CS Atlas GATE 2027
DBMS course
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DBMS/Phase 5 · Storage & Indexing

Checkpoint CP-5 · Item 95 of 119

Phase 5 Checkpoint (15 questions)

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15 questions · 25 marks · 45 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) — bfr

B = 2048, R = 100, unspanned. bfr?

Numerical answer: ________

Q2 · NAT · 2 mark(s) — Blocks

r = 50,000, R = 100, B = 2048, spanned. Number of blocks?

Numerical answer: ________

Q3 · NAT · 1 mark(s) — Latency

Disk at 10,000 rpm. Average rotational latency in ms?

Numerical answer: ________

Q4 · NAT · 1 mark(s) — Binary search

A sorted file has 2,500 blocks. Block accesses for binary search on the ordering key?

Numerical answer: ________

Q5 · MCQ · 1 mark(s) — File organisation

A heap file is searched for a non-key value. The number of block accesses is:

  1. (A) b/2
  2. (B) b
  3. (C) 1
  4. (D) ⌈log₂ b⌉

Q6 · NAT · 2 mark(s) — Primary index

A file with 100,000 records, 20 records per block, sorted on its key. Primary index with bfrᵢ = 100, single level. Block accesses to find a record?

Numerical answer: ________

Q7 · NAT · 2 mark(s) — Secondary index

Same file, secondary index on another key field, bfrᵢ = 100, single level. Block accesses?

Numerical answer: ________

Q8 · NAT · 2 mark(s) — Multilevel

The secondary index above made multilevel (fo = 100). Block accesses?

Numerical answer: ________

Q9 · MCQ · 1 mark(s) — Index types

How many clustering indexes can a file have?

  1. (A) exactly 2
  2. (B) 0 or 1
  3. (C) any number
  4. (D) one per non-key field

Q10 · NAT · 2 mark(s) — B-tree order

B = 512, K = 8, P = 4, Pr = 6. B-tree order p (max children)?

Numerical answer: ________

Q11 · NAT · 2 mark(s) — B-tree splits

Max 2 keys per node. Insert 1, 2, 3, 4, 5, 6, 7 into an empty B-tree. How many splits?

Numerical answer: ________

Q12 · NAT · 2 mark(s) — B-tree bounds

B-tree of order 5, height 4 levels. Minimum number of keys?

Numerical answer: ________

Q13 · NAT · 2 mark(s) — B+ order

B = 1024, K = 10, P = 6. B+ internal order p?

Numerical answer: ________

Q14 · NAT · 2 mark(s) — B+ capacity

p = 64, p_leaf = 60, height 3. Maximum records?

Numerical answer: ________

Q15 · MCQ · 2 mark(s) — B+ insertion

In a B+ tree, an internal node overflows during an insertion. Its middle key:

  1. (A) is deleted
  2. (B) moves up to the parent and is removed from both halves
  3. (C) is copied up and kept in the node
  4. (D) moves down to a leaf

✓Answer Vault

Reveal answers when you’re ready

Q1 Answer: 20 · 1 mark(s)

⌊2048/100⌋.

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

Q2 Answer: 2442 · 2 mark(s)

⌈5,000,000 / 2048⌉ = ⌈2441.4⌉.

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

Q3 Answer: 3 · 1 mark(s)

6 ms per rotation, half = 3.

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

Q4 Answer: 12 · 1 mark(s)

2¹² = 4096 ≥ 2500 > 2048.

Re-read: Lesson 5.2.1 — File organization: heap, sorted, hashed; cost of search/insert/delete

Q5 Answer: (B) · 1 mark(s)

All blocks must be read.

  • (A) is wrong: Average for an existing key.
  • (C) is wrong: Hashing.
  • (D) is wrong: Sorted.

Re-read: Lesson 5.2.1 — File organization: heap, sorted, hashed; cost of search/insert/delete

Q6 Answer: 7 · 2 mark(s)

5,000 blocks → 50 index blocks; ⌈log₂ 50⌉ + 1 = 6 + 1.

Re-read: Lesson 5.3.2 — Primary, clustering, secondary indexes; counting index blocks and block accesses

Q7 Answer: 11 · 2 mark(s)

100,000 entries → 1,000 blocks; ⌈log₂ 1000⌉ + 1 = 10 + 1.

Re-read: Lesson 5.3.2 — Primary, clustering, secondary indexes; counting index blocks and block accesses

Q8 Answer: 4 · 2 mark(s)

1000 → 10 → 1: 3 levels + 1.

Re-read: Lesson 5.3.3 — Multilevel indexes; fan-out; logarithm primer

Q9 Answer: (B) · 1 mark(s)

At most one ordering field.

  • (A) is wrong: No.
  • (C) is wrong: Secondary indexes only.
  • (D) is wrong: No.

Re-read: Lesson 5.3.2 — Primary, clustering, secondary indexes; counting index blocks and block accesses

Q10 Answer: 29 · 2 mark(s)

4p + 14(p − 1) ≤ 512 → 18p ≤ 526 → 29.

Re-read: Lesson 5.5.1 — B-tree structure and order (order conventions and the GATE trap)

Q11 Answer: 4 · 2 mark(s)

Same shape as the 10 … 70 trace.

Re-read: Lesson 5.5.2 — B-tree search and insertion with node splits

Q12 Answer: 53 · 2 mark(s)

2·3³ − 1.

Re-read: Lesson 5.5.3 — B-tree deletion basics; min/max keys; height bounds

Q13 Answer: 64 · 2 mark(s)

6p + 10(p − 1) ≤ 1024 → 16p ≤ 1034 → 64.

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

Q14 Answer: 245760 · 2 mark(s)

64² × 60.

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

Q15 Answer: (B) · 2 mark(s)

Internal splits move the separator.

  • (A) is wrong: No.
  • (C) is wrong: Copying happens for leaves.
  • (D) is wrong: No.

Re-read: Lesson 5.6.3 — B+ tree insertion and deletion traces; height and capacity; B vs B+

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