RelationalCalculus
Question 1 
Let R and S be relational schemes such that R={a,b,c} and S={c}. Now consider the following queries on the database:
Which of the above queries are equivalent?
I and II
 
I and III
 
II and IV  
III and IV 
Question 1 Explanation:
R={a,b,c}
S={c}
It looks like Division operator. Since a division operator in E(A,B)/ P(B) will be equal to
Now replacing A=RS P=r
B=S
E=r
we will get,
equivalent to I
∴ It is equivalent to division operator.
⇒ r(RS,S)/r(S)
This logical statement means that
① Select t(RS) from r such that
② for all tuples U in S,
③ there exists a tuple V in r, such that
④ U=V[S] & t=V[RS]
A(x,y) & B(y)
A/B = {(x)  ∃(x,y)∈A(y)∈B}
which means that A/B contains all x tuples, such that for every tuple in B, there is an xy tuple in A.
So, this is just equivalent to I.
This logical statement means that
① Select t(RS) from r such that
② for all tuples V in r,
③ there exists a tuple U in r, such that
④ U=V[S] & t=V[RS]
⇒ Select (RS) values from r, where the S value is in (r/r), which will be true only if S in r is a foreign key referring to S is r.
This selects (a,b) from all tuples from r which has an equivalent value in S.
S={c}
It looks like Division operator. Since a division operator in E(A,B)/ P(B) will be equal to
Now replacing A=RS P=r
B=S
E=r
we will get,
equivalent to I
∴ It is equivalent to division operator.
⇒ r(RS,S)/r(S)
This logical statement means that
① Select t(RS) from r such that
② for all tuples U in S,
③ there exists a tuple V in r, such that
④ U=V[S] & t=V[RS]
A(x,y) & B(y)
A/B = {(x)  ∃(x,y)∈A(y)∈B}
which means that A/B contains all x tuples, such that for every tuple in B, there is an xy tuple in A.
So, this is just equivalent to I.
This logical statement means that
① Select t(RS) from r such that
② for all tuples V in r,
③ there exists a tuple U in r, such that
④ U=V[S] & t=V[RS]
⇒ Select (RS) values from r, where the S value is in (r/r), which will be true only if S in r is a foreign key referring to S is r.
This selects (a,b) from all tuples from r which has an equivalent value in S.
Question 2 
I only  
II only  
III only  
III and IV only 
Question 2 Explanation:
Demorgan law:
∀xP(x)≡∼∃x(∼P(x))
∼∀x(∼P(x))≡∃x(P(x))
Given: ∀t ∈ r(P(t)) (1)
As per Demorgan law
(1) ⇒ ∼∃t ∈ r(∼P(t))
which is option (III).
∀xP(x)≡∼∃x(∼P(x))
∼∀x(∼P(x))≡∃x(P(x))
Given: ∀t ∈ r(P(t)) (1)
As per Demorgan law
(1) ⇒ ∼∃t ∈ r(∼P(t))
which is option (III).
Question 3 
Student (schoolid, schrollno, sname, saddress)
School (schoolid, schname, schaddress, schphone)
Enrolment(schoolid schrollno, erollno, examname)
ExamResult(erollno, examname, marks)
What does the following SQL query output?
SELECT schname, COUNT (*) FROM School C, Enrolment E, ExamResult R WHERE E.schoolid = C.schoolid AND E.examname = R.examname AND E.erollno = R.erollno AND R.marks = 100 AND S.schoolid IN (SELECT schoolid FROM student GROUP BY schoolid HAVING COUNT (*) > 200) GROUP By schoolid /* Add code here. Remove these lines if not writing code */ 
The empty set  
schools with more than 35% of its students enrolled in some exam or the other  
schools with a pass percentage above 35% over all exams taken together  
schools with a pass percentage above 35% over each exam 
Question 3 Explanation:
Query having the division with
{ x  x ∈ Enrolment ∧ x . schoolid = t }  * 100 > 35 }
This is school with enrollment % is 35 or above.
{ x  x ∈ Enrolment ∧ x . schoolid = t }  * 100 > 35 }
This is school with enrollment % is 35 or above.
Question 4 
Consider the relation employee(name, sex, supervisorName) with name as the key. supervisorName gives the name of the supervisor of the employee under consideration. What does the following Tuple Relational Calculus query produce?
Names of employees with a male supervisor.
 
Names of employees with no immediate male subordinates.  
Names of employees with no immediate female subordinates.  
Names of employees with a female supervisor.

Question 4 Explanation:
The given Tuple Relational calculus produce names of employees with no immediate female subordinates.
Question 5 
Which of the following relational calculus expressions is not safe?
{t∃u ∈ R_{1} (t[A] = u[A])∧ ¬∃s ∈ R_{2} (t[A] = s[A])}  
{t∀u ∈ R_{1} (u[A]= "x" ⇒ ∃s ∈ R_{2} (t[A] = s[A] ∧ s[A] = u[A]))}  
{t¬(t ∈ R_{1})}  
{t∃u ∈ R_{1} (t[A] = u[A])∧ ∃s ∈ R_{2} (t[A] = s[A])} 
There are 5 questions to complete.