Storage & RAID Management MCQs: 12 Solved Questions with Explanations

RAID questions compress striping, parity, redundancy and usable capacity into similar labels. This set also tests tape transfer, buffer caches, defragmentation and formatting. Answer each of the 12 questions before reading its explanation. These questions focus on storage hardware and management; File Systems & Allocation MCQs covers allocation methods and file-system structures.

Build the RAID level map before solving

Ask how data is split and where redundancy is kept. Use this map before comparing options.

Now take six identical 2 TB drives. RAID 5 gives (6 - 1) x 2 TB = 10 TB of usable capacity and tolerates one drive failure. RAID 6 gives (6 - 2) x 2 TB = 8 TB and tolerates two drive failures. RAID 10 makes three mirrored pairs, so usable capacity is 3 x 2 TB = 6 TB. RAID 10 can survive one failed drive in each mirrored pair, but two failures in the same pair lose the array. These are ideal raw-capacity calculations before file-system and vendor overhead. For a sequenced concept route, use GATE Guidance by Sanchit Sir.

Storage & RAID Management MCQs 1-2: speed and fault tolerance

Question 1 (ISRO 2007 and BEL 2007)

Which of the following RAID level provides the highest Data Transfer Rate (Read/Write)

• A. RAID 1

• B. RAID 3

• C. RAID 4

• D. RAID 5

Choose first.

Correct answer: B. RAID 3.

In this classic comparison, RAID 3 stripes data finely across synchronised disks. All data disks serve a large sequential read or write while a dedicated disk holds parity. That makes it the expected answer here, not a universal result.

Question 2 (ISRO 2008)

Raid configurations of the disks are used to provide

• A. Fault-tolerance

• B. High speed

• C. High data density

• D. A & B

Choose first.

Correct answer: D. A & B.

Striping serves data in parallel for speed; mirroring or parity adds fault tolerance. RAID levels combine them, so both A and B apply. Redundancy consumes raw capacity, so high data density is not the defining benefit.

Storage & RAID Management MCQs 3-5: RAID 5, RAID 6 and parity

Work one RAID 5 stripe first. Let D1 = 10110010, D2 = 01101100 and D3 = 11001001. Then D1 XOR D2 = 11011110, so P = D1 XOR D2 XOR D3 = 00010111. If D2 is lost, calculate D1 XOR D3 = 01111011, then 01111011 XOR P = 01101100. The missing byte is reconstructed exactly. RAID 5 rotates this single parity block among disks. RAID 6 adds a second independent parity value rather than storing the same XOR twice.

Question 3 (ISRO 2011)

Which RAID level gives block level striping with double distributed parity?

• A. RAID 10

• B. RAID 2

• C. RAID 6

• D. RAID 5

Choose first.

Correct answer: C. RAID 6.

RAID 6 distributes two independent parity calculations across block-striped disks, tolerating any two drive failures. RAID 5 has one distributed parity value and tolerates one failure. RAID 10 stripes mirrored pairs rather than using distributed parity.

Question 4 (ISRO May 2017)

Which one of these is characteristic of RAID 5?

• A. Dedicated parity

• B. Double parity

• C. Hamming code parity

• D. Distributed parity

Choose first.

Correct answer: D. Distributed parity.

RAID 5 stores one parity block per stripe and rotates its disk position. Dedicated block parity indicates RAID 4, double distributed parity RAID 6, and Hamming-code error correction RAID 2. The XOR stripe above demonstrates distributed single parity.

Question 5 (MPPSC 2025)

Which RAID level uses block-level striping with distributed parity across multiple disks but requires at least three disks?

• A. RAID 0

• B. RAID 1

• C. RAID 5

• D. RAID 10

Choose first.

Correct answer: C. RAID 5.

Three disks allow two data blocks and one rotating parity block per stripe. Usable capacity is (N - 1) x drive size, so three 2 TB drives give (3 - 1) x 2 TB = 4 TB before overhead. RAID 0 has no parity, RAID 1 mirrors, and RAID 10 needs mirrored pairs.

Storage & RAID Management MCQs 6-7: match each level to its organisation

Question 6 (UGC NET January 2025)

Match the List I to List II

List I List II

A. RAID Level 1 I. bit - interleaved parity organization

B. RAID Level 2 II. disk mirroring

C. RAID Level 3 III. block-interleaved parity organization

D. RAID Level 4 IV. ECC organization

Choose the correct answer from the options given below:

• A. (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

• B. (A)-(II), (B)-(IV), (C)-(I), (D)-(III)

• C. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

• D. (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Choose first.

Correct answer: B. (A)-(II), (B)-(IV), (C)-(I), (D)-(III).

RAID 1 mirrors disks; RAID 2 uses memory-style error correction; RAID 3 uses bit-interleaved parity on a dedicated disk; RAID 4 uses block-interleaved parity on a dedicated disk. This yields A-II, B-IV, C-I, D-III, only option B.

Question 7 (UGC NET June 2025)

Match List I with List II

List I List II

A. RAID Level 2 I. Block interleaved distribution parity

B. RAID Level 3 II. Also known as P+Q redundancy Scheme

C. RAID Level 5 III. Bit interleaved parity

D. RAID Level 6 IV. Also known as Memory style error correcting code organization

Choose the correct answer from the options given below:

• A. A → IV, B → III, C → I, D → II

• B. A → III, B → IV, C → I, D → II

• C. A → I, B → III, C → IV, D → II

• D. A → II, B → I, C → III, D → IV

Choose first.

Correct answer: A. A → IV, B → III, C → I, D → II.

RAID 2 uses memory-style error correction, RAID 3 bit-interleaved parity, RAID 5 block-interleaved distributed parity, and RAID 6 P+Q. This gives A-IV, B-III, C-I, D-II. Questions 6 and 7 permute the same map, so derive it instead of memorising an option.

Storage & RAID Management MCQs 8-9: mirroring, defragmentation and compression

Question 8 (KVS 2013)

The simplest, but most expensive, approach to introductory redundancy is duplicate to every disk. This technique is called

• A. Swap space

• B. Mirroring

• C. Page slots

• D. None of these

Choose first.

Correct answer: B. Mirroring.

Mirroring duplicates data on separate disks. With two 2 TB drives in RAID 1, only 2 TB is usable because one drive holds the copy. Either drive can fail without data loss. Swap space and page slots concern memory management, not disk redundancy.

Question 9 (DSSSB TGT 2023)

The process of organising the contents of each file into contiguous regions and creating larger areas of free space on the hard disk is called ______ and ______ is the process of making computer files take up less space on your computer by reducing the size of the file.

• A. Debugging; Assembly

• B. Defragmentation; Compression

• C. Compression; Formatting

• D. Defragmentation; Mirroring

Choose first.

Correct answer: B. Defragmentation; Compression.

Defragmentation rearranges extents into contiguous regions and consolidates free areas. Reorganising a 600 MB file split into 120 MB, 180 MB and 300 MB extents is defragmentation. Compression changes representation to use less space, such as encoding it into 420 MB. Neither operation is mirroring.

Storage & RAID Management MCQs 10-11: tape transfer and buffer cache

Question 10

Consider a tape of length 4m containing 80 parallel tracks and moving with a linear velocity of 32cm/sec and it employs the recording density of 512 KB/cm-track. If the tape is divided into records of size 8 cm and length of the gap between the records is 2cm, then calculate effective data transfer time.

• A. 1.48 GBPS

• B. 1.44 GBPS

• C. 1.34 GBPS

• D. 1.048 GBPS

Choose first.

Correct answer: D. 1.048 GBPS.

The stem says time, but the options give a rate. Convert 4 m = 400 cm. Each record and gap occupies 8 + 2 = 10 cm, giving 400/10 = 40 records and 40 x 8 = 320 cm recorded. Data is 320 x 80 x 512 = 13,107,200 KB. Travel time is 400/32 = 12.5 s. Thus 13,107,200/12.5 = 1,048,576 KB/s, matching the 1.048 GBPS convention.

Continue with Operating System MCQs.

Question 11 (Bihar STET 2025)

In an operating system, a "buffer cache" is used to:

• A. Store files in memory

• B. Store copies of frequently used disk blocks in memory

• C. Store the operating system kernel

• D. Store input/output devices

Choose first.

Correct answer: B. Store copies of frequently used disk blocks in memory.

A buffer cache keeps disk-block copies in RAM to avoid repeated physical reads. If a 4 KB block is read once and requested three more times while cached, those requests can use memory unless the entry is evicted or invalidated. It stores blocks, not devices or the kernel.

Storage & RAID Management MCQ 12: disk formatting

Question 12 (GATE 1998)

Formatting of a floppy disk refers to

• A. arranging the data on the disk in contiguous fashion

• B. writing the directory

• C. erasing the system area

• D. writing identification information on all tracks and sectors.

Choose first.

Correct answer: D. writing identification information on all tracks and sectors.

Here, physical formatting identifies tracks and sectors as addressable locations. Arranging existing file extents contiguously is defragmentation; writing a directory is a higher-level file-system step. Keep the historical floppy-disk context.

The short version and your next practice step

Misses on Questions 1 to 8 call for rebuilding the RAID map; a Question 9 miss means separating layout from representation. For Question 10, repeat every conversion and gap-utilisation step. Questions 11 and 12 test storage-path basics. Redraw the six-drive comparison and four-byte XOR stripe, then retry all 12 in 12 minutes, reserving 3 minutes for Question 10. Use GATE Test Series for timed practice. Revise with Disk Scheduling MCQs, or browse GATE CS Exam Preparation.