Question

Match List-I with List-II:

List-I (Aggregate function)	List-II (Description)
(A) count(marks)	(III) Count all non null values of marks column
(B) count()	(I) Count all rows
(C) avg(marks)	(II) Finding average of non null values of marks
(D) sum(marks)	(IV) Finding sum of all marks

• A. (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
• B. (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
• C. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
• D. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Hint:

Remember: - COUNT() counts all rows. - COUNT(column) ignores NULL values. - AVG() and SUM() both ignore NULL values while performing calculations.

Answer 1

The Correct Option is A

Step 1: Clarification of Aggregate Functions.
- count(marks): Tallies non-null entries in the "marks" column, aligning with (III).
- count(): Encompasses all rows, including those with NULL values, corresponding to (I).
- avg(marks): Computes the mean of all non-null "marks" values, matching (II).
- sum(marks): Provides the total of all non-null "marks" values, corresponding to (IV).

Step 2: Association.
The accurate pairings are as follows: (A) → (III), (B) → (I), (C) → (II), (D) → (IV).
Concluding Selection: \[\boxed{\text{Option (1)}}\]