Relational algebra defines the theoretical way of manipulating table contents using the eight relational functions. To be considered minimally relational, the DBMS must support the key relational functions SELECT, PROJECT and JOIN.

Durability indicates the performance of the database's consistent state. When a transaction is completed the database reaches a consistent state and that state cannot be lost, even in the event of a system failure.

Desktop database, workgroup database and enterprise database is the classification of DBMS according to number of users where a distributed DBMS is classification according to database site location.

Production database are also known as transaction databases. Although such database yield streams of useful information, there are specialized databases, known as data warehouses that are designed and managed specifically to meet information needs. The data warehouses derive their data from production databases.

A file system is said to exhibit structural dependence because a change in any file's structure such as addition or deletion of a field, requires modification of all programs using that file. That means access to a file is dependent on its structure.

In SQL, we can create virtual tables.

AVG is an aggregate function where INT, DATE, BOOLEAN are the data types in SQL.

Transactions such as product or service sates payments and supply purchases reflect critical day to day operations such transactions are time critical and must be recorded accurately and immediately.

SELECT yields for all attributes found in a table. SELECT can be used to list all of the row's values for each attribute, or it can yield selected rows values for each attribute. In other worlds SELECT yields a horizontal subset of a table.PROJECT produces a list of all values for selected attributes. In other words, PROJECT yields a vertical subset of a table.

I, II, IV and V corresponds to any string of at least 3 characters.

First we insert values in student table as “Hitesh”, 1928, “B.E”. Then we delete tuple from student but table is still known to database, if it is an empty relation.After that we again inserting tuple into student having values “Hitesh”, 1220, and “B.Tech”After that by using ALTER command we add one attribute as subject.After that we again use ALTER command to drop attribute subject.So we remain with student relation having values “Hitesh”, 1220,'B.Tech”

Only statement (iii) is correct. SQL does not allow use of distance with COUNT(*). If is legal to use DISTINCT MAX and MIN. By default the ‘ORDER BY’ clause list items in an ascending order.

Since, backslash () is an escape character, the string must be inserted after "abcd ". 

The most important advantages of hierarchical models are conceptual simplicity, database security, data independence, database integrity and efficiency. Hierarchical database model lacks structural independence.

SQL is case insensitive, meaning that it treats upper and lower case letters as the same letter.$\therefore$ Statements (I) and (III) are true.But patterns used for comparison of string are case sensitive. Instead all inside quotes is case sensitive. ^$\therefore$ Statements (II) and (IV) are also true hence, (4) is the answer.

Statement (2) will give the result of the requirement of the query. This statement will select the employees’ names from employee works, company and group them by same city, those who are having salary between 17000 and 35500.

This option is also incorrect which does not match with the correct query.

Since, the access is sequential greater the distance greater will be the access time. Since, all files are referenced with equal frequency overall access time can be reduced by arranging them as in option (1). 

SMALLINT - integer value up to 16 digits, option contains all the matching pairs.

The SQL statement isSelect customer name from deposit where (balance > 1000) $\therefore$Relational algebra query is $\pi_{customer name} (\sigma_{balance} > 1000(Deposits)$

When the system comes back, the operation redo to is performed since the record. Appears in the log on the disk after this operation is executed, the values of account A and B are $957 and $2055, respectively. The value of account C remains $720 because T₁ has not been committed yet.

The log at the time of crash will be option B, which reaches up to commit statement but not including commit statement. Option (2) gives the correct log at the time of crash.

The SQL statement is select client customer name, client customer city from client customer where (client customer name = customer. Customer name and banker name = “Agassi” $\therefore$Relational algebra query is$\pi_{client, customer\ name, customer\ city} (\sigma_{Client, customer\ name})= customer. customer name \\ (\sigma_{Banker name} = “Agassi” (client × customers)$

Since, each file is referenced with equal frequency and each record in a particular file can be referenced with equal frequency, average access time will be$\dfrac{25 + (50 +40 ) + (50 + 80 + 50) + ......}{6}$Approximately 268 units

The modifications in the relation schema are done by a statement that begins with the keywords ALTER TABLE and the name of the relation. We then have several options and one of the most important is ADD.Syntax: ALTER TABLE relation ADD new-attribute dates typeE.g. ALTER TABLE Employee ADD salary INTThe above statement will add a new attributes salary to employee relation.