Division operation in DBMS Theory - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to perform a division operation in SQL.

DBMS Theory

SELECT 10 [1] 2 AS result;

Drag options to blanks, or click blank then click option'

A+

B-

C/

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using * instead of / causes multiplication instead of division.

Using - or + will perform subtraction or addition, not division.

The division operator in SQL is /. This divides the first number by the second.

2fill in blank

medium

Complete the code to find the quotient of two columns in a table.

DBMS Theory

SELECT salary [1] bonus AS total FROM employees;

Drag options to blanks, or click blank then click option'

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using + or - will add or subtract instead of dividing.

Using * will multiply, not divide.

The division operator / divides the salary by bonus to find the quotient.

3fill in blank

hard

Fix the error in the division operation to avoid division by zero.

DBMS Theory

SELECT amount [1] CASE WHEN divisor = 0 THEN 1 ELSE divisor END AS result FROM transactions;

Drag options to blanks, or click blank then click option'

A*

B/

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using * or + will not perform division.

Not handling division by zero causes runtime errors.

The division operator / is used here with a CASE statement to avoid dividing by zero.

4fill in blank

hard

Fill both blanks to calculate the average price per unit.

DBMS Theory

SELECT total_price [1] total_units AS avg_price FROM sales WHERE total_units [2] 0;

Drag options to blanks, or click blank then click option'

A/

B>

C<

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of division for average.

Using < instead of > causes wrong filtering.

The division operator / calculates average price. The condition > 0 ensures no division by zero.

5fill in blank

hard

Fill all three blanks to select employees with salary per hour greater than 20.

DBMS Theory

SELECT name, salary [1] hours_worked AS salary_per_hour FROM employees WHERE salary [2] hours_worked [3] 20;

Drag options to blanks, or click blank then click option'

A/

B>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of division.

Mixing up comparison operators.

Use / (A for BLANK_1 and BLANK_2) to calculate salary per hour in both SELECT and WHERE clauses. Use > (B for BLANK_3) to filter employees where salary per hour is greater than 20.

Practice

1.

What does the division operation in a database primarily help to find?

easy

A. Entries that appear only once in a table

B. The sum of values in a column

C. The total count of rows in a table

D. Entries in one table related to all entries in another table

Solution

1. Step 1: Understand the division operation purpose

   The division operation is used to find records in one table that are related to every record in another table.

2. Step 2: Compare with other options

   The other options describe operations like finding unique entries, counting rows, or summing values, not the division operation.

3. Final Answer:

   Entries in one table related to all entries in another table -> Option D

4. Quick Check:

   Division operation = find all related entries [OK]

Hint: Division finds entries matching all in another set [OK]

Common Mistakes:

• Confusing division with counting or summing
• Thinking it finds entries related to some, not all
• Mixing division with join operations

2.

Which SQL keyword is commonly used to implement the division operation?

easy

A. GROUP BY

B. NOT EXISTS

C. HAVING

D. EXISTS

Solution

1. Step 1: Recall division implementation in SQL

   Division is often implemented using NOT EXISTS to check that no unmatched entries exist.

2. Step 2: Analyze other keywords

   GROUP BY groups rows, HAVING filters groups, and EXISTS checks existence but not for division logic.

3. Final Answer:

   NOT EXISTS -> Option B

4. Quick Check:

   Division uses NOT EXISTS for completeness check [OK]

Hint: Use NOT EXISTS to ensure all related entries exist [OK]

Common Mistakes:

• Using EXISTS instead of NOT EXISTS
• Confusing GROUP BY with division logic
• Assuming HAVING filters individual rows

3.

Consider two tables: Suppliers(supplier_id) and Parts(part_id), and a relation Supplies(supplier_id, part_id). Which SQL query finds suppliers who supply all parts?

SELECT supplier_id FROM Suppliers S WHERE NOT EXISTS (
  SELECT part_id FROM Parts P WHERE NOT EXISTS (
    SELECT * FROM Supplies SP WHERE SP.supplier_id = S.supplier_id AND SP.part_id = P.part_id
  )
);

medium

A. Finds parts not supplied by any supplier

B. Finds suppliers supplying no parts

C. Finds suppliers supplying all parts

D. Finds parts supplied by all suppliers

Solution

1. Step 1: Understand the nested NOT EXISTS logic

   The inner NOT EXISTS checks if a supplier does not supply a part. The outer NOT EXISTS ensures no such part exists, meaning the supplier supplies all parts.

2. Step 2: Match query purpose with options

   This query finds suppliers who supply every part, matching Finds suppliers supplying all parts.

3. Final Answer:

   Finds suppliers supplying all parts -> Option C

4. Quick Check:

   Double NOT EXISTS = all parts supplied [OK]

Hint: Double NOT EXISTS means 'for all' condition [OK]

Common Mistakes:

• Misreading NOT EXISTS as EXISTS
• Confusing suppliers with parts
• Ignoring nested query logic

4.

Identify the error in this SQL query intended to find students who have completed all courses:

SELECT student_id FROM Students S WHERE EXISTS (
  SELECT course_id FROM Courses C WHERE NOT EXISTS (
    SELECT * FROM Completed WHERE student_id = S.student_id AND course_id = C.course_id
  )
);

medium

A. Should use NOT EXISTS instead of EXISTS in outer query

B. Missing GROUP BY clause

C. Incorrect table alias usage

D. Should use INNER JOIN instead of subqueries

Solution

1. Step 1: Analyze the logic of EXISTS and NOT EXISTS

   The inner NOT EXISTS checks if a student hasn't completed a course. The outer EXISTS checks if such a course exists, meaning the student missed some courses.

2. Step 2: Correct the logic for 'all courses completed'

   To find students who completed all courses, the outer condition should be NOT EXISTS to ensure no missing courses.

3. Final Answer:

   Should use NOT EXISTS instead of EXISTS in outer query -> Option A

4. Quick Check:

   Use NOT EXISTS outside to confirm completeness [OK]

Hint: Outer NOT EXISTS ensures no missing entries [OK]

Common Mistakes:

• Using EXISTS instead of NOT EXISTS outside
• Assuming GROUP BY is needed here
• Confusing alias names

5.

You have two tables: Employees(emp_id) and Projects(proj_id), and a relation WorksOn(emp_id, proj_id). How would you find employees who work on every project that a specific manager manages, given a table Manages(manager_id, proj_id)?

hard

A. Use division operation with WorksOn and projects managed by the manager

B. Use a simple JOIN between Employees and Projects

C. Use GROUP BY on WorksOn without filtering projects

D. Use EXISTS to find employees working on any project

Solution

1. Step 1: Identify the problem as a division operation

   We want employees who work on all projects managed by a specific manager, which is a classic division problem.

2. Step 2: Apply division using WorksOn and filtered projects

   Filter projects by the manager in Manages, then find employees whose WorksOn covers all those projects.

3. Final Answer:

   Use division operation with WorksOn and projects managed by the manager -> Option A

4. Quick Check:

   Division finds employees working on all manager's projects [OK]

Hint: Filter projects by manager, then apply division on WorksOn [OK]

Common Mistakes:

• Ignoring filtering projects by manager
• Using JOIN without division logic
• Assuming GROUP BY alone solves the problem