[Type B] Chapter 14 Sumita Arora Solutions | Class 12 Computer Science

Solution:

(b) SELECT MAX(PRICE) FROM BOOK_INFORMATION;

Solution:

(c) SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES GROUP BY STORE ID;

Solution:

(c) SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES 
WHERE SUM(SALES_AMOUNT)>500 0GROUP BY STORE_ID;

Solution:

(b) SELECT COUNT(DISTINCT STORE_ID) FROM SALES;

Solution:

(d) SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES
WHERE STORE_ID=25 AND STORE_ID=45 GROUP BY STORE_ID;

Solution:

(c) SELECT AVG(EXAM_SCORE) FROM EXAM_RESULTS GROUP BY EXAM_ID HAVING EXAM_ID =1;

Solution:

(a) SELECT COUNT(DISTINCT STU_ID) FROM EXAM_RESULTS GROUP BY EXAM_ID;

Solution:

(a) SELECT * FROM EXAM_RESULTS WHERE LNAME LIKE 'L%';

Solution:

(d) 95

Solution:

(i) 4
(ii) 34
(iii)  35.33
(iv)  7800

Solution:

(i) 64.4
(ii) 1150.00
(iii)  475.00
(iv)  Error

Solution:

(i) SELECT employeeid, name, jobtitle from EMPLOYEE, JOB
WHERE EMPLOYEE.jobid = JOB.JOBID ;

(ii) SELECT name, sales, jobtitle from EMPLOYEE, JOB
WHERE EMPLOYEE.jobid = JOB.JOBID and EMPLOYEE.sales>1300000 ;

(iii) SELECT name, jobtitle from EMPLOYEE, JOB
WHERE EMPLOYEE.jobid = JOB.JOBID and EMPLOYEE.name LIKE "%SINGH%" ;

(iv) JOBID

(v) UPDATE EMPLOYEE
set jobid = 104
WHERE employeeid is "E4";

SELECT avg(salary), departments from JOB
where COUNT(DISTINCT departments) >3;

SELECT job from JOB
where max(salary) >= 3000 ;

SELECT COUNT(jobs) from JOB
where jobs is "Manager" ;

SELECT COUNT(employees) from EMPL
GROUP BY deptno ;

SELECT COUNT(salaries) from EMPL
GROUP BY department ;

SELECT MAX(salaries) from EMPL
GROUP BY dept_num ;

SELECT COUNT(order) from CUSTOMERS
GROUP BY id ;

SELECT name from CUSTOMERS;

SELECT SUM(orders) from CUSTOMERS
GROUP BY customer;

SELECT SUM(orders) from CUSTOMERS
where sum(orders)>1000
GROUP BY id
ORDER by slate;

SELECT name, order from CUSTOMERS;

SELECT name, sum(order) from CUSTOMERS;

SELECT DeptName from Dept;

SELECT Emp_name, DeptName,  from EMPL, Dept;