lossless decomposition and dependency-preserving decompositions in relational schemas (DBMS / GATE style).
Each question has four options (A–D), the correct answer, and a short step-by-step explanation (concise and exam-ready). Attribute names and numeric indices are varied across questions.
Questions 1–100
Q1.
Relation R(A,B,C) with F = {A → B}. Decompose into R1(A,B) and R2(A,C). Is the decomposition lossless and dependency-preserving?
A. Lossless & dependency-preserving
B. Lossless but not dependency-preserving
C. Dependency-preserving but not lossless
D. Neither
✅ Answer: A
Solution: Intersection = {A}. A → B means {A} → R1 so (R1∩R2)→R1 holds ⇒ lossless. Each FD (A→B) is in R1 ⇒ dependency preserved.
Q2.R(A,B,C) with F = {A → B}. Decompose into R1(A,C) and R2(B,C). Is it lossless and dependency-preserving?
A. Lossless & dependency-preserving
B. Lossless only
C. Dependency-preserving only
D. Neither
✅ Answer: D
Solution: Intersection = {C}. No FD with C on LHS ⇒ not lossless. A→B not contained in projections (neither R1 nor R2 has both A and B together) ⇒ not preserved.
Q3.R(A,B,C,D) with F = {A → B, B → C}. Decompose into R1(A,B) and R2(B,C,D). Lossless & dependency preserving?
A. Both yes
B. Lossless only
C. Dependency-preserving only
D. Neither
✅ Answer: A
Solution: Intersection {B}; B → C,D? B→C from F and B present in R2 LHS — (R1∩R2)={B}→R2? Actually need intersection → R1 or → R2. B → C exists (in R2) so {B} → R2’s attributes? Not fully; but for lossless we check (R1∩R2) → R1 or R2: {B} → B (trivial) insufficient. Better: use chase — since AB in R1 and B in R2 with B→C, decomposition is lossless because B is key for R2? Wait careful. Simpler view: Intersection {B}. If {B} → attributes of R1 or R2, we need {B} → (A,B) or {B} → (B,C,D). From given F, B→C but not B→D or B→A. So not guaranteed lossless. However AB is a key of R1? No. So decomposition is not necessarily lossless. Dependency preservation: A→B in R1, B→C in R2 ⇒ both preserved. So correct is C (dependency-preserving only).
Corrected Answer: C
Q4.R(A,B,C) with F = {AB → C}. Decompose into R1(A,B) and R2(A,C). Lossless & dependency-preserving?
A. Both yes
B. Lossless only
C. Dependency-preserving only
D. Neither
✅ Answer: A
Solution: Intersection {A}. Does {A} → R1 or R2? Not directly, but since AB→C and R1 contains AB, R2 contains AC; use chase: start row vectors, AB→C allows propagation ⇒ decomposition is lossless. AB→C is preserved (R1 has AB and R2 has AC — projection of AB→C appears as AB→C? Actually AB→C is present in R1∪R2 via R1 and R2 combined) ⇒ dependency preserved.
Q5.R(A,B,C,D) with F = {A → B, C → D}. Decompose into R1(A,B) and R2(C,D). Lossless & dependency-preserving?
A. Both yes
B. Lossless only
C. Dependency-preserving only
D. Neither
✅ Answer: A
Solution: The decomposed relations separate independent FDs; each FD preserved in respective relation; decomposition is lossless because original can be recovered by natural join (they share no attributes but decomposition is fine if original relation is cross product? Actually decomposition into disjoint attribute sets is lossless only when one relation contains a key — here none. But since no common attribute, lossless means R = R1 ⋈ R2 only if original had no cross constraints; standard test: for two-way decomposition with empty intersection, join is lossless iff one projection functionally determines the other — false. So actually decomposition is lossless iff R1∩R2 functionally determines R1 or R2. Here intersection empty ⇒ trivial? With empty intersection, natural join of projections will produce Cartesian product; unless constraints ensure uniqueness, lossless is false. But typical textbooks say decomposition into disjoint sets is lossless only if one side is empty? So answer should be C (dependency-preserving only).
Corrected Answer: C
Q6.R(A,B,C) with F = {A → B, B → A}. Decompose into R1(A,B) and R2(B,C). Lossless & dependency-preserving?
A. Both yes
B. Lossless only
C. Dependency-preserving only
D. Neither
✅ Answer: A
Solution: Intersection {B}. B ↔ A implies B→A so {B}→R1 holds ⇒ lossless. A↔B FDs available across R1 ⇒ preserved.
Q7.R(A,B,C) with F = {A → BC}. Decompose into R1(A,B) and R2(A,C). Lossless & dependency-preserving?
A. Both yes
B. Lossless only
C. Dependency-preserving only
D. Neither
✅ Answer: A
Solution: Intersection {A}; A→B and A→C mean {A}→R1 and {A}→R2 ⇒ lossless. Both FDs reside in decomposed relations ⇒ preserved.
Q8.R(A,B,C,D) with F = {AB → C, C → D}. Decompose into R1(A,B,C) and R2(C,D). Lossless & dependency-preserving?
A. Both yes
B. Lossless only
C. Dependency-preserving only
D. Neither
✅ Answer: A
Solution: Intersection {C} and C→D ⇒ {C}→R2 so lossless. AB→C in R1 and C→D in R2 ⇒ both preserved.
Q9.R(A,B,C) with F = {B → C}. Decompose into R1(A,B) and R2(B,C). Lossless & dependency-preserving?
A. Both yes
B. Lossless only
C. Dependency-preserving only
D. Neither
✅ Answer: A
Solution: Intersection {B}; B→C, so {B}→R2 holds ⇒ lossless. B→C is in R2 ⇒ preserved.
Q10.R(A,B,C,D) with F = {A → B, B → C}. Decompose into R1(A,B) and R2(C,D). Lossless & dependency-preserving?
A. Both yes
B. Lossless only
C. Dependency-preserving only
D. Neither
✅ Answer: D
Solution: Intersection empty ⇒ not lossless. A→B and B→C not preserved (B→C’s attributes C and B not together in any decomposed relation) ⇒ not preserved.
Q11.R(A,B,C) with F = {A → B}. Decompose into R1(A,B,C) and R2(B). Is decomposition lossless?
A. Yes
B. No
C. Depends on additional FDs
D. Cannot determine
✅ Answer: A
Solution: One component R1 is whole schema; decomposition trivial (R2 subset) so lossless trivially and FD preserved.
Q12.R(A,B,C) with F = {A → B}. Decompose into R1(B,C) and R2(A,C). Is A → B preserved?
A. Yes
B. No
C. Yes only if C→A
D. Yes only if B→C
✅ Answer: B
Solution: A→B requires A and B together in some relation — not present; cannot derive from projections ⇒ not preserved.
Q13.R(A,B,C,D) with F = {A → B, B → C, C → D}. Decompose into R1(A,B) and R2(B,C) and R3(C,D). Is decomposition dependency-preserving?
A. Yes
B. No
C. Yes for some FDs only
D. Only if A is a key
✅ Answer: A
Solution: Each FD appears in some component: A→B in R1, B→C in R2, C→D in R3 ⇒ preserved.
Q14.R(A,B,C) with F = {A → B, B → C}. Decompose into R1(A,C) and R2(B,C). Is decomposition lossless?
A. Yes
B. No
C. Only if A→C holds
D. Only if B is key
✅ Answer: A
Solution: Intersection = {C}. Check if {C}→R1 or R2: from B→C we cannot get C→B; but use chase: since B→C and R2 contains BC, and R1 contains AC; chase shows A determines C via transitivity A→B→C, so join recovers original ⇒ lossless. (Alternate: because A→C (from A→B and B→C) gives {A}→C present in R1 making join lossless)
Q15.R(A,B,C,D) with F = {A → B, C → D}. Decompose into R1(A,C,B) and R2(C,D). Is decomposition lossless?
A. Yes
B. No
C. Only if A→C
D. Only if B→C
✅ Answer: A
Solution: Intersection {C,B}? Actually R1∩R2 = {C}. C→D in F ⇒ {C}→R2 ⇒ lossless.
Q16.R(A,B,C) with F = {A → B}. Decompose into R1(A) and R2(B,C). Is decomposition dependency-preserving?
A. Yes
B. No
C. Only if A→C
D. Only if B→A
✅ Answer: B
Solution: A→B needs A and B together in some relation — not present ⇒ not preserved.
Q17.R(A,B,C,D) with F = {AB → C, C → D}. Decompose into R1(A,B) and R2(B,C,D). Is decomposition lossless?
A. Yes
B. No
C. Only if B → D
D. Only if A → D
✅ Answer: A
Solution: Intersection {B}. Need {B}→R1 or {B}→R2. From AB→C and C→D, may not have B→… but chase shows AB in R1 provides C in R2 via AB→C when joining, but formal: since R2 contains BC and C→D, AB→C exists in R1 => AB determines C in R2 when joined — standard result: because AB is key for whole R? Actually AB→C and together AB in R1 → combining yields lossless. So yes lossless.
Q18.R(A,B,C,D) with F = {A → C, B → D}. Decompose into R1(A,B,C) and R2(B,D). Dependency-preserving?
A. Yes
B. No
C. Only if C→D
D. Only if A→B
✅ Answer: A
Solution: A→C in R1, B→D in R2 ⇒ both preserved.
Q19.R(A,B,C) with F = {A → B}. Decompose into R1(A,B) and R2(A). Is it lossless?
A. Yes
B. No
C. Only if B→A
D. Cannot determine
✅ Answer: A
Solution: R2 is subset of R1 projecting on A; trivial lossless.
Q20.R(A,B,C,D) with F = {A → B, B → C}. Decompose into R1(A,B,C) and R2(A,D). Lossless & dependency-preserving?
A. Lossless & preserved
B. Lossless only
C. Preserved only
D. Neither
✅ Answer: C
Solution: A→B and B→C both in R1 so preserved. Intersection {A} -> R1? A→B exists but does {A}→(A,B,C)? Not {A}→C unless transitive considered; but since R1 contains A,B,C and F implies A→C, {A}→R1 holds ⇒ lossless actually. On careful check, A→B and B→C implies A→C so {A}→R1 true ⇒ decomposition is lossless and preserved.
Corrected Answer: A
Q21.R(A,B,C,D) with F = {A → B, C → D}. Decompose into R1(A,C) and R2(B,D). Is decomposition dependency-preserving?
A. Yes
B. No
C. Only if A→D
D. Only if C→B
✅ Answer: B
Solution: A→B not preserved (A and B not together), C→D not preserved (C and D not together) ⇒ not preserved.
Q22.R(A,B,C) with F = {A→B, A→C}. Decompose into R1(A,B) and R2(A,C). Lossless & preserved?
A. Both yes
B. Lossless only
C. Preserved only
D. Neither
✅ Answer: A
Solution: Intersection {A} and A→B,A→C ⇒ {A}→R1 and R2 ⇒ lossless; both FDs present in decompositions ⇒ preserved.
Q23.R(A,B,C) with F = {A→B, B→C}. Decompose into R1(A,B) and R2(A,C). Is decomposition dependency-preserving?
A. Yes
B. No
C. Only if B→A
D. Only if C→B
✅ Answer: A
Solution: A→B in R1, B→C can be checked? B→C not in any component directly, but A→C is in R2 and since A→B in R1 and A→C in R2, original B→C can be inferred in F+? However dependency preservation requires union of projections imply original FD. Here B→C may not be preserved directly — but because A→B and A→C exist, B→C is not necessarily derivable. So preserve? Actually B→C is lost. So answer is B.
Corrected Answer: B
Q24.R(A,B,C,D) with F = {A→B, B→C}. Decompose into R1(A,B) and R2(C,D). Is decomposition dependency preserving?
A. Yes
B. No
C. Only if B→D
D. Only if A→C
✅ Answer: B
Solution: B→C requires B and C together; not present in any relation, so not preserved.
Q25.R(A,B,C,D) with F = {AB→C, C→A}. Decompose into R1(A,B) and R2(C,A). Is decomposition lossless?
A. Yes
B. No
C. Only if B→C
D. Only if A→B
✅ Answer: A
Solution: Intersection {A}. Since C→A, {A} is determined by C hence {A}→R2? Use property: C→A implies {C}→(C,A) so join is lossless; equivalently C in R2 determines A in intersection ⇒ lossless.
Section 2 — Questions 26–50
Q26.R(A,B,C,D) with F = {A→B, B→C}. Decompose into R1(A,B) and R2(A,C,D). Is decomposition dependency-preserving?
A. Yes
B. No
C. Only if C→D
D. Only if A→D
✅ Answer: A
Solution: A→B in R1; B→C in R2? R2 has A,C,D; B absent — but A→C (from A→B and B→C) is in R2 so B→C not preserved; however A→B is preserved, but original B→C lost → not fully preserved. So correct answer B.
Corrected Answer: B
Q27.R(A,B,C) with F = {A→B}. Decompose into R1(A,B) and R2(B). Is dependency preserved?
A. Yes
B. No
C. Only if B→A
D. Only if A→B holds in R2
✅ Answer: A
Solution: A→B is in R1 ⇒ preserved.
Q28.R(A,B,C,D) with F = {A→B, B→C}. Decompose into R1(A,B) and R2(B,C) and R3(B,D). Is decomposition lossless?
A. Yes
B. No
C. Only with additional FD B→D
D. Only if A→D
✅ Answer: A
Solution: Pairwise join of relations sharing B and having B→C ensures lossless when B forms intersection and determines others across components.
Q29.R(A,B,C,D) with F = {A→B, A→C}. Decompose into R1(A,B) and R2(A,C,D). Is decomposition lossless?
A. Yes
B. No
C. Only if A→D
D. Only if B→D
✅ Answer: A
Solution: Intersection {A}; A→B and A→C ⇒ {A}→R1 and {A}→R2 ⇒ lossless.
Q30.R(A,B,C) with F = {A→B, B→A}. Decompose into R1(A,C) and R2(B,C). Is decomposition dependency-preserving?
A. Yes
B. No
C. Only if C→A
D. Only if A→C
✅ Answer: A
Solution: A↔B means A→B and B→A; neither FD needs both A and B in same relation for preservation? Actually A→B requires A and B together somewhere: neither R1 nor R2 contains A and B together, so A→B not directly preserved. But because B→A is symmetric, again not preserved. So answer is B.
Corrected Answer: B
Q31.R(A,B,C,D) with F = {AB→C, C→D}. Decompose into R1(A,B) and R2(A,C,D). Is AB→C preserved?
A. Yes
B. No
C. Only if A→C
D. Only if B→D
✅ Answer: A
Solution: AB present in R1, but C in R2; AB→C becomes preserved when reconstructing using join? Dependency preservation requires AB→C to be in projection onto some relation; AB not with C in same relation ⇒ not preserved directly. So answer B.
Corrected Answer: B
Q32.R(A,B,C,D) with F = {A→B, B→C}. Decompose into R1(A,B) and R2(C,D) and R3(B,C). Is decomposition dependency-preserving?
A. Yes
B. No
C. Only if A→D
D. Only if D→B
✅ Answer: A
Solution: A→B in R1, B→C in R3 ⇒ preserved.
Q33.R(A,B,C) with F = {A→B}. Decompose into R1(A,B) and R2(A,C). Is A→B preserved?
A. Yes
B. No
C. Only if C→B
D. Only if B→A
✅ Answer: A
Solution: A→B resides in R1 ⇒ preserved.
Q34.R(A,B,C) with F = {AB→C}. Decompose into R1(A,C) and R2(B,C). Is decomposition lossless?
A. Yes
B. No
C. Only if A→B
D. Only if C→A
✅ Answer: B
Solution: Intersection {C}; C→? No FD with C on LHS ⇒ not lossless.
Q35.R(A,B,C,D) with F = {A→BC}. Decompose into R1(A,B) and R2(A,C,D). Is decomposition dependency-preserving?
A. Yes
B. No
C. Only if A→D
D. Only if C→D
✅ Answer: A
Solution: A→B in R1; A→C in R2; original A→BC preserved.
Q36.R(A,B,C) with F = {A→B}. If we decompose into R1(A,B) and R2(A), what is the nature of decomposition?
A. Lossless and preserved
B. Lossless only
C. Preserved only
D. Neither
✅ Answer: A
Solution: R2 subset relation ⇒ trivial lossless and A→B in R1 ⇒ preserved.
Q37.R(A,B,C,D) with F = {B→C}. Decompose into R1(A,B) and R2(B,C) and R3(B,D). Is decomposition lossless?
A. Yes
B. No
C. Only if B→D
D. Only if C→D
✅ Answer: A
Solution: All components share B and B→C, so intersection B determines components ⇒ lossless.
Q38.R(A,B,C) with F = {A→B, B→C}. Decompose into R1(A,B) and R2(B,C). Are all original FDs preserved?
A. Yes
B. No
C. Only A→C preserved
D. Only B→C preserved
✅ Answer: A
Solution: A→B in R1, B→C in R2, A→C derivable but original FDs are preserved.
Q39.R(A,B,C,D) with F = {A→B, B→C}. Decompose into R1(A,B,C) and R2(A,D). Is decomposition lossless?
A. Yes
B. No
C. Only if A→D
D. Only if C→D
✅ Answer: A
Solution: R1 includes A,B,C and F implies A→C, so {A}→R1 holds ⇒ lossless.
Q40.R(A,B,C) with F = {A→B}. Decompose into R1(B,C) and R2(A,B). Is the decomposition dependency-preserving?
A. Yes
B. No
C. Only if B→C
D. Only if C→A
✅ Answer: A
Solution: A→B present in R2 ⇒ preserved.
Q41.R(A,B,C) with F = {A→B, B→C}. Decompose into R1(A,B) and R2(A,C). Is A→C preserved?
A. Yes (directly)
B. Yes (via inference)
C. No
D. Only if B→A
✅ Answer: B
Solution: A→C is not directly in any relation but follows from A→B and B→C preserved across relations, so A→C ∈ F+ (preserved by inference).
Q42.R(A,B,C,D) with F = {AB→C, C→D}. Decompose into R1(A,B,C) and R2(A,D). Is decomposition dependency-preserving?
A. Yes
B. No
C. Only if C→A
D. Only if B→D
✅ Answer: B
Solution: AB→C is in R1, C→D is in R2? R2 has AD only; C→D is not in any projection ⇒ not preserved.
Q43.R(A,B,C) with F = {B→A}. Decompose into R1(A,B) and R2(B,C). Lossless?
A. Yes
B. No
C. Only if A→C
D. Only if C→A
✅ Answer: A
Solution: Intersection {B}; B→A ⇒ {B}→R1 ⇒ lossless.
Q44.R(A,B,C) with F = {A→BC}. Decompose into R1(A,B) and R2(B,C). Dependency preservation?
A. Yes
B. No
C. Only if B→A
D. Only if C→A
✅ Answer: B
Solution: A→BC requires A and both B and C together; none of the relations contain both A and C ⇒ A→C lost ⇒ not preserved.
Q45.R(A,B,C,D) with F = {C→D}. Decompose into R1(A,B) and R2(B,C,D). Is decomposition lossless?
A. Yes
B. No
C. Only if B→C
D. Only if C→B
✅ Answer: A
Solution: Intersection {B}. No FD with B→… but since R2 includes C→D, intersection doesn’t determine R1 or R2; however since R1 is AB, R2 is BCD, we need {B}→R1 or {B}→R2. No such FD ⇒ not lossless. So correct answer B.
Corrected Answer: B
Q46.R(A,B,C) with F = {A→B}. Decompose into R1(A,C) and R2(B,C). Is A→B preserved?
A. Yes
B. No
C. Only if C→A
D. Only if B→A
✅ Answer: B
Solution: A and B not together in any relation; A→B not preserved.
Q47.R(A,B,C,D) with F = {A→B, B→C}. Decompose into R1(A,B) and R2(B,C) and R3(A,D). Is original F preserved?
A. Yes
B. No
C. Only some FDs preserved
D. Only if A→D
✅ Answer: A
Solution: A→B in R1; B→C in R2 ⇒ preserved.
Q48.R(A,B,C) with F = {A→B}. Decompose into R1(A,B) and R2(C). Is decomposition lossless?
A. Yes
B. No
C. Only if A→C
D. Only if B→C
✅ Answer: A
Solution: R2 is single attribute; R1 contains A,B; since R2 shares nothing, lossless? Trivial decomposition where R2 is independent: natural join of R1 and R2 reconstructs only when original allowed — typically not lossless unless R2 attributes determined by R1. With no dependency, join is Cartesian product ⇒ not lossless. So correct answer B.
Corrected Answer: B
Q49.R(A,B,C,D) with F = {A→B, C→D}. Decompose into R1(A,B) and R2(B,C,D). Is decomposition lossless?
A. Yes
B. No
C. Only if B→C
D. Only if A→D
✅ Answer: B
Solution: Intersection {B} and we have no FD with B on LHS determining either side ⇒ not lossless.
Q50.R(A,B,C) with F = {A→B, B→C}. Decompose into R1(A,C) and R2(B,C). Is B→C preserved?
A. Yes
B. No
C. Only if A→B
D. Only if C→B
✅ Answer: A
Solution: B,C appear together in R2 so B→C preserved.
Section 3 — Questions 51–75
Q51.R(A,B,C,D) with F = {A→B, B→C}. Decompose into R1(A,B) and R2(A,C) and R3(A,D). Is the decomposition lossless?
A. Yes
B. No
C. Only if A is key
D. Only if C→D
✅ Answer: A
Solution: Each component contains A; A→B and A→C imply A determines components ⇒ lossless.
Q52.R(A,B,C) with F = {A→B}. Decompose into R1(A,B) and R2(A,C). Is decomposition dependency preserving?
A. Yes
B. No
C. Only if B→C
D. Only if C→B
✅ Answer: A
Solution: A→B present in R1 ⇒ preserved.
Q53.R(A,B,C) with F = {B→A}. Decompose into R1(A,B) and R2(A,C). Is B→A preserved?
A. Yes
B. No
C. Only if A→C
D. Only if C→B
✅ Answer: A
Solution: B→A in R1 (A,B together) ⇒ preserved.
Q54.R(A,B,C,D) with F = {A→B, D→C}. Decompose into R1(A,B) and R2(D,C) and R3(A,D). Is the decomposition dependency preserving?
A. Yes
B. No
C. Only if A→C
D. Only if D→B
✅ Answer: A
Solution: A→B in R1, D→C in R2 ⇒ preserved.
Q55.R(A,B,C) with F = {A→BC}. Decompose into R1(A,B) and R2(B,C). Is decomposition lossless?
A. Yes
B. No
C. Only if B→A
D. Only if C→A
✅ Answer: B
Solution: Intersection {B}; need B→… but no B→A ⇒ not lossless.
Q56.R(A,B,C,D) with F = {AB→C, C→D}. Decompose into R1(A,B) and R2(B,C) and R3(C,D). Are FDs preserved?
A. Yes
B. No
C. Only AB→C preserved
D. Only C→D preserved
✅ Answer: A
Solution: AB→C in R1∪R2? AB present in R1; C in R3 — but AB→C is preserved in R1? R1 has AB but not C. However R2 has BC; AB→C not directly in any single relation — but union of projections may imply it. Typically such decomposition preserves AB→C via R1 and R2 combined — but strict dependence preservation requires each original FD appear in some projected FDs; AB→C is not in any single component ⇒ not preserved. So answer B.
Corrected Answer: B
Q57.R(A,B,C,D) with F = {A→B, B→C}. Decompose into R1(A,B) and R2(B,C) and R3(A,D). Is AB→C preserved?
A. Yes
B. No
C. Only if A→C
D. Only if D→C
✅ Answer: B
Solution: AB→C not in any single relation; original AB→C is derivable from A→B and B→C but AB→C may not be required — question ambiguous. Conservative answer: original FDs are preserved (A→B and B→C present) so AB→C ∈ F+ but projection may not contain AB→C ⇒ not directly preserved.
Q58.R(A,B,C) with F = {A→B, B→C}. Decompose into R1(A,C) and R2(B,C). Is decomposition lossless?
A. Yes
B. No
C. Only if A→C
D. Only if C→B
✅ Answer: A
Solution: A→C via transitivity exists and R1 has A,C ⇒ {A}→R1 so lossless.
Q59.R(A,B,C) with F = {A→B}. Decompose into R1(A,C) and R2(B,C). Are all FDs preserved?
A. No
B. Yes
C. Only if C→A
D. Only if B→A
✅ Answer: A
Solution: A→B not in any component ⇒ not preserved.
Q60.R(A,B,C,D) with F = {A→B, B→C, C→D}. Decompose into R1(A,B) and R2(B,C) and R3(C,D). Is decomposition dependency preserving?
A. Yes
B. No
C. Only if A→D
D. Only if B→D
✅ Answer: A
Solution: Each FD appears in some component ⇒ preserved.
Q61.R(A,B,C) with F = {A→B, A→C}. Decompose into R1(A,B) and R2(B,C). Is decomposition dependency preserving?
A. No
B. Yes
C. Only if B→A
D. Only if C→A
✅ Answer: A
Solution: A→C not present in any component (A and C not together) ⇒ not preserved.
Q62.R(A,B,C,D) with F = {A→B, B→C, D→A}. Decompose into R1(A,B,C) and R2(D,A). Is decomposition lossless?
A. Yes
B. No
C. Only if D→C
D. Only if A→D
✅ Answer: A
Solution: Intersection {A}; D→A ⇒ {A}? Wait {A} is in intersection and D→A gives D->A but that is not {A}->… For two-way decomposition R1(A,B,C) and R2(D,A), intersection = {A}. If {A}→R1 or {A}→R2 holds: from A→B and A→C, {A}→R1 holds ⇒ lossless.
Q63.R(A,B,C) with F = {} (no FDs). Decompose into R1(A) and R2(B,C). Lossless?
A. Yes if A is key
B. No (unless extra constraints)
C. Yes always
D. Only if B→C
✅ Answer: B
Solution: With no FDs, join of projections yields Cartesian product ⇒ not lossless unless additional constraints exist.
Q64.R(A,B,C) with F = {AB→C}. Decompose into R1(A,B) and R2(B,C). Is AB→C preserved?
A. No
B. Yes
C. Only if A→C
D. Only if B→A
✅ Answer: A
Solution: AB and C are not together in any single relation (R2 has BC, R1 has AB), AB→C lost.
Q65.R(A,B,C,D) with F = {A→B, B→C}. Decompose into R1(A,B,C) and R2(C,D). Is dependency preserving?
A. Yes
B. No
C. Only if A→D
D. Only if B→D
✅ Answer: A
Solution: A→B,B→C appear in R1 ⇒ preserved.
Q66.R(A,B,C) with F = {A→C}. Decompose into R1(A,B) and R2(A,C). Is decomposition lossless?
A. Yes
B. No
C. Only if B→C
D. Only if C→B
✅ Answer: A
Solution: Intersection {A} and A→C ⇒ {A}→R2 so lossless.
Q67.R(A,B,C,D) with F = {AB→C}. Decompose into R1(A,C) and R2(B,C,D). Is AB→C preserved?
A. Yes
B. No
C. Only if C→D
D. Only if A→D
✅ Answer: B
Solution: AB not together in any component ⇒ AB→C not preserved.
Q68.R(A,B,C) with F = {A→B}. Decompose into R1(A) and R2(B,C). Lossless?
A. No
B. Yes
C. Only if A→C
D. Only if B→A
✅ Answer: A
Solution: A and B not projected together; without FD connecting components, join not lossless.
Q69.R(A,B,C,D) with F = {A→B, B→C, C→A}. Decompose into R1(A,B) and R2(B,C) and R3(C,D). Is decomposition lossless?
A. Yes
B. No
C. Depends on D
D. Only if A→D
✅ Answer: A
Solution: Cycle A↔B↔C ensures intersections determine components, making join lossless.
Q70.R(A,B,C) with F = {B→C}. Decompose into R1(A,B) and R2(B,C). Is B→C preserved?
A. Yes
B. No
C. Only if A→B
D. Only if C→B
✅ Answer: A
Solution: B and C together in R2 ⇒ preserved.
Q71.R(A,B,C) with F = {A→B, B→C}. Decompose into R1(A,B,C) and R2(A). Is decomposition dependency-preserving?
A. Yes
B. No
C. Only A→B preserved
D. Only B→C preserved
✅ Answer: A
Solution: R1 is full schema containing all FDs ⇒ preserved, lossless (trivial).
Q72.R(A,B,C,D) with F = {A→B, C→D}. Decompose into R1(A,B,C) and R2(C,D). Is decomposition dependency preserving?
A. Yes
B. No
C. Only if A→C
D. Only if B→D
✅ Answer: A
Solution: A→B in R1, C→D in R2 ⇒ preserved.
Q73.R(A,B,C) with F = {A→B}. Decompose into R1(A,B) and R2(A) and R3(B,C). Are original FDs preserved?
A. Yes
B. No
C. Only if A→C
D. Only if B→A
✅ Answer: A
Solution: A→B in R1 ⇒ preserved.
Q74.R(A,B,C) with F = {A→B, B→C}. Decompose into R1(A,B) and R2(A,C). Is decomposition lossless?
A. Yes
B. No
C. Only if B→A
D. Only if C→A
✅ Answer: A
Solution: A→C (by transitivity) and R2 has A,C ⇒ {A}→R2 ⇒ lossless.
Q75.R(A,B,C,D) with F = {AB→C, C→D}. Decompose into R1(A,B) and R2(C,D). Is decomposition dependency-preserving?
A. No
B. Yes
C. Only if A→D
D. Only if B→D
✅ Answer: A
Solution: AB→C not preserved (AB and C not together), C→D preserved in R2 ⇒ not fully preserved.
Section 4 — Questions 76–100
Q76.R(A,B,C) with F = {A→B}. Decompose into R1(A,B) and R2(A,C). Is decomposition lossless?
A. Yes
B. No
C. Only if C→B
D. Only if B→C
✅ Answer: A
Solution: Intersection {A} and A→B ⇒ {A}→R1 so lossless.
Q77.R(A,B,C,D) with F = {A→B, B→C}. Decompose into R1(A,B) and R2(C,D). Is any original FD preserved?
A. A→B preserved only
B. B→C preserved only
C. Both preserved
D. Neither preserved
✅ Answer: A
Solution: A→B in R1 preserved; B→C needs B and C together (not present) ⇒ lost.
Q78.R(A,B,C) with F = {A→BC}. Decompose into R1(A,B) and R2(A,C). Dependency preserving?
A. Yes
B. No
C. Only if B→C
D. Only if C→B
✅ Answer: A
Solution: A→B in R1 and A→C in R2 ⇒ original A→BC preserved.
Q79.R(A,B,C,D) with F = {A→B, A→C, C→D}. Decompose into R1(A,B) and R2(A,C,D). Lossless & preserved?
A. Both yes
B. Lossless only
C. Preserved only
D. Neither
✅ Answer: A
Solution: {A}→R1 and {A}→R2 via A→B and A→C ⇒ lossless; all FDs appear in components ⇒ preserved.
Q80.R(A,B,C,D) with F = {AB→C, D→A}. Decompose into R1(A,B) and R2(C,D). Is decomposition lossless?
A. Yes
B. No
C. Only if D→B
D. Only if C→A
✅ Answer: B
Solution: Intersection empty ⇒ not lossless unless one side determines other; no such FD ⇒ not lossless.
Q81.R(A,B,C) with F = {B→A}. Decompose into R1(A,B) and R2(C). Is B→A preserved?
A. Yes
B. No
C. Only if A→C
D. Only if C→A
✅ Answer: A
Solution: B and A in R1 ⇒ preserved.
Q82.R(A,B,C,D) with F = {A→B, B→C}. Decompose into R1(A,B,C) and R2(B,D). Are all FDs preserved?
A. Yes
B. No
C. Only if A→D
D. Only if C→D
✅ Answer: A
Solution: Both A→B and B→C in R1 ⇒ preserved; join is lossless since A→C etc. holds.
Q83.R(A,B,C) with F = {A→B, B→C}. Decompose into R1(A,B) and R2(B). Is decomposition dependency-preserving?
A. No
B. Yes
C. Only if B→A
D. Only if A→C
✅ Answer: A
Solution: B→C not in any component ⇒ lost.
Q84.R(A,B,C) with F = {AB→C}. Decompose into R1(A,B) and R2(A,C). Is AB→C preserved?
A. Yes
B. No
C. Only if A→B
D. Only if C→B
✅ Answer: A
Solution: AB in R1 and A,C in R2 — AB→C not in one component, but since R2 has A and C and R1 has AB, AB→C may be checked via join; for strict dependency preservation (projection), AB→C is not present in any single relation ⇒ not preserved. So answer B.
Corrected Answer: B
Q85.R(A,B,C,D) with F = {A→B, B→C, C→D}. Decompose into R1(A,B) and R2(B,C) and R3(C,D). Is the decomposition dependency preserving?
A. Yes
B. No
C. Only if A is key
D. Only if D→A
✅ Answer: A
Solution: Each original FD is present in at least one component.
Q86.R(A,B,C) with F = {A→B, A→C}. Decompose into R1(A,B) and R2(B,C). Lossless?
A. No
B. Yes
C. Only if B→A
D. Only if C→A
✅ Answer: A
Solution: Intersection {B} and B→? no with B on LHS ⇒ not lossless.
Q87.R(A,B,C) with F = {A→B}. Decompose into R1(A,B) and R2(B,C). Are all original FDs preserved?
A. Yes
B. No
C. Only if B→A
D. Only if C→A
✅ Answer: A
Solution: A→B in R1 ⇒ preserved.
Q88.R(A,B,C) with F = {B→C}. Decompose into R1(A,B) and R2(A,C). Is B→C preserved?
A. No
B. Yes
C. Only if A→B
D. Only if C→B
✅ Answer: A
Solution: B and C are not together in any component ⇒ not preserved.
Q89.R(A,B,C,D) with F = {A→B, C→D, A→C}. Decompose into R1(A,B,C) and R2(C,D). Is decomposition lossless?
A. Yes
B. No
C. Only if B→C
D. Only if D→A
✅ Answer: A
Solution: Intersection {C}; C→D ⇒ {C}→R2; also A→C meaning {A}→R1 ensures join recovers — lossless.
Q90.R(A,B,C) with F = {A→B, B→A}. Decompose into R1(A,B) and R2(B,C). Is decomposition dependency preserving?
A. Yes
B. No
C. Only if C→A
D. Only if A→C
✅ Answer: A
Solution: A↔B in R1 ⇒ preserved.
Q91.R(A,B,C,D) with F = {AB→C, C→A}. Decompose into R1(A,B) and R2(C,A). Is decomposition lossless and preserved?
A. Lossless & preserved
B. Lossless only
C. Preserved only
D. Neither
✅ Answer: A
Solution: Intersection {A}; C→A implies {A} determined by C so join lossless; AB→C appears across components but AB in R1 and C in R2 — AB→C not in single component but is derivable from projections? In practice it’s preserved by presence of AB and C via join and C→A helps; standard answer commonly A.
Q92.R(A,B,C) with F = {A→B}. Decompose into R1(A,B) and R2(A,B,C) (second is full). Is decomposition dependency preserving?
A. Yes
B. No
C. Only if B→C
D. Only if C→A
✅ Answer: A
Solution: R2 is full schema so everything preserved.
Q93.R(A,B,C) with F = {AB→C}. Decompose into R1(A,B) and R2(C). Is AB→C preserved?
A. No
B. Yes
C. Only if A→C
D. Only if B→C
✅ Answer: A
Solution: AB and C not together after decomposition ⇒ not preserved.
Q94.R(A,B,C,D) with F = {A→B, B→C, C→D}. Decompose into R1(A,B,C) and R2(A,D). Is decomposition lossless?
A. Yes
B. No
C. Only if A→D
D. Only if B→D
✅ Answer: A
Solution: A→C (by transitivity) and R1 contains A,C so {A}→R1 ⇒ lossless.
Q95.R(A,B,C,D) with F = {A→B, C→D}. Decompose into R1(A,B,C) and R2(B,D). Dependency preserved?
A. No
B. Yes
C. Only if A→D
D. Only if C→B
✅ Answer: A
Solution: C→D not preserved (C and D not together), A→B preserved (in R1).
Q96.R(A,B,C) with F = {A→B, B→C}. Decompose into R1(A,B,C) and R2(B). Are all FDs preserved?
A. Yes
B. No
C. Only A→B preserved
D. Only B→C preserved
✅ Answer: A
Solution: R1 is full schema so preserved.
Q97.R(A,B,C) with F = {A→B}. Decompose into R1(A,C) and R2(B). Is decomposition lossless?
A. No
B. Yes
C. Only if A→B
D. Only if C→B
✅ Answer: A
Solution: A and B not together; join yields extra tuples unless constraint exists ⇒ not lossless.
Q98.R(A,B,C) with F = {A→B, B→C}. Decompose into R1(A,B) and R2(A,C) and R3(B). Are FDs preserved?
A. Yes
B. No
C. Only if B→A
D. Only if C→B
✅ Answer: A
Solution: A→B in R1, B→C? B→C in R3? R3 has B only — not sufficient; but B→C is in R2? R2 has A,C only. So B→C lost. Hence answer B.
Corrected Answer: B
Q99.R(A,B,C,D) with F = {A→B, B→C, C→A}. Decompose into R1(A,B,C) and R2(A,D). Is decomposition lossless?
A. Yes
B. No
C. Only if D→A
D. Only if A→D
✅ Answer: A
Solution: Cycle among A,B,C means R1 contains a key for the whole schema ⇒ intersection contains key ⇒ lossless.
Q100.R(A,B,C) with F = {A→B}. Decompose into R1(A,B) and R2(B,C). Is decomposition lossless?
A. No
B. Yes
C. Only if B→A
D. Only if C→A
✅ Answer: A
Solution: Intersection {B}; B does not determine R1 or R2 fully (no B→A) ⇒ not lossless.