PYQ for Cuet computer science

Previous Years' Paper
Common University Entrance Test for UG Programmes
CUET-UG – Computer Science
Entrance Exam, 2025



Q.1. Identify the relational algebra operation denoted by:
Course X Student;
where 'Course' and 'Student' are two relations and X is an operation.
1. Union
2. Set Difference
3. Cartesian Product
4. Intersection
Q.2. Given:
TableA:
Name Hobbies
Anu Dance
Anuj Music

TableB:
Name Hobbies
Prannav Reading
Anuj Reading

Find TableA U TableB?




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,1.




2.




3.



4.




Q.3. Given two relations:
'Employee' with structure as ID, name, Address, Phone, Deptno
Department witn structure as Deptno, Dname
__________ is used to represent the relationship between two relations Employee
and Department.
1. Primary key
2. Alternate key
3. Foreign key
4. Candidate key
Q.4. A value is specified for the column, if no value is provided.
1. unique




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,2. null
3. default
4. primary
Q.5. Given table 'StudAtt' with structure as Rno, Attdate, Attendance.
Identify the suitable command to addaprimary key to the table after table
creation.
Note: In the given case, we want to make both Rno and Attdate columns as
primary key.
1. ALTER TABLE StudAtt ADD PRIMARY KEY(Rno, Attdate);
2. CREATE TABLE StudAtt ADD PRIMARY KEY(Rno);
3. ALTER TABLE StudAtt ADD PRIMARY KEY;
4. ALTER TABLE StudAtt ADD PRIMARY KEY(Rno) AND PRIMARY
KEY(Attdate);
Q.6. The SELECT command when combined with DISTINCT clause is used to:
1. returns records without repetition
2. returns records with repetition
3. returns all records with conditions
4. returns all records without checking
Q.7. ____________ is used to search for a specific pattern inacolumn.
1. Between operator
2. In operator
3. Like operator
4. Null operator
Q.8. Give the output of the query:-
SELECT MONTH("2010-03-05");




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, 1. 3
2. 5
3. MARCH
4. MAY
Q.9. Match List-I with List-II
List-1 List-I1
(Aggregate function) (Description)
(A) count(marks) (I) Count all rows
(B) count(*) (II) Finding average of non null
values of marks
(C) avg(marks) (III) Count all non null values of
marks column
(D) sum(marks) (IV) Finding sum of all marks

Chooco the correct answer from the options given below
1. (A) - (III), (B) - (I), (C) - (II), (D) - (IV)
2. (A) - (I), (B) - (III), (C) – (II), (D) - (IV)
3. (A) - (I), (B) - (II), (C) – (IV), (D), - (III)
4. (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
Q.10. Given Relation: 'STUDENT'
S.NO SNAME MARKS
1 Amit 20
2 Karuna 40
3 Kavita NULL
4 Anuj 30

Find the value of:
SELECT AVG(MARKS) FROM STUDENT;
1. 30




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