Keys and Candidate-Key Constraints for StudentName,\ StudentAge: Solving the “X cannot be …” Problem | AI Research

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

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Sep 12, 2026

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

StudentID	StudentName	StudentE-mail	StudentAge	CPI
2345	Shankar	shankar@math	X	9.4
1287	Swati	swati@ee	19	9.5
7853	Shankar	shankar@cse	19	9.4
9876	Swati	swati@mech	18	9.3
8765	Ganesh	ganesh@civil	19	8.7

We are asked: for $(StudentName,\ StudentAge)$ to be the key for this instance, the value $X$ 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 $(StudentName, StudentAge)$ pair.

Key terms:

Key
Candidate Key
Uniqueness Constraint
Functional Dependency

Key idea

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

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

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

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

A collision can only happen if:

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

No other existing tuple has $StudentName = Shankar$ .

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

Ensure $(StudentName, StudentAge)$ is a Key

1
Step 1
Extract $(StudentName, StudentAge)$ from each tuple; the only unknown is the first tuple’s age $X$ .
2
Step 2
You obtain: $(Swati,19)$ , $(Shankar,19)$ , $(Swati,18)$ , $(Ganesh,19)$ , plus $(Shankar,X)$ .
3
Step 3
A collision happens if $(Shankar,X)$ equals an existing pair.
4
Step 4
Only $(Shankar,19)$ has the same StudentName, so equality requires $X=19$ .
5
Step 5
Therefore, to keep uniqueness, $X$ should not be $19$ (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 $StudentID$ uniqueness to decide keyness. Even if $StudentID$ is unique, $(StudentName, StudentAge)$ 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 $X$ ), 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 $X$ with Existing Pair $(Shankar,19)$

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

Why other values are fine

Knowledge Check

Question 1 of 4

Q1Single choice

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

Their $(StudentName, StudentAge)$ values must be different

Their $StudentID$ values must be different

Their CPI values must be different

Their $StudentE-mail$ values must be different

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