GATE CS 2012 Q50 — Relational Model: Relational Algebra

First, we find the relation $R = A ext{ } \cup \text{ } B$. Since union eliminates duplicate tuples, we get:

$$R = A \cup B = \begin{array}{|c|c|c|} \hline \text{ID} & \text{Name} & \text{Age} \\ \hline 12 & \text{Arun} & 60 \\ \hline 15 & \text{Shreya} & 24 \\ \hline 25 & \text{Hari} & 40 \\ \hline 98 & \text{Rohit} & 20 \\ \hline 99 & \text{Rohit} & 11 \\ \hline \end{array}$$

Relation $C$ is given as:

$$C = \begin{array}{|c|c|c|} \hline \text{ID} & \text{Phone} & \text{Area} \\ \hline 10 & 2200 & 02 \\ \hline 99 & 2100 & 01 \\ \hline \end{array}$$

The join condition is $R.\text{ID} > 40 \lor C.\text{ID} < 15$. We evaluate this condition for each pair in the Cartesian product $R \times C$:

1. For the tuple in $C$ with $\text{ID} = 10$: Since $10 < 15$, the condition $C.\text{ID} < 15$ is True. Thus, the disjunction is satisfied for all $5$ tuples of $R$. This contributes $5$ tuples to the result.
2. For the tuple in $C$ with $\text{ID} = 99$: Since $99 \not< 15$, the condition $C.\text{ID} < 15$ is False. The overall condition simplifies to $R.\text{ID} > 40$. The tuples in $R$ satisfying this are those with $\text{ID} = 98$ and $\text{ID} = 99$. This contributes $2$ tuples to the result.

Total number of tuples in the result = $5 + 2 = 7$.