collegeCurriculum

Sem 2

C++ Record

1. Program to perform all arithematic operations using default arguments.

2. Program to find the volume of cube, cylinder and rectangular box using function overloading within a class.

3. Program to create class student with 3 data members and member functions.

4. Program to create a class employee having data members employee id, name, designation and salary using the concept array of objects and print the salaryship of employee using nesting of member functions.

5. Program to implement the concept of static data members and member functions.

6. Program to implement friend function with two classes.

7. Program to perform time addition using object as function arguments and returning objects.

8. Program to implement constructor overloading for complex numbers.

9. Program to implement the concept of copy constructor.

10. Program to implement the concept of dynamic constructor.

11. Program with class rectangle with data members length & breadth and member functions to calculate the area. The class has 2 constructors having one parameter(length = breadth) and two parameters.

12. Program to overload unary minus operator with member function.

13. Program with class distance with data members km and m. C++ program to add 2 distances using the concept of binary operator overloading(+ operator).

14. Program to find the difference between 2 vector objects with the concept of operator overloading using friend functions.

15. Program to perform addition of 2 matrix object by operator overloading using friend function.

16. Program to implement the concept of single inheritance.

17. Program to implement the concept of multiple inheritance.

18. Program to implement the concept of multilevel inheritance.

19. Program to implement the concept of hierarchical inheritance.

20. Program to implement the concept of hybrid inheritance.

21. Program to implement the concept of virtual base class.

22. Program to implement constructors in derived class.

23. Program to implement the concept of ‘this’ pointer.

24. Program to implement the concept of virtual function.

25. Program to implement the concept of run time polymorphism.

Previous Year Questions and Answers

Consolidated previous year question papers

C++ Previous Year Questions with Answers

1. Create a class Student with three data members name, age and address and member functions read(), display() and getAge(). Write a C++ program to read the data of n students and display the details of students whose age greater than the average age.

2. Create a class Time with data members hour, minute and second and member functions readTime(), displayTime() and an overloaded operator ‘+’. Write a C++ program to find the sum of two Time objects. (use 24 hours format).

3. Create a class Rectangle with data members length and breadth and member function to calculate the Area (length*breadth). The class has two constructors (1) having one parameter (assumme same values for both length and breadth) and (2) having two parameters. Write C++ program to create objects of the Rectangle class having one and two parameters and print their area.

4. Design a base class Person with data members name, address and phone_number. Derive a class Employee with data member emp_number from Person. Derive a class Manager with data members designation, department_name and basic_salary from Employee. Write a CPP program to accept all details of ‘n’ managers.

5. Write a C++ program to design a class Shape with data member volume and three overloaded member functions to find the volume of cube, cuboid and cone. Use another member function display to print the volume of three shapes.

   volume of cube = side^3 
   volume of cone = 3.14*radius*radius*height/3
   volume of cuboid = length*width*height
   

6. Create a class Vector contains data members a, b, c as coefficients of i, j, k and member functions to read and display vector objects. Write a C++ program to find the difference between two vector objects using the concept operator overloading with friend function. Eg: (a1i + b1j + c1k) - (a2i + b2j + c2k) = (a1 - a2)i + (b1 - b2)j + (c1 - c2)k

7. Create classes
   
   Student
   Data members : stud_id, stud_name, stud_age
   Member functions: readstud(), displaystud()
   Course</ins>
   Data members : course_name, department
   Member functions : readcourse(), displaycourse()
   Admission</ins>
   Data members : adm_no, adm_date
   Member functions: readadm(), displayadm().
   
   Write a CPP program to inherit student and course classes to admission.

8. Write a CPP program to perform addition of two Matrix objects by operator overloading using friend function.

DBMS Previous Year Questions with Answers

Question 1

TABLE 1: BOOK COLUMN NAME | DATA TYPE | CONSTRAINTS — | — | — BookId | Varchar(5) | Primary Key Book_Name | Varchar(20) | Not Null Author_Name | Varchar(20) | Price | Number(5) |

TABLE 2: PUBLISHER COLUMN NAME | DATA TYPE | CONSTRAINTS — | — | — Pub_Id | Varchar(5) | Primary Key Pub_Name | Varchar(20) | Not Null BookId | Varchar(5) | Foreign Key

Write SQL queries for the following:

1. Create the above tables. ```sql CREATE TABLE book(bookid VARCHAR(5) PRIMARY KEY, book_name VARCHAR(20) NOT NULL, author_name VARCHAR(20), price NUMERIC(5, 0));

CREATE TABLE publisher(pub_id VARCHAR(5) PRIMARY KEY, pub_name VARCHAR(20) NOT NULL, bookid VARCHAR(5) REFERENCES book(bookid));

2. Insert 5 records into each table.
```sql
INSERT INTO book (bookid, book_name, author_name, price)
VALUES
    ('B001', 'Book 1', 'Author 1', 2000),
    ('B002', 'Book 2', 'Author 2', 3500),
    ('B003', 'Book 3', 'Author 3', 5000),
    ('B004', 'Book 4', 'Author 4', 6000),
    ('B005', 'Book 5', 'Author 5', 4500);

INSERT INTO publisher (pub_id, pub_name, bookid)
VALUES
    ('P001', 'PHI', 'B001'),
    ('P002', 'Publisher 2', 'B002'),
    ('P003', 'Publisher 3', 'B003'),
    ('P004', 'Publisher 4', 'B004'),
    ('P005', 'Publisher 5', 'B005');

1. Display the details of book published by “PHI”. ```sql SELECT * FROM book WHERE bookid IN (SELECT bookid FROM publisher WHERE pub_name = ‘PHI’);

– or –

SELECT b.bookid, b.book_name, b.author_name, b.price FROM book AS b JOIN publisher AS p ON b.bookid = p.bookid WHERE p.pub_name = ‘PHI’;

4. Delete all books having price greater than 5000.
```sql
-- Delete records from the PUBLISHER table first
DELETE FROM publisher
WHERE bookid IN (SELECT bookid FROM book WHERE price > 5000);

-- Now, delete records from the BOOK table
DELETE FROM book
WHERE price > 5000;

1. List the details of books and its publishers in ascending order of their price.

   SELECT b.bookid, b.book_name, b.author_name, b.price, p.pub_name 
   FROM book AS b
   JOIN publisher AS p ON b.bookid = p.bookid 
   ORDER BY price ASC;
   

Question 2

TABLE 1: COLLEGE COLUMN NAME | DATA TYPE | CONSTRAINTS — | — | — COLL_CODE | NUMBER(5) | PRIMARY KEY COLL_NAME | VARCHAR(25) | NOT NULL COLL_PLACE | VARCHAR(25) | STARTING_DATE | DATE CONTACT_NUMBER | VARCHAR(10)

TABLE 2: COURSE COLUMN NAME | DATA TYPE | CONSTRAINTS — | — | — COURSE_CODE | NUMBER(5) | PRIMARY KEY COURSE_NAME | VARCHAR(25) | NOT NULL DEPARTMENT | VARCHAR(25) C_CODE | NUMBER(5) | FOREIGN KEY

Write SQL queries for the following:

1. Create the above tables. ```sql CREATE TABLE college(coll_code INT, coll_name VARCHAR(25) NOT NULL, coll_place VARCHAR(25), starting_date DATE, contact_number VARCHAR(10), PRIMARY KEY(coll_code));

CREATE TABLE course(course_code INT PRIMARY KEY, course_name VARCHAR(25) NOT NULL, department VARCHAR(25), c_code INT, FOREIGN KEY(c_code) REFERENCES college(coll_code));

2. Insert 5 records into each table.
```sql
INSERT INTO college (coll_code, coll_name, coll_place, starting_date, contact_number)
VALUES
    (1, 'ABC College', 'City A', '1990-05-15', '1234567890'),
    (2, 'XYZ College', 'City B', '2000-09-20', '9876543210'),
    (3, 'PQR College', 'City A', '1985-03-10', '5555555555'),
    (4, 'LMN College', 'City C', '2005-11-28', '1112223333'),
    (5, 'EFG College', 'City B', '1995-07-08', '4447778888');

INSERT INTO course (course_code, course_name, department, c_code)
VALUES
    (101, 'BCA', 'Computer Science', 1),
    (102, 'BBA', 'Management', 2),
    (103, 'BSc', 'Science', 3),
    (104, 'BCA', 'Computer Science', 4),
    (105, 'MCA', 'Computer Science', 4),
    (106, 'MBA', 'Management', 5);

1. Display the details of all colleges offering BCA course in the ascending order of college name.

   SELECT * FROM college WHERE coll_code
   IN 
   (SELECT c_code FROM course WHERE course_name = 'BCA')
   ORDER BY coll_name ASC;
   

2. Delete records of colleges started on or before 01/01/1990. ```sql DELETE FROM course WHERE c_code IN (SELECT coll_code FROM college WHERE starting_date <= ‘1990-01-01’);

DELETE FROM college WHERE starting_date <= ‘1990-01-01’;

5. Display the Count of colleges in the same place.
```sql
SELECT coll_place, COUNT(coll_name) 
AS college_count FROM college
GROUP BY coll_place;

Question 3

TABLE 1: DEPARTMENT COLUMN NAME | DATA TYPE | CONSTRAINTS — | — | — DEPT_ID | VARCHAR(10) | PRIMARY KEY DEPT_NAME | VARCHAR(15) | NOT NULL

TABLE 2: TEACHER COLUMN NAME | DATA TYPE | CONSTRAINTS — | — | — TEACH_ID | VARCHAR(10) | PRIMARY KEY NAME | VARCHAR(15) | NOT NULL DEPT_ID |VARCHAR(10) | FOREIGN KEY SUBJECT | VARCHAR(15) |

Write SQL queries for the following:

1. Create the above tables. ```sql CREATE TABLE department ( dept_id VARCHAR(10) PRIMARY KEY, dept_name VARCHAR(15) NOT NULL );

CREATE TABLE teacher ( teach_id VARCHAR(10), name VARCHAR(15) NOT NULL, dept_id VARCHAR(10), subject VARCHAR(15), FOREIGN KEY (dept_id) REFERENCES department(dept_id), PRIMARY KEY (teach_id) );

2. Insert 5 records into each table.
```sql
INSERT INTO department (dept_id, dept_name)
VALUES
    ('D001', 'Computer Application'),
    ('D002', 'Mathematics'),
    ('D003', 'Physics'),
    ('D004', 'English'),
    ('D005', 'History');

INSERT INTO teacher (teach_id, name, dept_id, subject)
VALUES
    ('TI01', 'John Smith', 'D001', 'Database Management'),
    ('TI02', 'Jane Doe', 'D002', 'Calculus'),
    ('TI03', 'Michael Johnson', 'D001', 'Software Engineering'),
    ('TI04', 'Emily Williams', 'D003', 'Physics I'),
    ('TI05', 'Robert Brown', 'D001', 'Networking');

1. Update subject of the teacher to ‘COMPUTER NETWORK’ whose TEACH_ID is “TI05”.

   UPDATE teacher SET subject='Computer Network'
   WHERE teach_id='TI05';
   

2. Display the number of teachers in each department.

   SELECT dept_name, COUNT(name) AS teacher_count FROM department JOIN 
   teacher ON department.dept_id = teacher.dept_id
   GROUP BY dept_name;
   

3. Display the details of all teachers in the Department of Computer Application.

   SELECT * FROM teacher WHERE dept_id 
   IN
   (SELECT dept_id FROM department WHERE dept_name = 'Computer Application');
   

Question 4

TABLE 1: MANAGER COLUMN NAME | DATA TYPE | CONSTRAINTS — | — | — MGR_ID | VARCAHR(5) | PRIMARY KEY NAME | VARCHAR(25) | NOT NULL DEPT_ID | NUMBER(5) | FOREIGN KEY CONTACT_NO | NUMBER(10) | SALARY | NUMBER(10) |

TABLE 2: DEPARTMENT COLUMN NAME | DATA TYPE | CONSTRAINTS — | — | — DEPT_ID | NUMER(5) | PRIMARY KEY DEPT_NAME | VARACHAR(10) | NOT NULL

Write SQL queries for the following:

1. Create the above tables. ```sql CREATE TABLE department(dept_id INT PRIMARY KEY, dept_name VARCHAR(10) NOT NULL);

CREATE TABLE manager(mgr_id VARCHAR(5) PRIMARY KEY, name VARCHAR(25) NOT NULL, dept_id INT, contact_no INT, salary INT, FOREIGN KEY(dept_id) REFERENCES department(dept_id));

2. Insert 5 records into each table.
```sql
INSERT INTO department (dept_id, dept_name)
VALUES
    (1, 'Sales'),
    (2, 'Marketing'),
    (3, 'Finance'),
    (4, 'HR'),
    (5, 'Operations');

INSERT INTO manager (mgr_id, name, dept_id, contact_no, salary)
VALUES
    ('M001', 'John Smith', 1, 1234567890, 55000),
    ('M002', 'Jane Doe', 2, 9876543210, 60000),
    ('M003', 'Michael Johnson', 1, 5555555555, 48000),
    ('M004', 'Emily Williams', 3, 1112223333, 72000),
    ('M005', 'Robert Brown', 1, 4447778888, 49000);

1. Allow an increment of Rs. 2500 for managers whose salary is less than 50000

   UPDATE manager SET salary = salary + 2500 
   WHERE salary < 50000;
   

2. Display the details of managers who is getting maximum salary.

   SELECT * FROM manager WHERE salary = 
   (SELECT MAX(salary) FROM manager);
   

3. Display the details of manager who are working in sales department.

   SELECT * FROM manager WHERE dept_id IN
   (SELECT dept_id FROM department WHERE dept_name = 'Sales');
   

Question 5

TABLE 1: SAILOR COLUMN NAME | DATA TYPE | CONSTRAINTS — | — | — SAIL_ID | NUMBER | PRIMARY KEY SAILOR_NAME | VARCHAR(15) | NOT NULL AGE | NUMBER |

TABLE 2: RESERVE_BOAT COLUMN NAME | DATA TYPE | CONSTRAINTS — | — | — BOAT_ID | VARCHAR(10) | PRIMARY KEY BOAT_NAME | VARCHAR(15) | NOT NULL BOAT_COLOR | VARCHAR(10) | RED, BLUE, GREEN SAIL_ID | NUMBER | FOREIGN KEY

Write SQL queries for the following:

1. Create the above tables. ```sql CREATE TABLE sailor ( sail_id INT PRIMARY KEY, sailor_name VARCHAR(15) NOT NULL, age INT );

CREATE TABLE reserve_boat ( boat_id VARCHAR(10) PRIMARY KEY, boat_name VARCHAR(15) NOT NULL, boat_color VARCHAR(10) CHECK (boat_color IN (‘red’, ‘blue’, ‘green’)), sail_id INT, FOREIGN KEY (sail_id) REFERENCES sailor(sail_id) );

2. Insert 5 records into each table.
```sql
INSERT INTO sailor (sail_id, sailor_name, age)
VALUES
    (1, 'John', 30),
    (2, 'Jane', 28),
    (3, 'Michael', 35),
    (4, 'Emily', 22),
    (5, 'Robert', 40);
   
INSERT INTO reserve_boat (boat_id, boat_name, boat_color, sail_id)
VALUES
    ('B001', 'Boat 1', 'red', 1),
    ('B002', 'Boat 2', 'blue', 2),
    ('B003', 'Boat 3', 'green', 3),
    ('B004', 'Boat 4', 'red', 4),
    ('B005', 'Boat s', 'green', 5);

1. Display all the boat names ending with “s”.

   SELECT * FROM reserve_boat 
   WHERE boat_name LIKE '%s';
   

2. Display the details of sailors who have reserved a ‘red’ and ‘green’ boat.

   SELECT * FROM sailor WHERE sail_id IN
   (SELECT sail_id FROM reserve_boat 
   WHERE boat_color IN ('red', 'green'));
   

3. Create a view contains list of all sailors whose age in between 25 and 45.

   CREATE VIEW sailors_age_view AS
   SELECT * FROM sailor 
   WHERE age BETWEEN 25 AND 45;
   

Question 6

TABLE 1: PRODUCT COLUMN NAME | DATA TYPE | CONSTRAINTS — | — | — PRODUCT_NUM | VARCHAR(6) | PRIMARY KEY PRODUCT_NAME | VARCHAR(15) | NOT NULL QTY_IN_HAND | NUMBER(8) | UNIT_PRICE | NUMBER(8, 2) | NOT NULL, CANNOT BE ZERO

TABLE 2: SALES COLUMN NAME | DATA TYPE | CONSTRAINTS — | — | — BILL_NUM | VARCHAR(6) | PRIMARY KEY QUANTITY | NUMBER(8) | PRODUCT_NUM | VARCHAR(6) |FOREIGN KEY

Write SQL queries for the following:

1. Create the above tables. ```sql CREATE TABLE product(product_num VARCHAR(6) PRIMARY KEY, product_name VARCHAR(15) NOT NULL, qty_in_hand INT, unit_price NUMERIC(8, 2) NOT NULL, CHECK(unit_price != 0));

CREATE TABLE sales(bill_num VARCHAR(6) PRIMARY KEY, quantity INT, product_num VARCHAR(6), FOREIGN KEY(product_num) REFERENCES product(product_num));

2. Insert 5 records into each table.
```sql
INSERT INTO product (product_num, product_name, qty_in_hand, unit_price)
VALUES
    ('P001', 'Product 1', 100, 50.00),
    ('P002', 'Product 2', 75, 30.00),
    ('P003', 'Product 3', 200, 20.00),
    ('P004', 'Product 4', 50, 100.00),
    ('P005', 'Product 5', 300, 10.00);


INSERT INTO sales (bill_num, quantity, product_num)
VALUES
    ('K101', 5, 'P001'),
    ('K102', 3, 'P002'),
    ('K103', 10, 'P003'),
    ('K104', 2, 'P004'),
    ('K105', 8, 'P005');

1. List the Product_num, name and Quantity for the Bill_num “K101”.

   SELECT p.product_num, p.product_name, s.quantity 
   FROM product p -- alias can be used without the keyword 'AS'
   JOIN sales AS s ON p.product_num = s.product_num 
   WHERE s.bill_num = 'K101';
   

2. Increase the product price of all products by 10%.

   UPDATE product SET unit_price = unit_price + (unit_price / 10);
   

3. Display the details of product having highest unit price. ```sql SELECT * FROM product WHERE unit_price = (SELECT MAX(unit_price) FROM product);

– or –

SELECT * FROM product ORDER BY unit_price DESC LIMIT 1;

#### Question 7

TABLE 1: CUSTOMER
COLUMN NAME | DATA TYPE | CONSTRAINTS
--- | --- | ---
CUST_ID | VARCHAR(10) | PRIMARY KEY
NAME | VARCHAR(15) | NOT NULL
LOCATION | VARCHAR(15) |
AGE | NUMBER |

TABLE 2: ORDERS
COLUMN NAME | DATA TYPE | CONSTRAINTS
--- | --- | ---
ORDER_ID | VARCHAR(10) | PRIMARY KEY
PRODUCT | VARCHAR(15) | NOT NULL
CUST_ID | VARCHAR(10) | FOREIGN KEY
ORDER_DATE | DATE 
ORDER_AMT | NUMBER


##### Write SQL queries for the following:

1. Create the above tables.
```sql
CREATE TABLE customer(cust_id VARCHAR(10) PRIMARY KEY,
name VARCHAR(15) NOT NULL,
location VARCHAR(15),
age INT);

CREATE TABLE orders(order_id VARCHAR(10) PRIMARY KEY,
product VARCHAR(15) NOT NULL,
cust_id VARCHAR(10),
order_date DATE,
order_amt INT,
FOREIGN KEY(cust_id) REFERENCES customer(cust_id));

1. Insert 5 records into each table. ```sql INSERT INTO customer (cust_id, name, location, age) VALUES (‘C1100’, ‘John Doe’, ‘New York’, 30), (‘C1200’, ‘Jane Smith’, ‘Los Angeles’, 25), (‘C1300’, ‘Michael Johnson’, ‘Chicago’, 22), (‘C1400’, ‘Emily Williams’, ‘Miami’, 28), (‘C1500’, ‘Robert Brown’, ‘San Francisco’, 35);

INSERT INTO orders (order_id, product, cust_id, order_date, order_amt) VALUES (‘O101’, ‘Product 1’, ‘C1100’, ‘2023-08-01’, 1000), (‘O102’, ‘Product 2’, ‘C1200’, ‘2023-08-02’, 750), (‘O103’, ‘Product 1’, ‘C1300’, ‘2023-08-03’, 1200), (‘O104’, ‘Product 3’, ‘C1400’, ‘2023-08-04’, 500), (‘O105’, ‘Product 2’, ‘C1500’, ‘2023-08-05’, 1800), (‘O106’, ‘Product 1’, ‘C1100’, ‘2023-08-06’, 800), (‘O107’, ‘Product 3’, ‘C1300’, ‘2023-08-07’, 1500), (‘O108’, ‘Product 2’, ‘C1200’, ‘2023-08-08’, 900), (‘O109’, ‘Product 1’, ‘C1500’, ‘2023-08-09’, 1600), (‘O110’, ‘Product 2’, ‘C1400’, ‘2023-08-10’, 700);

3. Change the location of customer to Kochi whose cust_id is "C1100".
```sql
UPDATE customer SET location='Kochi' WHERE cust_id='C1100'; 

1. Display the details of younger customer in the group. ```sql SELECT * FROM customer ORDER BY age ASC LIMIT 1;

– or –

SELECT * FROM customer WHERE age = (SELECT MIN(age) FROM customer);

7. Display name, location, order_amt and order_date of customer who placed the highest order.
```sql
SELECT name, location, order_amt, order_date FROM customer 
JOIN orders ON orders.cust_id = customer.cust_id 
ORDER BY orders.order_amt DESC
LIMIT 1;

Question 8

TABLE 1: EMPLOYEE COLUMN NAME | DATA TYPE | CONSTRAINTS — | — | — EMP_NUM | NUMBER(5) | PRIMARY KEY EMP_NAME | VARCHAR(25) | NOT NULL DESIGNATION | VARCHAR(20) | SALARY | NUMBER(5)

TABLE 2: PROJECT COLUMN NAME | DATA TYPE | CONSTRAINTS — | — | — PROJ_ID | VARCHAR(5) | PRIMARY KEY PROJ_tfNAME | VARCHAR(25) | EMP_NUM | NUMBER(5) | FOREIGN KEY

Write SQL queries for the following:

1. Create the above tables. ```sql CREATE TABLE employee (emp_num INT PRIMARY KEY, emp_name VARCHAR(25) NOT NULL, designation VARCHAR(20), salary INT);

CREATE TABLE project (proj_id VARCHAR(5) PRIMARY KEY, proj_name VARCHAR(25), emp_num INT REFERENCES employee(emp_num)); – foreign key

2. Insert 5 records into each table.
```sql
INSERT INTO employee (emp_num, emp_name, designation, salary)
VALUES
    (101, 'John Kumar', 'Manager', 60000),
    (102, 'Jane Smith', 'Developer', 45000),
    (103, 'Michael Kumar', 'Analyst', 55000),
    (104, 'Emily Kumar', 'Designer', 50000),
    (105, 'Robert Brown', 'Tester', 48000),
    (106, 'Kumar Shekar', 'Debugger', 50000),
    (107, 'Shone Binu', 'Manager', 40000);

INSERT INTO project (proj_id, proj_name, emp_num)
VALUES
    ('P001', 'Project A', 101),
    ('P002', 'Project B', 102),
    ('P003', 'Project C', 103),
    ('P004', 'Project D', 104),
    ('P005', 'Project E', 105),
    ('P006', 'Project X', 107),
    ('P007', 'Project S', 101);

1. Display the details of employees in alphabetical order of EMP_NAME whose name contains “kumar”.

   SELECT * FROM employee
   WHERE LOWER(emp_name) LIKE '%kumar%'
   ORDER BY emp_name;
   

2. List EMP_NUM, EMP_NAME and DESIGNATION of employees who have not assigned any projects.

   SELECT emp_num, emp_name, designation FROM employee
   WHERE emp_num NOT IN (SELECT emp_num FROM project);
   

3. Display the details of project in which MANAGER with highest pay works. ```sql SELECT * FROM project WHERE emp_num = (SELECT emp_num FROM employee WHERE salary = (SELECT MAX(salary) FROM employee));

– or –

SELECT * FROM project WHERE emp_num = (SELECT emp_num FROM employee ORDER BY salary DESC LIMIT 1);

– or –

SELECT p.* FROM project p – ‘AS’ not compulsory for alias JOIN ( SELECT emp_num, salary – salary not required here FROM employee ORDER BY salary DESC LIMIT 1 ) AS e ON p.emp_num = e.emp_num; ```