Indexing MCQ Quiz in বাংলা - Objective Question with Answer for Indexing - বিনামূল্যে ডাউনলোড করুন [PDF]
Last updated on Mar 19, 2025
Latest Indexing MCQ Objective Questions
Top Indexing MCQ Objective Questions
Indexing Question 1:
Consider a B+ tree in which the minimum number of keys in a node is 15. What is the maximum number of keys in any non-root node?
Answer (Detailed Solution Below)
Indexing Question 1 Detailed Solution
Data:
Minimum number of keys = MIN = 15
order of a B+ tree = p
Formula:
MIN = ⌈p ÷ 2⌉ – 1
Maximum number of keys = MAX = p – 1
Calculation:
15 = ⌈p ÷ 2⌉ – 1
∴ p = 32
MAX = 32 – 1 = 31
Indexing Question 2:
Suppose that we have an ordered file with r = 30000 records stored on a disk with block size B = 1024 bytes. If file records are of fixed size and are unspanned with record length R = 100 bytes. What is/are correct about the blocking factor for the file and the number of blocks needed for the file?
Answer (Detailed Solution Below)
Indexing Question 2 Detailed Solution
Concept:
Blocking factor: Number of records that can be stored in one block. It is calculated by dividing the block size by the length of each record.
Calculation:
Block size = 1024 bytes
Number of records = 30000
Record length or size = 100 bytes
File records are of fixed size and unspanned.
So, blocking factor = 1024/ 100 = ⌊10.24⌋ = 10
To find number of blocks needed:
10 records require = 1 block
1 record requires = 1/ 10 block
30000 records require = \(\frac{1}{{10}}\; \times \;30000 = 3000\) blocksIndexing Question 3:
With reference to the B+ tree index of order 1 shown below, the minimum number of nodes (including the Root node) that must be fetched in order to satisfy the following query: "Get all records with a search key greater than or equal to 7 and less than 15" is ____________.
Answer (Detailed Solution Below) 5
Indexing Question 3 Detailed Solution
Traversal: To get all records with a search key greater than or equal to 7 and less than 15
- Start from root node (9).
- search for the key 7 (as 7 < 9, search will continue to the left subtree) 2 nodes are visited in this case 5 and 7.
- Then to get the records greater than and equal to 7 and less than 15. For greater than 7, visit towards right side linearly until 15 is found.
In this way, total 5 nodes are traversed to get all records with a search key greater than or equal to 7 and less than 15.
Indexing Question 4:
In a B+ tree, if the search-key value is 8 bytes long the block size is 512 bytes and the block pointer size is 2 bytes, then the maximum order of the B+ tree is ______.
Answer (Detailed Solution Below) 52
Indexing Question 4 Detailed Solution
Data for B+ Tree:
Disk Block size = D= 512 byte
Block pointer = B = 2 byte
Key field = K = 8 byte.
Order = p
Formula:
p × 2 + (p – 1) (8) ≤ 512
Calculation:
p × 2 + (p – 1) (8) ≤ 512
2p + 8p ≤ 520
p ≤ 52
∴ p = 52
Indexing Question 5:
Consider a B+ tree in which the minimum number of keys in a node is 10. What is the maximum number of keys in any non-root node?
Answer (Detailed Solution Below) 21
Indexing Question 5 Detailed Solution
Data:
Minimum number of keys = MIN = 10
order of a B+ tree = p
Formula:
MIN = ⌈p ÷ 2⌉ – 1
Maximum number of keys = MAX = p – 1
Calculation:
10 = ⌈p ÷ 2⌉ – 1
∴ p = 21 or p = 22
If p = 21
MAX = 21 – 1 = 20
If p = 22
MAX = p – 1 = 22 – 1 = 21
For p = 22 we have maximum number of keys and the number of maximum keys is 21Indexing Question 6:
A file is organized so that the ordering of data records is the same as or close to the ordering of data entries in some index. Then that index is called
Answer (Detailed Solution Below)
Indexing Question 6 Detailed Solution
Indexing Question 7:
The order of a leaf node in a B+ tree is the maximum number of pairs (key ,record pointer)that it can hold. Given that the block size is 512 bytes, data record pointer is 7 bytes long and the key size is 9 bytes. If block pointer size is 6 bytes then what will be the order of leaf node?
Answer (Detailed Solution Below)
Indexing Question 7 Detailed Solution
Data for B+ Tree:
Disk Block size = D = 512 byte
Block pointer = B = 6 byte
Key field = K = 9 byte.
Record or data pointer = R = 7 byte
Order of leaf = p
Formula:
B + (p) (K + R) ≤ D
Calculation:
6 + (p) (9 + 7) ≤ 512
16p ≤ 512 - 6
16p ≤ 506
p ≤ 31.625
∴ p = 31
Indexing Question 8:
Construct a B+ tree with key sequence 10, 7, 3, 9, 5, 3, 4, 11, 8. The order of the B+ tree is 3. What is the key value at root node [N, 1, 35. File structure Indexing]
Answer (Detailed Solution Below) 5
Indexing Question 8 Detailed Solution
The B+ tree will be as follows
Indexing Question 9:
________ is the process of preparing the disk for its initial use and create new file systems.
Answer (Detailed Solution Below)
Indexing Question 9 Detailed Solution
The correct answer is Formatting.
Key Points
- Disk formatting is the configuring process of a data storage media such as a hard disk drive, floppy disk, or flash drive for initial usage.
- Any existing files on the drive would be erased with disk formatting.
Important Points
- Disk formatting is usually required when a new operating system is used by the user. When we format the disk, the existing files within the disk are also erased.
- We can perform disk formatting on both magnetic platter hard drives and solid-state drives.
- Disk formatting has also the capability to erase bad applications and various sophisticated viruses.
- Formatting a disk does not erase the data on the disk, only the address tables. It makes it much more difficult to recover the files.
Additional Information
- Disc address:- The disks are comprised of tracks and further sectors, each of which has a unique address.
- Disc header:- Disks that belong to a disk group, that is, disks that have a disk group name in the disk header, show a header status of the member.
Indexing Question 10:
The block of size 1000 bytes, records are of size 100 bytes of which 12 bytes is the key field pointer. And record pointer and black pointer takes 8 bytes. The file consists of 10,000 records.
Find the number of blocks required for a sparse index on file is?
Answer (Detailed Solution Below) 20
Indexing Question 10 Detailed Solution
The correct answer is 20.
Concept:
The given data,
Block size B= 1000 bytes
Record size = 100 bytes
key pointer size =12 bytes
Record pointer size = 8 bytes
Block pointer size =8 bytes
Number of data records = 10000
Explanation:
A blocking factor is an average number of records that can be stored in a block.
Blocking factor = Block size / Record size
Blocking factor = 1000 / 100 records per block.
Blocking factor = 10 records per block.
Sparse index:
It has an index record only for some search key values.
Number of index records = number of blocks in a data file.
Number of data blocks = 10000 /10 = 1000 blocks = Number of index records
Size of index records = 12+ 8 =20 bytes
Blocking factor for index file = 1000/20 index records per block.
Blocking factor for index file = 50 index records per block.
Number of index records = 1000
Number of spare index blocks = 1000 / 50 = 20
Hence the correct answer is 20.