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Exam Questions Papers - Engineering

Q1:

Given f = GEF + KHIJ + LMON + TUVWXYZ and
  1. E + F + G = LMN
  2. V + W = U . Z . Y . X . T
  3. NAND B
  4. (UVW ⊕ WVU) = KJ(XY ⊕ XY)

Then f is equivalent to

A 0

B 1

C
EF + UZ + HI

D None of these

ANS:B - 1

(UVW ⊕ WVU) = 1 = KJ(XY ⊕ XY) ∴ KJ = 1 NAND B ∴ HI = 1 Since f = GEF + 1 + LMON + TUVWXYZ = 1.