ISRO 2013 Answer Key and Solutions Computer Science Scientist Engineer

The ISRO Computer Science (CS) Scientist Engineer test held on May 12, 2013 (Sunday). Total questions were 80 and test duration were 90 Minutes. Here is the solution of paper according to me of set A.

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23. Which of the following is the highest isolation level in transaction management?
(a) Serializable
(b) Repeated Read
(c) Committed Read
(d) Uncommitted Read

Correct Answer: (a)

Explanation: Isolation means the the querying user must be able to get a complete, consistent picture of the data, unaffected by other user's actions. There are four isolation levels:

• Serializable: It is highest isolation level. It requires read and write locks to be released at the end of the transaction. The range-locks must be acquired when a SELECT query uses a ranged WHERE clause, especially to avoid the phantom reads.
• Repeatable reads: It keeps read and write locks until the end of the transaction. However, range-locks are not managed, so phantom reads (two identical queries are executed, and the collection of rows returned by the second query is different from the first) can occur.
• Read committed: It restricts the reader from seeing any intermediate, uncommitted, dirty read. Keeps write locks until the end of the transaction, but read locks are released as soon as the SELECT operation is performed. Phantom reads can occur.
• Read uncommitted: It is lowest isolation level. Dirty reads (read data that has been not yet committed) are allowed. One transaction may see not-yet-committed changes made by other transactions.

Reference: https://en.wikipedia.org/wiki/Isolation_(database_systems)

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24. Consider the following relational schema:
Suppliers (sid:integer, sname:string, saddress:string)
Parts (pid:integer, pname:string, pcolor:string)
Catalog (sid:integer, pid:integer, pcost:real)

What is the result of the following query?

(SELECT Catalog.pid from suppliers, Catalog
WHERE Suppliers.sid = Catalog.pid)
MINUS
(SELECT Catalog.pid from suppliers, Catalog
WHERE Suppliers.sname <> 'Sachin' and Suppliers.sid = Catalog.sid)

(a) Pid of parts supplied by all except Sachin
(b) Pid of parts supplied only by Sachin
(c) Pid of parts available in catalog supplied by Sachin
(d) Pid of parts available in catalog supplied by all except Sachin

Correct Answer: None of These
Explanation: There is printing error in the question, so no any option is correct. Lets do solution by correcting the error. In 2nd line read like this "WHERE Supplier.sid = Catalog.sid". The 1st query (query before MINUS) gives list of all the suppliers who supply a part. And 2nd query (query after MINUS) gives all the parts that supplied by all except Sachin. So complete query gives Pids of parts available in catalog supplied by Sachin (option c).
Reference: http://gateoverflow.in/43857/isro-2013-24

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25. Consider the following dependencies and the Book table in a relational database design.
Determine the normal form of the given relation.
ISBN -> Title
ISBN -> Publisher
Publisher -> Address
(a) First Normal Form
(b) Second Normal Form
(c) Third Normal Form
(d) BCNF

Correct Answer: (b)

Explanation: First find candidate key, for this follow easy approach suggested by stinaq.me, draw diagram of dependencies. The diagram shows that we need ISBN to reach every attribute of relation. So ISBN is only the candidate key in this relation. To find normal form, follow easy approach suggested by stinaq.me.

• First normal form: Are all the attributes atomic? Yes.
• Second normal form: Is there a non primary attribute that’s determined by parts of a candidate key? No.
• Third normal form: Does a non primary attribute exists that is determined by another non primary attribute? Yes.

So, given relation is in Second Normal Form.

Reference: http://gateoverflow.in/16128/which-of-the-following-is-true

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26. Calculate the order of leaf (P_leaf) and non leaf (P) nodes of a B⁺ tree based on the information given below
Search key field = 12 bytes
Record pointer = 10 bytes
Block pointer = 8 bytes
Block size = 1 KB

(a) P_leaf = 51 & P = 46
(b) P_leaf = 47 & P = 52 
(c) P_leaf = 46 & P = 51
(d) P_leaf = 52 & P = 47

Correct Answer: (c)

Explanation: A B⁺ tree is a balanced binary search tree that follows a multi-level index format. Index is stored on the disk along with the actual database files. Memory block size is given 1 KB (1024 bytes), any node size will not be more than this. Order of a leaf and non-leaf node helps to determine number total size of node. Here we need to find order n using total size of node 1 KB.

• Any intermediary non-leaf node have n-1 search keys and n block pointers.
  ((n-1)*Search Key Field) + (n*Block Pointer) <= 1024 bytes
  ((n-1)*12) + (n*8) <= 1024 bytes
  12n - 12 + 8n <= 1024
  n <= 51.8
  So order of non-leaf node is 51  
• The leaf node have n search key fields, n record pointers, block pointer to right leaf node.
  (n*Search Key Field) + (n*Record Pointer) + (1*Block Pointer) <= 1024 bytes
  (n*12) + (n*10) + (1*8) <= 1024 bytes
  12n + 10n + 8 <= 1024
  n <= 46.18
  So order of non-leaf node is 46

Note: Similar question had asked in GATE CS 2015 exam.
Reference: http://gateoverflow.in/8555/gate2015-3_46

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27. The physical location of a record determined by a formula that transforms a file key into a record location is
(a) Hashed file
(b) B-Tree file
(c) Indexed file
(d) Sequential file

Correct Answer: (a)
Reference: http://gateoverflow.in/16991/physical-location-determined-formula-transforms-location

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61. Embedded pointer provides

(a) An inverted index

(b) A secondary access path

(c) A physical record key

(d) A primary key

Correct Answer: (b)

Explanation: Embedded pointer is a pointer set in a data record that provides secondary access path.
Note: Same question had asked in ISRO 2008 CS exam.

Reference: http://gateoverflow.in/44407/isro-2013-61-isro2008-35

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