. (A) one table with flags
  2. (B) superclass table only
  3. (C) one table with a type attribute
  4. (D) subclass tables only

✓Answer Vault

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Q1 Answer: 4 · 1 mark(s)

Four attributes.

Re-read: Lesson 2.1.1 — Relation, tuple, attribute, domain, degree, cardinality, schema; properties of relations

Q2 Answer: (A) · 1 mark(s)

Atomic values (1NF), NULL allowed outside the PK.

  • (B) is wrong: No duplicates.
  • (C) is wrong: Any domain.
  • (D) is wrong: No order.

Re-read: Lesson 2.1.1 — Relation, tuple, attribute, domain, degree, cardinality, schema; properties of relations

Q3 Answer: 2 · 2 mark(s)

x = 3 is FALSE for 8 and 1 → NOT gives TRUE for 2 rows; NULL rows stay UNKNOWN.

Re-read: Lesson 2.1.2 — NULL: meaning and basic behaviour

Q4 Answer: 24 · 2 mark(s)

Contain AB: 2⁴ = 16; contain AC: 16; contain ABC: 2³ = 8. 16 + 16 − 8 = 24.

Re-read: Lesson 2.2.1 — Superkey, candidate key, primary key, alternate key; counting superkeys

Q5 Answer: (C) · 2 mark(s)

B and C each repeat; (B, C) pairs are all different. {A, B} is unique but not minimal if {A} is unique ({A} is unique here).

  • (A) is wrong: Repeats.
  • (B) is wrong: Repeats.
  • (D) is wrong: Contains {A}, which is already unique here, so {A, B} is not minimal.

Re-read: Lesson 2.2.1 — Superkey, candidate key, primary key, alternate key; counting superkeys

Q6 Answer: (B) (C) (D) · 2 mark(s)

Only dangling references are forbidden.

  • (A) is wrong: Violates referential integrity.

Re-read: Lesson 2.2.2 — Foreign key; referencing vs referenced relations

Q7 Answer: (A) · 1 mark(s)

Only domain.

  • (B) is wrong: Not a key.
  • (C) is wrong: Not the PK.
  • (D) is wrong: Not an FK.

Re-read: Lesson 2.3.1 — Domain, key, entity and referential integrity; which operations violate what

Q8 Answer: (B) (D) · 2 mark(s)

New parents and removed children are always safe.

  • (A) is wrong: The new FK may match nothing.
  • (C) is wrong: Children may be orphaned.

Re-read: Lesson 2.3.1 — Domain, key, entity and referential integrity; which operations violate what

Q9 Answer: 5 · 2 mark(s)

2 → 4 → 5 deleted. 8 − 3 = 5.

Re-read: Lesson 2.3.2 — Referential actions: restrict, cascade, set null; tracing cascading deletes

Q10 Answer: (A) · 2 mark(s)

Only direct children are changed.

  • (B) is wrong: 4 and 6 still refer to existing tuples.
  • (C) is wrong: That is cascade.
  • (D) is wrong: That is restrict.

Re-read: Lesson 2.3.2 — Referential actions: restrict, cascade, set null; tracing cascading deletes

Q11 Answer: (C) · 1 mark(s)

Flatten.

  • (A) is wrong: No.
  • (B) is wrong: Only multivalued attributes need tables.
  • (D) is wrong: Not allowed.

Re-read: Lesson 2.4.1 — Mapping strong and weak entities, composite and multivalued attributes

Q12 Answer: 4 · 1 mark(s)

E1, E2, R, multivalued.

Re-read: Lesson 2.4.2 — Mapping 1:1, 1:N, M:N with participation; minimum number of tables

Q13 Answer: 2 · 2 mark(s)

E1 (1). E2 and E3 merge (1:1 both total) into one table that also holds E1's key: 2.

Re-read: Lesson 2.4.2 — Mapping 1:1, 1:N, M:N with participation; minimum number of tables

Q14 Answer: (A) · 2 mark(s)

All keys form the PK.

  • (B) is wrong: Cannot be stored as FKs.
  • (C) is wrong: Not unique.
  • (D) is wrong: Pairwise tables lose the three-way meaning.

Re-read: Lesson 2.4.3 — Ternary relationships; specialization/generalization mapping

Q15 Answer: (D) · 1 mark(s)

Total: nobody is lost; disjoint: nobody repeated.

  • (A) is wrong: NULLs.
  • (B) is wrong: Loses subclass attributes.
  • (C) is wrong: NULLs in subclass columns.

Re-read: Lesson 2.4.3 — Ternary relationships; specialization/generalization mapping

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