PYQs / GATE DA / 2026 / Set 1 / Q42 GATE DA 2026 Set 1 — Question 42 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +2 / -0.67 Medium Selection & Projection Relational Algebra Database Management & Warehousing Join & Division
Database Management & Warehousing → Relational Algebra → Join & Division
Last updated 5 September 2026
Question Consider the given relations
X X X ,
Y Y Y and
Z Z Z . The relation
X X X has three columns
P P P ,
Q Q Q and
R R R . The relation
Y Y Y has three columns
P P P ,
Q Q Q and
S S S . The relation
Z Z Z has two columns
P P P and
T T T .
Relation X
Relation Y
Relation Z
Consider the relational algebra expression
Π P R S [ ( σ ( Q = Q 3 ∨ R = R 2 ) [ X ⋈ Y ] ) ⋈ ( σ ( S > 1 ) [ Y ⋈ Z ] ) ] \Pi_{PRS} [ (\sigma_{(Q=Q3 \lor R=R2)} [X \bowtie Y]) \bowtie (\sigma_{(S>1)} [Y \bowtie Z]) ] Π P R S [( σ ( Q = Q 3 ∨ R = R 2 ) [ X ⋈ Y ]) ⋈ ( σ ( S > 1 ) [ Y ⋈ Z ])] where
⋈ \bowtie ⋈ denotes natural join operation.
Which of the following options is the correct output for the given expression?
Correct answer (D) Zero rows
Solution Let's evaluate the expression step by step.
Step 1: Evaluate Left Side L = σ ( Q = Q 3 ∨ R = R 2 ) [ X ⋈ Y ] L = \sigma_{(Q=Q3 \lor R=R2)} [X \bowtie Y] L = σ ( Q = Q 3 ∨ R = R 2 ) [ X ⋈ Y ] 1. X ⋈ Y X \bowtie Y X ⋈ Y P , Q P, Q P , Q (P1, Q1) matches: X ( P 1 , Q 1 , R 1 ) X(P1, Q1, R1) X ( P 1 , Q 1 , R 1 ) Y ( P 1 , Q 1 , 2 ) → ( P 1 , Q 1 , R 1 , 2 ) Y(P1, Q1, 2) \to (P1, Q1, R1, 2) Y ( P 1 , Q 1 , 2 ) → ( P 1 , Q 1 , R 1 , 2 ) (P2, Q2) in X, but Y has (P2, Q1). No match. (P3, Q3) matches: X ( P 3 , Q 3 , R 2 ) X(P3, Q3, R2) X ( P 3 , Q 3 , R 2 ) Y ( P 3 , Q 3 , 1 ) → ( P 3 , Q 3 , R 2 , 1 ) Y(P3, Q3, 1) \to (P3, Q3, R2, 1) Y ( P 3 , Q 3 , 1 ) → ( P 3 , Q 3 , R 2 , 1 ) Result of Join: { ( P 1 , Q 1 , R 1 , 2 ) , ( P 3 , Q 3 , R 2 , 1 ) } \{ (P1, Q1, R1, 2), (P3, Q3, R2, 1) \} {( P 1 , Q 1 , R 1 , 2 ) , ( P 3 , Q 3 , R 2 , 1 )} 2. Filter
σ ( Q = Q 3 ∨ R = R 2 ) \sigma_{(Q=Q3 \lor R=R2)} σ ( Q = Q 3 ∨ R = R 2 ) :
Row ( P 1 , Q 1 , R 1 , 2 ) (P1, Q1, R1, 2) ( P 1 , Q 1 , R 1 , 2 ) Q = Q 1 ≠ Q 3 Q=Q1 \neq Q3 Q = Q 1 = Q 3 R = R 1 ≠ R 2 R=R1 \neq R2 R = R 1 = R 2 Row ( P 3 , Q 3 , R 2 , 1 ) (P3, Q3, R2, 1) ( P 3 , Q 3 , R 2 , 1 ) Q = Q 3 Q=Q3 Q = Q 3 L = { ( P 3 , Q 3 , R 2 , 1 ) } L = \{ (P3, Q3, R2, 1) \} L = {( P 3 , Q 3 , R 2 , 1 )}
Step 2: Evaluate Right Side R = σ ( S > 1 ) [ Y ⋈ Z ] R = \sigma_{(S>1)} [Y \bowtie Z] R = σ ( S > 1 ) [ Y ⋈ Z ] 1. Y ⋈ Z Y \bowtie Z Y ⋈ Z P P P P 1 P1 P 1 Y Y Y ( P 1 , Q 1 , 2 ) , ( P 1 , Q 2 , 5 ) (P1, Q1, 2), (P1, Q2, 5) ( P 1 , Q 1 , 2 ) , ( P 1 , Q 2 , 5 ) Z Z Z ( P 1 , T 1 ) (P1, T1) ( P 1 , T 1 ) Matches: ( P 1 , Q 1 , 2 , T 1 ) , ( P 1 , Q 2 , 5 , T 1 ) (P1, Q1, 2, T1), (P1, Q2, 5, T1) ( P 1 , Q 1 , 2 , T 1 ) , ( P 1 , Q 2 , 5 , T 1 ) P 2 P2 P 2 Y Y Y ( P 2 , Q 1 , 6 ) (P2, Q1, 6) ( P 2 , Q 1 , 6 ) Z Z Z P 2 P2 P 2 P 3 P3 P 3 Y Y Y ( P 3 , Q 3 , 1 ) (P3, Q3, 1) ( P 3 , Q 3 , 1 ) Z Z Z ( P 3 , T 2 ) (P3, T2) ( P 3 , T 2 ) Match: ( P 3 , Q 3 , 1 , T 2 ) (P3, Q3, 1, T2) ( P 3 , Q 3 , 1 , T 2 ) Result of Join: { ( P 1 , Q 1 , 2 , T 1 ) , ( P 1 , Q 2 , 5 , T 1 ) , ( P 3 , Q 3 , 1 , T 2 ) } \{ (P1, Q1, 2, T1), (P1, Q2, 5, T1), (P3, Q3, 1, T2) \} {( P 1 , Q 1 , 2 , T 1 ) , ( P 1 , Q 2 , 5 , T 1 ) , ( P 3 , Q 3 , 1 , T 2 )} 2. Filter
σ ( S > 1 ) \sigma_{(S>1)} σ ( S > 1 ) :
( P 1 , Q 1 , 2 , T 1 ) (P1, Q1, 2, T1) ( P 1 , Q 1 , 2 , T 1 ) S = 2 > 1 S=2 > 1 S = 2 > 1 ( P 1 , Q 2 , 5 , T 1 ) (P1, Q2, 5, T1) ( P 1 , Q 2 , 5 , T 1 ) S = 5 > 1 S=5 > 1 S = 5 > 1 ( P 3 , Q 3 , 1 , T 2 ) (P3, Q3, 1, T2) ( P 3 , Q 3 , 1 , T 2 ) S = 1 ≯ 1 S=1 \ngtr 1 S = 1 ≯ 1 R = { ( P 1 , Q 1 , 2 , T 1 ) , ( P 1 , Q 2 , 5 , T 1 ) } R = \{ (P1, Q1, 2, T1), (P1, Q2, 5, T1) \} R = {( P 1 , Q 1 , 2 , T 1 ) , ( P 1 , Q 2 , 5 , T 1 )}
Step 3: Final Join L ⋈ R L \bowtie R L ⋈ R L = { ( P 3 , Q 3 , R 2 , 1 ) } L = \{ (P3, Q3, R2, 1) \} L = {( P 3 , Q 3 , R 2 , 1 )} R = { ( P 1 , Q 1 , 2 , T 1 ) , ( P 1 , Q 2 , 5 , T 1 ) } R = \{ (P1, Q1, 2, T1), (P1, Q2, 5, T1) \} R = {( P 1 , Q 1 , 2 , T 1 ) , ( P 1 , Q 2 , 5 , T 1 )} Natural Join on common attributes P , Q , S P, Q, S P , Q , S Row in L L L P = P 3 P=P3 P = P 3 R R R P = P 1 P=P1 P = P 1 No intersection. The result is empty.
Output: Zero rows.
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Correct answer (D) Zero rows
Solution Let's evaluate the expression step by step.
Step 1: Evaluate Left Side L = σ ( Q = Q 3 ∨ R = R 2 ) [ X ⋈ Y ] L = \sigma_{(Q=Q3 \lor R=R2)} [X \bowtie Y] L = σ ( Q = Q 3 ∨ R = R 2 ) [ X ⋈ Y ] 1. X ⋈ Y X \bowtie Y X ⋈ Y P , Q P, Q P , Q (P1, Q1) matches: X ( P 1 , Q 1 , R 1 ) X(P1, Q1, R1) X ( P 1 , Q 1 , R 1 ) Y ( P 1 , Q 1 , 2 ) → ( P 1 , Q 1 , R 1 , 2 ) Y(P1, Q1, 2) \to (P1, Q1, R1, 2) Y ( P 1 , Q 1 , 2 ) → ( P 1 , Q 1 , R 1 , 2 ) (P2, Q2) in X, but Y has (P2, Q1). No match. (P3, Q3) matches: X ( P 3 , Q 3 , R 2 ) X(P3, Q3, R2) X ( P 3 , Q 3 , R 2 ) Y ( P 3 , Q 3 , 1 ) → ( P 3 , Q 3 , R 2 , 1 ) Y(P3, Q3, 1) \to (P3, Q3, R2, 1) Y ( P 3 , Q 3 , 1 ) → ( P 3 , Q 3 , R 2 , 1 ) Result of Join: { ( P 1 , Q 1 , R 1 , 2 ) , ( P 3 , Q 3 , R 2 , 1 ) } \{ (P1, Q1, R1, 2), (P3, Q3, R2, 1) \} {( P 1 , Q 1 , R 1 , 2 ) , ( P 3 , Q 3 , R 2 , 1 )} 2. Filter
σ ( Q = Q 3 ∨ R = R 2 ) \sigma_{(Q=Q3 \lor R=R2)} σ ( Q = Q 3 ∨ R = R 2 ) :
Row ( P 1 , Q 1 , R 1 , 2 ) (P1, Q1, R1, 2) ( P 1 , Q 1 , R 1 , 2 ) Q = Q 1 ≠ Q 3 Q=Q1 \neq Q3 Q = Q 1 = Q 3 R = R 1 ≠ R 2 R=R1 \neq R2 R = R 1 = R 2 Row ( P 3 , Q 3 , R 2 , 1 ) (P3, Q3, R2, 1) ( P 3 , Q 3 , R 2 , 1 ) Q = Q 3 Q=Q3 Q = Q 3 L = { ( P 3 , Q 3 , R 2 , 1 ) } L = \{ (P3, Q3, R2, 1) \} L = {( P 3 , Q 3 , R 2 , 1 )}
Step 2: Evaluate Right Side R = σ ( S > 1 ) [ Y ⋈ Z ] R = \sigma_{(S>1)} [Y \bowtie Z] R = σ ( S > 1 ) [ Y ⋈ Z ] 1. Y ⋈ Z Y \bowtie Z Y ⋈ Z P P P P 1 P1 P 1 Y Y Y ( P 1 , Q 1 , 2 ) , ( P 1 , Q 2 , 5 ) (P1, Q1, 2), (P1, Q2, 5) ( P 1 , Q 1 , 2 ) , ( P 1 , Q 2 , 5 ) Z Z Z ( P 1 , T 1 ) (P1, T1) ( P 1 , T 1 ) Matches: ( P 1 , Q 1 , 2 , T 1 ) , ( P 1 , Q 2 , 5 , T 1 ) (P1, Q1, 2, T1), (P1, Q2, 5, T1) ( P 1 , Q 1 , 2 , T 1 ) , ( P 1 , Q 2 , 5 , T 1 ) P 2 P2 P 2 Y Y Y ( P 2 , Q 1 , 6 ) (P2, Q1, 6) ( P 2 , Q 1 , 6 ) Z Z Z P 2 P2 P 2 P 3 P3 P 3 Y Y Y ( P 3 , Q 3 , 1 ) (P3, Q3, 1) ( P 3 , Q 3 , 1 ) Z Z Z ( P 3 , T 2 ) (P3, T2) ( P 3 , T 2 ) Match: ( P 3 , Q 3 , 1 , T 2 ) (P3, Q3, 1, T2) ( P 3 , Q 3 , 1 , T 2 ) Result of Join: { ( P 1 , Q 1 , 2 , T 1 ) , ( P 1 , Q 2 , 5 , T 1 ) , ( P 3 , Q 3 , 1 , T 2 ) } \{ (P1, Q1, 2, T1), (P1, Q2, 5, T1), (P3, Q3, 1, T2) \} {( P 1 , Q 1 , 2 , T 1 ) , ( P 1 , Q 2 , 5 , T 1 ) , ( P 3 , Q 3 , 1 , T 2 )} 2. Filter
σ ( S > 1 ) \sigma_{(S>1)} σ ( S > 1 ) :
( P 1 , Q 1 , 2 , T 1 ) (P1, Q1, 2, T1) ( P 1 , Q 1 , 2 , T 1 ) S = 2 > 1 S=2 > 1 S = 2 > 1 ( P 1 , Q 2 , 5 , T 1 ) (P1, Q2, 5, T1) ( P 1 , Q 2 , 5 , T 1 ) S = 5 > 1 S=5 > 1 S = 5 > 1 ( P 3 , Q 3 , 1 , T 2 ) (P3, Q3, 1, T2) ( P 3 , Q 3 , 1 , T 2 ) S = 1 ≯ 1 S=1 \ngtr 1 S = 1 ≯ 1 R = { ( P 1 , Q 1 , 2 , T 1 ) , ( P 1 , Q 2 , 5 , T 1 ) } R = \{ (P1, Q1, 2, T1), (P1, Q2, 5, T1) \} R = {( P 1 , Q 1 , 2 , T 1 ) , ( P 1 , Q 2 , 5 , T 1 )}
Step 3: Final Join L ⋈ R L \bowtie R L ⋈ R L = { ( P 3 , Q 3 , R 2 , 1 ) } L = \{ (P3, Q3, R2, 1) \} L = {( P 3 , Q 3 , R 2 , 1 )} R = { ( P 1 , Q 1 , 2 , T 1 ) , ( P 1 , Q 2 , 5 , T 1 ) } R = \{ (P1, Q1, 2, T1), (P1, Q2, 5, T1) \} R = {( P 1 , Q 1 , 2 , T 1 ) , ( P 1 , Q 2 , 5 , T 1 )} Natural Join on common attributes P , Q , S P, Q, S P , Q , S Row in L L L P = P 3 P=P3 P = P 3 R R R P = P 1 P=P1 P = P 1 No intersection. The result is empty.
Output: Zero rows.
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