CS Atlas GATE 2027
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DBMS/Phase 7 · Integration & Final Test

Revision 7.2.3 · Item 117 of 119

Mixed practice set 3

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Third mixed set. Same running data as Set 1.

12 questions · 20 marks · 35 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 · MCQ · 1 mark(s) — (min,max)

WORKS_IN(EMPLOYEE, DEPT) shows EMPLOYEE (1, 1) and DEPT (0, N). Which is true?

  1. (A) Every employee works in exactly one department; a department may have none.
  2. (B) Every department has at least one employee.
  3. (C) The relationship is M:N.
  4. (D) An employee may work in many departments.

Q2 · NAT · 2 mark(s) — ER mapping

EMPLOYEE has a recursive 1:N relationship SUPERVISES and a 1:1 relationship (both total) with PARKING_SPOT. How many tables at minimum?

Numerical answer: ________

Q3 · NAT · 2 mark(s) — Outer join + NULL

SELECT COUNT(*) FROM STUDENT S LEFT JOIN ENROLL E ON S.roll = E.roll WHERE E.grade IS NULL OR E.grade <> 'A';

Numerical answer: ________

Q4 · NAT · 1 mark(s) — INTERSECT

How many rows: SELECT roll FROM ENROLL INTERSECT SELECT roll FROM STUDENT WHERE dept = 'CS';

Numerical answer: ________

Q5 · MCQ · 2 mark(s) — Aggregates

SELECT dept, AVG(cgpa) FROM STUDENT GROUP BY dept HAVING COUNT(*) > COUNT(cgpa); returns:

  1. (A) no rows
  2. (B) (CS, 8.75)
  3. (C) (CS, 5.83)
  4. (D) (CS, 8.75), (EE, 7.8)

Q6 · MCQ · 2 mark(s) — Normal forms

R(A, B, C, D, E), F = {AB → C, C → D, D → A}. Highest normal form?

  1. (A) 1NF
  2. (B) 2NF
  3. (C) BCNF
  4. (D) 3NF

Q7 · NAT · 2 mark(s) — 3NF synthesis

How many relations does 3NF synthesis produce for R(A, B, C, D, E, F) with F = {A → B, C → D, E → F, AC → E}?

Numerical answer: ________

Q8 · NAT · 1 mark(s) — Spanned blocks

r = 80,000 records, R = 300 bytes, B = 4096, spanned. Number of blocks?

Numerical answer: ________

Q9 · NAT · 2 mark(s) — B+ capacity

A B+ tree has internal order p = 50 and leaf order p_leaf = 40. Maximum records with 3 levels?

Numerical answer: ________

Q10 · MCQ · 2 mark(s) — Serializability

S = R3(X) R2(X) W3(X) R1(X) W1(X). Equivalent serial order?

  1. (A) T2, T3, T1
  2. (B) T3, T2, T1
  3. (C) T1, T2, T3
  4. (D) not CS

Q11 · MCQ · 1 mark(s) — Thomas rule

TS(T) = 8, RTS(X) = 5, WTS(X) = 12. T issues write(X). With the Thomas write rule:

  1. (A) T waits
  2. (B) the write is ignored
  3. (C) the write executes
  4. (D) T aborts

Q12 · MSQ · 2 mark(s) — Recovery

Immediate update. Log: ⟨T1 start⟩ ⟨T1, A, 1, 2⟩ ⟨T1 commit⟩ ⟨checkpoint⟩ ⟨T2 start⟩ ⟨T2, B, 3, 4⟩ ⟨T3 start⟩ ⟨T3, A, 2, 7⟩ ⟨T3 commit⟩ — crash. Which actions are needed?

  1. (A) Redo T3
  2. (B) Undo T2
  3. (C) Undo T3
  4. (D) Redo T1

✓Answer Vault

Reveal answers when you’re ready

Q1 Answer: (A) · 1 mark(s)

Each pair counts its own entity's links.

  • (B) is wrong: DEPT's min is 0.
  • (C) is wrong: It is N:1.
  • (D) is wrong: EMPLOYEE's max is 1.

Re-read: Lesson 1.2.2 — Cardinality ratios; total vs partial participation; (min,max) notation

Q2 Answer: 1 · 2 mark(s)

Supervisor id as an FK inside EMPLOYEE; PARKING_SPOT merges with EMPLOYEE (1:1 both total).

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

Q3 Answer: 5 · 2 mark(s)

Non-A rows: (1, C2, B), (2, C3, B), (3, C1, B), (4, C4, C), plus Esha's padded row.

Re-read: Lesson 3.3.6 — Joins in SQL: INNER, NATURAL, LEFT/RIGHT/FULL OUTER

Q4 Answer: 2 · 1 mark(s)

Rolls 1 and 3 (Esha has no enrolment).

Re-read: Lesson 3.3.7 — UNION / INTERSECT / EXCEPT and their ALL variants

Q5 Answer: (B) · 2 mark(s)

Only CS has a NULL cgpa; AVG ignores it: (9.1 + 8.4)/2.

  • (A) is wrong: CS qualifies.
  • (C) is wrong: That divides by 3.
  • (D) is wrong: EE has no NULL.

Re-read: Lesson 3.3.5 — Aggregates, GROUP BY, HAVING; COUNT(*) vs COUNT(col); NULL handling

Q6 Answer: (D) · 2 mark(s)

Keys ABE, BCE, BDE: every attribute is prime. C → D has a non-superkey left side.

  • (A) is wrong: Higher.
  • (B) is wrong: Higher.
  • (C) is wrong: C is not a superkey.

Re-read: Lesson 4.4.3 — 3NF (transitive dependency + formal condition)

Q7 Answer: 4 · 2 mark(s)

Minimal cover = F; tables AB, CD, EF, ACE. ACE contains the key AC: no extra table.

Re-read: Lesson 4.4.6 — Algorithms: 3NF synthesis, BCNF decomposition

Q8 Answer: 5860 · 1 mark(s)

⌈24,000,000 / 4096⌉ = ⌈5859.4⌉.

Re-read: Lesson 5.1.1 — Storage basics: disk, blocks, block access cost, records, blocking factor, spanned vs unspanned

Q9 Answer: 100000 · 2 mark(s)

50² × 40.

Re-read: Lesson 5.6.2 — B+ tree order from block size, key size and pointer size

Q10 Answer: (A) · 2 mark(s)

R2(X) before W3(X) and W1(X): T2 → T3, T2 → T1. W3(X) before R1(X): T3 → T1.

  • (B) is wrong: T2 must precede T3.
  • (C) is wrong: Reversed.
  • (D) is wrong: Acyclic.

Re-read: Lesson 6.4.2 — Conflict serializability: conflicting ops, precedence graph, equivalent serial orders

Q11 Answer: (B) · 1 mark(s)

TS < WTS but ≥ RTS: obsolete write.

  • (A) is wrong: TO never waits.
  • (C) is wrong: Breaks timestamp order.
  • (D) is wrong: Basic TO.

Re-read: Lesson 6.7.1 — Timestamp ordering protocol; Thomas write rule

Q12 Answer: (A) (B) · 2 mark(s)

T1 committed before the checkpoint: nothing. T2 uncommitted: undo. T3 committed after the checkpoint: redo.

  • (C) is wrong: T3 committed.
  • (D) is wrong: Before the checkpoint.

Re-read: Lesson 6.8.1 — Multiple granularity, validation-based protocol, log-based recovery

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