Question 1
Question
Define a degree of a relation
Answer
-
How many rows a table has
-
how long each tuple is, or how many columns the table has
-
how many different tuples there are
-
how many different datatypes table has
Question 2
Question
Define a cardinality of a relation
Answer
-
how long each tuple is
-
how many columns the table has
-
how many different tuples there are, how many rows a table has
-
how many different datatypes table has
Question 3
Question
Which of the following refers to union-compatibility requirements ?
Answer
-
Same number of columns
-
Same number of rows
-
Same number of tuples
-
Different domains
Question 4
Question
Which of the following refers to union-compatibility requirements ?
Question 5
Answer
-
restrict the possible values a tuple can assign to each attribute
-
relations to each other
-
uniquely identifies each tuple that appears in a relation
-
minimality of the attribute
Question 6
Question
Define a foreign key
Answer
-
restrict the possible values a tuple can assign to each attribute
-
relations to each other
-
uniquely identifies each tuple that appears in a relation
-
minimality of the attribute
Question 7
Question
Define a primary key
Question 8
Answer
-
stop the user from doing it
-
let the changes flow on
-
make referencing values the default for their column
-
make referencing values null
Question 9
Answer
-
stop the user from doing it
-
let the changes flow on
-
make referencing values the default for their column
-
make referencing values null
Question 10
Question
Define set default
Answer
-
stop the user from doing it
-
let the changes flow on
-
make referencing values the default for their column
-
make referencing values null
Question 11
Answer
-
stop the user from doing it
-
let the changes flow on
-
make referencing values the default for their column
-
make referencing values null
Question 12
Question
Define Data Definition Language
Answer
-
Specify database format
-
Specify access controls (privileges)
-
Specify and retrieve database contents
-
Specify table attribute uniqueness
Question 13
Question
Define Data Control Language
Answer
-
Specify access controls (privileges)
-
Specify database format
-
Specify and retrieve database contents
-
Specify table attribute uniqueness
Question 14
Question
Define Data Manipulation Language
Answer
-
Specify table attribute uniqueness
-
Specify database format
-
Specify access controls (privileges)
-
Specify and retrieve database contents
Question 15
Question
Which of the following does not refer to DBMS tools
Answer
-
Oracle
-
PostgreSQL
-
MySQL
-
Python
Question 16
Question
Which of the following is used to provide privilege to only a particular attribute?
Answer
-
Grant select on employee to finance
-
Grant update(salary, rate) on employee to finance
-
Grant update(salary) on employee to finance
-
Grant delete to finance
Question 17
Question
Which of the following statement is used to remove the privilege from the user finance?
Answer
-
Remove update on employee from finance
-
Delete select on employee from finance
-
Revoke update on employee from finance
-
Grant update on employee from finance
Question 18
Question
Which of the following is true regarding views?
Answer
-
The user who creates a view cannot be given update authorization on a view without having update authorization on the relations used to define the view.
-
If a user creates a view on which no authorization can be granted, the system will allow the view creation request.
-
A user who creates a view receives all privileges on that view.
Question 19
Question
If we wish to grant a privilege and to allow the recipient to pass the privilege on to other users, we append the __________ clause to the appropriate grant command.
Answer
-
With grant
-
Grant user
-
With grant option
-
Grant pass privelege
Question 20
Question
Which of the following is used to avoid cascading of authorizations from the user?
Answer
-
Granted by current role
-
Revoke grant option for select on department from finance;
-
Revoke select on employee from finance, cashier restrict;
-
Revoke select on department from finance, cashier cascade;
Question 21
Question
Privileges are granted over some specified parts of a database, such as a
Answer
-
Schema
-
Environment
-
Relation Or view
-
Query statement
Question 22
Question
Prevention of access to the database by unauthorized users is referred to as:
Answer
-
Integrity
-
Productivity
-
Security
-
Reliability
Question 23
Question
Database Authentication refers to:
Answer
-
methods of restricting user access to system
-
controlling access to portions of database
-
all of the answers mentioned
-
controlling the operation on the data
Question 24
Question
A set of possible data values is called
Answer
-
attribute
-
degree
-
domain
-
tuple
Question 25
Question
A functional dependency between two or more non-key attributes is called
Question 26
Question
__ refers to the correctness and completeness of the data in a database.
Answer
-
Database security
-
Data constraint
-
Data integrity
-
Data independence
Question 27
Question
Which of the following creates a virtual relation for storing the query?
Answer
-
Function
-
Procedure
-
View
-
None of the mentioned
Question 28
Question
Which of the following is the syntax for views where v is view name?
Answer
-
Create view v as “query name”;
-
Create “query expression” as view;
-
Create view v as “query expression”;
-
Create view “query expression”;
Question 29
Question
Updating the value of the view
Answer
-
Will not change the view definition
-
Will not affect the relation from which it is defined
-
Will affect the relation from which it is defined
-
Cannot determine
Question 30
Question
Create view faculty as: Select ID, name, dept name from instructor; Find the error in this query.
Answer
-
Instructor
-
Select
-
None of the mentioned
-
View …as
Question 31
Question
Which of the following is a basic form of grant statement?
Answer
-
Grant ‘privilege list’ on ‘user/role list’ to ‘relation name or view name’;
-
Grant ‘privilege list’ to ‘user/role list’;
-
Grant ‘privilege list’ on ‘relation name or view name’ to ‘user/role list’;
-
Grant ‘privilege list’ on ‘relation name or view name’ on ‘user/role list’;
Question 32
Question
Retrieve all data from the table OFFICE { id,room, name}
Answer
-
Select *from office;
-
Select from office;
-
Select name from office;
-
Select *form office;
Question 33
Question
Retrieve office name from the table OFFICE {id, room, name}
Answer
-
Select *from office;
-
Select from office;
-
Select name from office;
-
Select *form office;
Question 34
Question
Retrieve office id from the table OFFICE {id, room, name}
Answer
-
Select *from office;
-
Select id from office;
-
Select name from office;
-
Select id form office;
Question 35
Question
Retrieve office id and room from the table OFFICE {id, room, name}
Question 36
Question
Retrieve quantity of offices in the office table - OFFICE {id, rom, name}
Answer
-
Select avg (id) from office;
-
Select count (id) from office;
-
Select sum (id) from office;
-
Select max (id) from office;
Question 37
Question
Retrieve total number scholarship in the students table – STUDENTS {id, name, scholarship, registereddate, tutorid}
Answer
-
Select avg (scholarship) from office;
-
Select count (scholarship) from office;
-
Select sum (scholarship) from office;
-
Select max (scholarship) from office;
Question 38
Question
Retrieve quantity of students in the students table – STUDENTS {id, name, scholarship, registereddate, tutorid}
Answer
-
Select avg (id) from office;
-
Select count (id) from office;
-
Select sum (id) from office;
-
Select max (id) from office;
Question 39
Question
Sort students name by descending order - STUDENTS {id, name, scholarship, registereddate, tutorid}
Answer
-
Select name from students order by desc;
-
Select name from students group by desc;
-
Select *from students order by desc;
-
Select name form students order by desc;
Question 40
Question
Find an average scholarship of students - STUDENTS {id, name, scholarship, registereddate, tutorid}
Answer
-
Select avg (scholarship) from students;
-
Select count (scholarship) from students;
-
Select sum (scholarship) from students;
-
Select max (scholarship) from students;
Question 41
Question
Find maximum scholarship of students - STUDENTS {id, name, scholarship, registereddate, tutorid}
Answer
-
Select avg (scholarship) from students;
-
Select count (scholarship) from students;
-
Select sum (scholarship) from students;
-
Select max (scholarship) from students;
Question 42
Question
Find minimum scholarship of students - STUDENTS {id, name, scholarship, registereddate, tutorid}
Answer
-
Select avg (scholarship) from students;
-
Select count (scholarship) from students;
-
Select min (scholarship) from students;
-
Select max (scholarship) from students;
Question 43
Question
Retrieve offices’ name and tutors’, who work in CSSE department. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, OFFICEID (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Answer
-
Select o.name, t.name from office o join tutor t on o.id=t.officeid where o.name = ‘CSSE’
-
Select o.name, t.name from office o join tutor t on o.id=t.id where o.name = ‘CSSE’
-
Select o.name, t.name from office o join tutor t on o.id=t.officeid where o.name = ‘csse’
-
Select o.name, t.name from office o join tutor t on o.id=t.officeid where o.name ‘CSSE’
Question 44
Question
Retrieve all offices’ name and tutors’, who work in departments. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Answer
-
Select o.name, t.name from office o left join tutor t on o.id=t.officeid;
-
Select o.name, t.name from office o left join tutor t on o.id=t.id ;
-
Select o.name, t.name from office o right join tutor t on o.id=t.officeid;
-
Select o.name, t.name from office o right join tutor t on o.id=t.id ;
Question 45
Question
Retrieve students’ name, who have more than average scholarship. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Answer
-
select s.name from students s where s.scholarship > (select avg (scholarship)from students);
-
select s.name from students s where s.scholarship > (select scholarship from students);
-
select s.name from students s where s.scholarship > avg (scholarship);
-
select s.name from students s where s.scholarship < avg (scholarship);
Question 46
Question
Retrieve students’ name, who registered in 10th of Jan 2016. Tables: STUDENTS {id (PK), name, scholarship, registereddate, tutorid }.
Answer
-
select name from students where registereddate = '2016-01-10';
-
select name from students where registereddate = '2016-10-10';
-
select name from students where registereddate = 2016-10-10;
-
select name from students where registereddate = '2016-01-01';
Question 47
Question
Retrieve tutors’ name, who have more than experience others along with their students name . Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Answer
-
select t.name, s.name from tutor t join students s on t.id=s.tutorid where experience = (select max (experience) from tutor);
-
select t.name, s.name, max(experience) from tutor t join students s on t.id=s.tutorid;
-
select t.name, s.name from tutor t join students s on t.id=s.id where experience = (select max (experience) from tutor);
Question 48
Question
Retrieve tutors’ name, who have less experience than others along with their students name. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Answer
-
select t.name, s.name from tutor t join students s on t.id=s.tutorid where experience = (select min (experience) from tutor);
-
select t.name, s.name, min(experience) from tutor t join students s on t.id=s.tutorid;
-
select t.name, s.name from tutor t join students s on t.id=s.id where experience = (select min (experience) from tutor);
Question 49
Question
Increase tutors experience and students’ scholarship twice. Retrieve experience and scholarship along with their names. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Answer
-
select t.name,t.experience*2, s.name, s.scholarship*2 from tutor t join students s on t.id=s.tutorid;
-
select t.name,t.experience*2, s.name, s.scholarship*2 from tutor t join students s on t.id=s.id;
-
select t.name,t.experience*2, s.name, s.scholarship*2 from tutor t join tutor s on t.id=s.tutorid;
-
select t.name,t.experience*2, s.name, s.scholarship*2 from tutor t join tutor s on t.id=s.id;
Question 50
Question
Retrieve office, tutor and students name, but tutors name should have y value. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Answer
-
select o.name, t.name, st.name from tutor t join office o on o.id=t.officeid join students st on t.id=st.tutorid where t.name like '%y%';
-
select o.name, t.name, st.name from tutor t join office o o.id=t.officeid join students st on t.id=st.tutorid where t.name like '%y%';
-
select o.name, t.name, st.name from tutor t join office o on o.id=t.officeid join students st t.id=st.tutorid where t.name like '%y%';
-
select o.name, t.name, st.name from tutor t join office o on o.id=t.officeid join students st on t.id=st.id where t.name like '%y%';
Question 51
Question
Retrieve office, tutor and students name, but tutors name should not have y value. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutored (FK references tutor (id))}.
Answer
-
select o.name, t.name, st.name from tutor t join office o on o.id=t.officeid join students st on t.id=st.tutorid where t.name not like '%y%';
-
select o.name, t.name, st.name from tutor t join office o o.id=t.officeid join students st on t.id=st.tutorid where t.name not like '%y%';
-
select o.name, t.name, st.name from tutor t join office o on o.id=t.officeid join students st t.id=st.tutorid where t.name not like '%y%';
-
select o.name, t.name, st.name from tutor t join office o on o.id=t.officeid join students st on t.id=st.id where t.name not like '%y%';
Question 52
Question
Retrieve students name and scholarship along with their department name, who have scholarship between 4000 and 5000. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Answer
-
select s.name, s.scholarship, o.name from office o join tutor t on o.id=t.officeid join students s on t.id=s.tutorid where s.scholarship between 4000 and 5000;
-
select s.name, s.scholarship, o.name from office o join tutor t on o.id=t.officeid join students s on t.id=s.tutorid where s.scholarship 4000 and 5000;
-
select s.name, s.scholarship, o.name from tutor t join students s on t.id=s.tutorid where s.scholarship between 4000 and 5000;
-
select s.name, s.scholarship, o.name from tutor t join students s on t.id=s.tutorid where s.scholarship >=4000
Question 53
Question
Retrieve information about office and their tutors as well. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.d
Answer
-
select *from office o join tutor t on o.id=t.officeid;
-
select *from office o join tutor t in o.id=t.officeid;
-
select *form office o join tutor t on o.id=t.officeid;
-
select *from office o join tutor t on o.id=t.id;
Question 54
Question
Retrieve office name and tutor name, which have maximum experienced tutors. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Answer
-
select o.name, t.name from office o join tutor t on o.id=t.officeid where t.experience = (select max (experience) from tutor);
-
select o.name, t.name from office o join tutor t on o.id=t.officeid where t.experience = (select min (experience) from tutor);
-
select o.name, t.name from office o join tutor t on o.id=t.officeid where t.experience = (select high (experience) from tutor);
-
select o.name, t.name from office o join tutor t on o.id=t.officeid where t.experience = (select low (experience) from tutor);
Question 55
Question
Retrieve all tutors and students names. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Answer
-
select t.name, s.name from tutor t full join students s on t.id=s.tutorid;
-
select t.name, s.name from tutor t full join students s t.id=s.tutorid;
-
select t.name, s.name from tutor t full join students s t.id=s.id;
-
select t.name, s.name from tutor t full join students s on t.id=s.id;
Question 56
Question
Retrieve tutors’ name and the number of students for each of them. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Answer
-
select count (s.name) as numberofstudents, t.name from tutor t join students s on t.id=s.tutoridgroup by t.name;
-
select count (s.name) as numberofstudents, t.name from tutor t join students s on t.id=s.tutorid;
-
select count (s.name) as numberofstudents, t.name from tutor t join students s on t.id=s.idgroup by t.name;
-
select count (s.name) as numberofstudents, t.name from tutor t join students s on t.id=s.id;
Question 57
Question
Retrieve office name and the number of students for each of them. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))}.
Answer
-
select count (s.name) as numberofstudents, o.name from tutor t join students s on t.id=s.tutoridjoin office o on o.id=t.officeid group by o.name;
-
select count (s.name) as numberofstudents, o.name from students s on join office o on o.id=s.id group by o.name;
-
select count (s.name) as numberofstudents, o.name from tutor t join students s on t.id=s.tutoridjoin office o o.id=t.officeid group by o.name;
-
select count (s.name) as numberofstudents, o.name from tutor t join students s on t.id=s.idjoin office o on o.id=t.id group by o.name;
Question 58
Question
Retrieve all students’ name and scholarship, tutors name except students who have scholarship 3000$. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))
Answer
-
select s.name, t.name from students s join tutor t on s.tutorid=t.id where s.scholarship <> 3000;
-
select s.name, t.name from students s join tutor t on s.tutorid=t.id where s.scholarship = 3000;
-
select s.name, t.name from students s join tutor t on s.tutorid=t.id where s.scholarship < 3000;
-
select s.name, t.name from students s join tutor t on s.tutorid=t.id where s.scholarship is 3000;
Question 59
Question
How many types of anomalies exist?
Question 60
Question
Which of the following anomaly does not exist?
Answer
-
Creation anomaly
-
Deletion anomaly
-
Insertion anomaly
-
Modification anomaly
Question 61
Question
When an insertion anomaly occurs?
Answer
-
we are prevented from inserting some data into a relation until other data can be supplied
-
deletion leads to an unintended loss of data
-
it is possible that not all data needs to be changed will always be changed
-
it does not occur
Question 62
Question
When a modification anomaly occurs?
Answer
-
it is possible that not all data needs to be changed will always be changed
-
we are prevented from inserting some data into a relation until other data can be supplied
-
deletion leads to an unintended loss of data
-
it does not occur
Question 63
Question
When a deletion anomaly occurs?
Answer
-
deletion leads to an unintended loss of data
-
it is possible that not all data needs to be changed will always be changed
-
we are prevented from inserting some data into a relation until other data can be supplied
-
it does not occur
Question 64
Question
Define inner join
Answer
-
selects all rows from both tables as long as there is a match between the columns in both tables
-
returns all rows from the left table (1), with the matching rows in the right table (2)
-
returns all rows from the right table (2), with the matching rows in the right table (1)
-
returns all rows from the left table (1) and from the right table (2)
Question 65
Question
Define left join
Answer
-
returns all rows from the left table (1), with the matching rows in the right table (2)
-
selects all rows from both tables as long as there is a match between the columns in both tables
-
returns all rows from the right table (2), with the matching rows in the right table (1)
-
returns all rows from the left table (1) and from the right table (2)
Question 66
Question
Define right join
Answer
-
returns all rows from the right table (2), with the matching rows in the left table (1)
-
returns all rows from the left table (1), with the matching rows in the right table (2)
-
selects all rows from both tables as long as there is a match between the columns in both tables
-
returns all rows from the left table (1) and from the right table (2)
Question 67
Question
Define full join
Answer
-
returns all rows from the left table (1) and from the right table (2)
-
returns all rows from the right table (2), with the matching rows in the right table (1)
-
returns all rows from the left table (1), with the matching rows in the right table (2)
-
selects all rows from both tables as long as there is a match between the columns in both tables
Question 68
Answer
-
сombines the result set of two or more select statements
-
returns all rows from the left table (1) and from the right table (2)
-
returns all rows from the right table (2), with the matching rows in the right table (1)
-
returns all rows from the left table (1), with the matching rows in the right table (2)
Question 69
Question
INNER JOIN and JOIN are the same
Question 70
Question
Define self join
Answer
-
returns all rows from the table that references to yourself
-
returns all rows from the right table (2), with the matching rows in the right table (1)
-
selects all rows from both tables as long as there is a match between the columns in both tables
-
returns all rows from the left table (1) and from the right table (2)
Question 71
Question
How many types of functional dependencies exist
Question 72
Question
Which of the following does not to the functional dependency
Answer
-
equational
-
full
-
transitive
-
partial
Question 73
Question
Which of the following represent a full dependency?
Answer
-
it exists in relation if there is no attribute A that can be removed from X and the dependency still holds
-
if there exists an attribute A that is part of X that can be removed from X and the dependency still holds
-
x->y dependency in relation R and x, y , z are columns in R. X->Y and Y>Z in R. Final: X->Y
-
all above mentioned
Question 74
Question
Which of the following represent a partial dependency?
Answer
-
if there exists an attribute A that is part of X that can be removed from X and the dependency still holds
-
it exists in relation if there is no attribute A that can be removed from X and the dependency still holds
-
x->y dependency in relation R and x, y , z are columns in R. X->Y and Y>Z in R. Final: X->Y
-
all above mentioned
Question 75
Question
Which of the following represent a transitive dependency?
Answer
-
x->y dependency in relation R and x, y , z are columns in R. X->Y and Y->Z in R. Final: X->Y
-
if there exists an attribute A that is part of X that can be removed from X and the dependency still holds
-
it exists in relation if there is no attribute A that can be removed from X and the dependency still holds
-
all above mentioned
Question 76
Question
When was first normalization developed?
Question 77
Question
When was second normalization developed?
Question 78
Question
When was Boyce–Codd normalization developed?
Question 79
Question
Who is an inventor of relational model?
Answer
-
Edgar F.Codd
-
Raymond Boyce
-
Marine Jone
-
John Salamondor
Question 80
Question
Which of the following refers to the requirement of 1NF
Question 81
Question
Which of the following refers to the requirement of 2NF
Answer
-
All attributes (non-key columns) dependent on the key
-
Each cell should be single valued
-
All fields (columns) can be determined only by the key in the table and no other column
-
All above mentioned
Question 82
Question
Which of the following refers to the requirement of 3NF
Answer
-
All fields (columns) can be determined only by the key in the table and no other column
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All attributes (non-key columns) dependent on the key
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Each cell should be single valued
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All above mentioned
Question 83
Question
Define avg function
Answer
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Returns average value
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Returns total value
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Returns the first value
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Converts to lowercase
Question 84
Question
Which function is used to retrieve quantity of rows
Question 85
Question
Retrieve avg scholarship of students. Tables: office {id (PK), locations, name}, tutor {id (PK), name, officeid (FK references office (id)), experience}, students {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))
Answer
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select avg (scholarship) from students;
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select aveg (scholarship) from students;
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select avr (scholarship) from students;
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select avgr (scholarship) from students;
Question 86
Question
Retrieve students’, scholarship and teacher's’ name, who have more than average scholarship. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))
Answer
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select s.name, s.scholarship, t.name from students s, tutor t where t.id=s.tutorid and s.scholarship> (select avg(scholarship) from students)
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select s.name, s.scholarship, t.name from students s, tutor t where t.id=s.tutorid and s.scholarship> avg(scholarship)
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select s.name, s.scholarship, t.name from students s, tutor t where s.scholarship> (select avg(scholarship) from students)
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select s.name, avg(s.scholarship), t.name from students s, tutor t
Question 87
Question
Retrieve students’, scholarship and teacher's’ name, who have more than average scholarship AND increase those students twice. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))
Answer
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select s.name, s.scholarship*2, t.name from students s, tutor t where t.id=s.tutorid and s.scholarship> (select avg(scholarship) from students)
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select s.name, s.scholarship*2, t.name from students s, tutor t where t.id=s.tutorid and s.scholarship> avg(scholarship)
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select s.name, s.scholarship*2, t.name from students s, tutor t where s.scholarship> (select avg(scholarship) from students)
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select s.name, avg(s.scholarship)*2, t.name from students s, tutor t
Question 88
Question
Retrieve students and mentors name, but mentors registration date should be before students registration date. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))
Answer
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select s.name, m.name from students s join students m on s.tutorid=m.tutorid where s.registereddate<m.registereddate;
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select s.name, m.name from students s join students m on s.tutorid=m.tutorid where s.registereddate>m.registereddate;
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select s.name, m.name from students s join students m on s.tutorid=m.tutorid;
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select s.name, m.name from students s join students m on s.registereddate=m.registereddate;
Question 89
Question
Retrieve the highest experience from tutor. Tables: OFFICE {id (PK), locations, name}, TUTOR {id (PK), name, officeid (FK references office (id)), experience}, STUDENTS {id (PK), name, scholarship, registereddate, tutorid (FK references tutor (id))
Answer
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select max (experience) from tutor;
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select min (experience) from tutor;
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select max (t.experience) from tutor t group by t.name;
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select min (t.experience) from tutor t group by t.name;