ABS and MOD functions in SQL - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ query_result

intermediate

Output of ABS function with negative input

What is the output of the following SQL query?

SELECT ABS(-15) AS result;

SQL

SELECT ABS(-15) AS result;

A 15

B -15

C 0

D NULL

💡 Hint

ABS returns the positive value of a number.

The ABS function returns the absolute value, so ABS(-15) is 15.

❓ query_result

intermediate

Output of MOD function with positive numbers

What is the output of this SQL query?

SELECT MOD(29, 5) AS remainder;

SQL

SELECT MOD(29, 5) AS remainder;

A 1

B 5

C 0

D 4

💡 Hint

MOD returns the remainder after division.

29 divided by 5 is 5 with remainder 4, so MOD(29,5) is 4.

📝 Syntax

advanced

Identify the syntax error in MOD usage

Which option contains a syntax error in using the MOD function?

A SELECT MOD(10 3) AS remainder;

B SELECT MOD(10, 3) AS remainder;

C SELECT MOD(10,3) AS remainder;

D SELECT MOD(10 % 3) AS remainder;

💡 Hint

Check the commas separating function arguments.

Option A misses the comma between arguments, causing a syntax error.

❓ query_result

advanced

Result of MOD with negative dividend

What is the output of this SQL query?

SELECT MOD(-17, 5) AS remainder;

SQL

SELECT MOD(-17, 5) AS remainder;

A -2

B 3

C -3

D 2

💡 Hint

MOD returns the remainder with the sign of the divisor in many SQL dialects.

In standard SQL, MOD(-17,5) returns 3 because -17 = ( -4 * 5 ) + 3.

🧠 Conceptual

expert

Understanding ABS and MOD combined

Given the query:

SELECT ABS(MOD(-23, 7)) AS result;

What is the value of result?

SQL

SELECT ABS(MOD(-23, 7)) AS result;

A 1

B -2

C 5

D 2

💡 Hint

Calculate MOD first, then apply ABS.

MOD(-23,7) is 5 because -23 = (-4*7) + 5. Standard SQL returns 5. ABS(5) is 5.