Consider the following relations $A$, $B$ and $C$: How many tuples does the result of the following relational algebra expression contain? Assume that the schema of $A \cup B$ is the same as that of $A$. $(A \cup B) \bowtie_{A.\text{Id} > 40 \lor C.\text{Id} < 15} C$

1. Find $A \cup B$: Relational algebra operations work on sets, so duplicates are eliminated.

   • Tuples in $A$: $(12, \text{Arun}, 60)$, $(15, \text{Shreya}, 24)$, $(99, \text{Rohit}, 11)$

   • Tuples in $B$: $(15, \text{Shreya}, 24)$, $(25, \text{Hari}, 40)$, $(98, \text{Rohit}, 20)$, $(99, \text{Rohit}, 11)$

   • Set union $A \cup B$ contains $5$ unique tuples:

     1. $(12, \text{Arun}, 60)$

     2. $(15, \text{Shreya}, 24)$

     3. $(25, \text{Hari}, 40)$

     4. $(98, \text{Rohit}, 20)$

     5. $(99, \text{Rohit}, 11)$

2. Tuples in $C$:

   • $(10, 2200, 02)$

   • $(99, 2100, 01)$

   Total $2$ tuples.

3. Evaluate the Theta-Join Condition $(A.\text{Id} > 40 \lor C.\text{Id} < 15)$:

   • For tuple with $C.\text{Id} = 10$ in $C$: Since $C.\text{Id} = 10 < 15$ is true, the OR condition is satisfied for all $5$ tuples of $A \cup B$. (Yields $5$ tuples)

   • For tuple with $C.\text{Id} = 99$ in $C$: Here $C.\text{Id} < 15$ is false, so we need $A.\text{Id} > 40$. The tuples from $A \cup B$ satisfying this are $(98, \text{Rohit}, 20)$ and $(99, \text{Rohit}, 11)$. (Yields $2$ tuples)

4. Total tuples in the result:

   $$5 + 2 = 7$$

Hence, the correct option is (A).