(b) Write the output of the queries (i) to (iv) based on the table,

TECH_COURSE given below:

Table: TECH_COURSE

 

CID CNAME FEES STARTDATE TID
C201 Animation and VFX 12000 2022-07-02 101
C202 CADD 15000 2021-11-15 NULL
C203 DCA 10000 2020-10-0 102
C204 DCA 9000 2021-09-15 104
C205

Mobile Application

Development
18000 2022-11-01 101
C206 Digital marketing 16000 2022-07-25 103

(i) SELECT DISTINCT TID FROM TECH_COURSE;

(ii) SELECT TID, COUNT(*), MIN(FEES) FROM

TECH_COURSE GROUP BY TID HAVING COUNT(TID)>1;

(iii) SELECT CNAME FROM TECH_COURSE WHERE

FEES>15000 ORDER BY CNAME;

(iv) SELECT AVG(FEES) FROM TECH_COURSE WHERE

FEES BETWEEN 15000 AND 17000;

 

Question 26(b)(i)

Answer:

Output:

DISTINCT TID

101

NULL

102

104

103

Explanation:

The SELECT statement when combined with DISTINCT clause, returns records without repetition. 

So the query will return all the values in the column TID without repetition.

Question 26(b)(ii)

Answer:

Output:

TID

COUNT(*)

MIN(FEES)

101

2

12000

Explanation:

The above query groups the records based on the value in the column TID and then checks if the count is greater than 1 . If the count is greater than 1 for any TID value, then that TID, the count and the minimum value in the column FEES for all records with that particular TID are displayed.

Question 26(b)(iii)

Answer:

Output:

CNAME

Digital marketing 

Mobile Application Development

Explanation:

The above query displays CNAME of records in the table with FEES value greater than 15000 , arranged in ascending order of CNAME .

Question 26(b)(iv)

Answer:

Output:

AVG(FEES)

15500.00

Explanation:

The above query will display the average of values in the column FEES for records with FEES values between 15000 and 17000 , both 15000 and 17000 included.

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