Electronics and Communication Engineering - Exam Questions Papers - Discussion

Discussion Forum : Exam Questions Papers - Exam Paper 8 (Q.No. 47)
47.
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
0
1
EF + UZ + HI
None of these
Answer: Option
Explanation:

(UVW ⊕ WVU) = 1 = KJ(XY ⊕ XY)

KJ = 1

NAND B

HI = 1

Since f = GEF + 1 + LMON + TUVWXYZ = 1.

Discussion:
1 comments Page 1 of 1.

Pranati Sharma said:   8 years ago
How (UWV+WVU)'=1?

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