Keys and Candidate-Key Constraints for (StudentName, StudentAge)(StudentName,\ StudentAge): Solving the “XX cannot be …” Problem

Keys and Candidate-Key Constraints for (StudentName, StudentAge)(StudentName,\ StudentAge): Solving the “XX cannot be …” Problem

Verified Sources
Sep 12, 2026

We are given a relation Students(StudentID, StudentName, StudentE-mail, StudentAge, CPI) with tuples:

StudentIDStudentNameStudentE-mailStudentAgeCPI
2345Shankarshankar@mathX9.4
1287Swatiswati@ee199.5
7853Shankarshankar@cse199.4
9876Swatiswati@mech189.3
8765Ganeshganesh@civil198.7

We are asked: for (StudentName, StudentAge)(StudentName,\ StudentAge) to be the key for this instance, the value XX should not be equal to: (i) 18, (ii) 19, (iii) 15, (iv) 20.

A key (and specifically the combination given) must satisfy uniqueness for that attribute set: no two different rows may have the same (StudentName,StudentAge)(StudentName, StudentAge) pair.

Key terms:

  • Key
  • Candidate Key
  • Uniqueness Constraint
  • Functional Dependency

Key idea

Check the existing (StudentName,StudentAge)(StudentName, StudentAge) pairs and see when a collision occurs.

Current pairs (ignoring StudentIDStudentID and CPI for key-check):

  • Tuple 1: (Shankar,X)(Shankar, X)
  • Tuple 2: (Swati,19)(Swati, 19)
  • Tuple 3: (Shankar,19)(Shankar, 19)
  • Tuple 4: (Swati,18)(Swati, 18)
  • Tuple 5: (Ganesh,19)(Ganesh, 19)

So, to keep (StudentName,StudentAge)(StudentName, StudentAge) as a key, (Shankar,X)(Shankar, X) must not equal any other existing pair.

A collision can only happen if:

  • (Shankar,X)=(Shankar,19)(Shankar, X) = (Shankar, 19), which implies X=19X = 19.

No other existing tuple has StudentName=ShankarStudentName = Shankar.

Thus, the only forbidden value among the options is X19X \ne 19, i.e., option (ii) 19.

Ensure $(StudentName, StudentAge)$ is a Key

  1. 1
    Step 1

    Extract (StudentName,StudentAge)(StudentName, StudentAge) from each tuple; the only unknown is the first tuple’s age XX.

  2. 2
    Step 2

    You obtain: (Swati,19)(Swati,19), (Shankar,19)(Shankar,19), (Swati,18)(Swati,18), (Ganesh,19)(Ganesh,19), plus (Shankar,X)(Shankar,X).

  3. 3
    Step 3

    A collision happens if (Shankar,X)(Shankar,X) equals an existing pair.

  4. 4
    Step 4

    Only (Shankar,19)(Shankar,19) has the same StudentName, so equality requires X=19X=19.

  5. 5
    Step 5

    Therefore, to keep uniqueness, XX should not be 1919 (option (ii)).

Pro Tip

When checking if an attribute set is a key, ignore all non-key attributes and focus purely on whether the key-attribute values are unique across rows. If any two rows share the same key-attribute values, it’s not a key.

Common Mistake

Don’t use StudentIDStudentID uniqueness to decide keyness. Even if StudentIDStudentID is unique, (StudentName,StudentAge)(StudentName, StudentAge) still must be unique by its own values.

Mental Workflow for Key-Constraint Problems

Extract

Step A

Write the value pairs/tuples for the proposed key attributes."

Compare

Step B

Look for identical pairs among rows."

Solve Constraint

Step C

If an unknown exists (here XX), compute the values that would cause a collision."

Pick the Option

Step D

Select the option value that must be excluded to maintain uniqueness."

Collision Check for XX with Existing Pair (Shankar,19)(Shankar,19)

If X=19X=19, then (Shankar,X)=(Shankar,19)(Shankar,X)=(Shankar,19) collides and violates key uniqueness.

Why other values are fine

Knowledge Check

Question 1 of 4
Q1Single choice

For (StudentName,StudentAge)(StudentName, StudentAge) to be a key, what must be true for any two tuples?

Explore Related Topics

1

Evaluating ER-to-Relational Mapping Statements

The content explains how standard ER‑to‑relational mapping rules validate three statements about weak entities, partial keys, and many‑to‑many relationships, and shows that the fourth claim—every generated relation has only one candidate key—is false.

  • Weak entities depend on a strong owner and are identified by the owner’s primary key plus a partial (discriminator) key.
  • A partial key is allowed for weak entities, forming a composite primary key with the owner key.
  • Binary M:NM:N relationships are mapped to a separate associative relation containing the participating primary keys.
  • A generated relation may have multiple candidate keys; one is chosen as the primary key, so statement (iv) is false.
2

The Banker's Algorithm: Deadlock Avoidance in Operating Systems

The Banker's Algorithm is a deadlock‑avoidance method that keeps a system in a safe state by checking each resource request against the maximum declared needs of processes.

  • Maintains Available, Max, Allocation, and Need matrices, where Need[i][j]=Max[i][j]Allocation[i][j]\text{Need}[i][j]=\text{Max}[i][j]-\text{Allocation}[i][j].
  • The Safety Algorithm uses vectors Work and Finish to find an execution order; if all processes finish, the state is safe.
  • The Resource‑Request Algorithm simulates allocation, runs the safety check, and commits only if the resulting state remains safe.
  • Time complexity of the safety check is O(m×n2)O(m \times n^{2}).
  • In practice the algorithm is rarely used because processes must predeclare maximum needs and the algorithm’s overhead is high.
3

Designing a Four-Input Logic Circuit with an Equality Condition