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  9. Q177

ibps-so 2016 · Question 177 of 394

The redesign is being reviewed against a second relation R, whose attributes have been lettered A, B, C, D and E for the purposes of the review, and upon which exactly four functional dependencies are known to hold and no others: # Functional dependency asserted upon R f1 A → B, C f2 C, D → E f3 B→D f4 E→A The reviewer must state how many candidate keys R possesses and how far up the sequence of normal forms it already stands. The number of candidate keys of R, and the highest of the normal forms R already meets, are respectively:

  1. ATwo keys, namely A and E; and BCNF, which R meets in full
  2. BThree keys, namely A, E and CD; and 2NF, R falling short of 3NF on account of f2
  3. CFour keys, namely A, E, BC and CD; and 3NF, which R meets because every one of its ve attributes is prime, while BCNF fails at f3, whose determinant B is no superkeyCorrect
  4. DFour keys, namely A, E, BC and CD; and BCNF, since each of the four determinants is itself a key
  5. EOne key, namely A by itself; and 1NF alone, every other form failing

Answer: C. Four keys, namely A, E, BC and CD; and 3NF, which R meets because every one of its ve attributes is prime, while BCNF fails at f3, whose determinant B is no superkey

Official answer key verified. Detailed explanation coming soon.

Source: IBPS SO 2016 IT Officer (held 28-29 Jan 2017, single-stage CWE) - Prepp · memory-based

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