Gate CS 1996 Question no. 47

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f implements,

A) (ABC)' + A'BC' + ABC

B) A + B + C

C) A ⊕ B ⊕ C

D) AB + BC + CA

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I think options are wrong . Above function will generate C as o/p :

The SOP equation : A'B'C + A'BC + AB'C + ABC = C(A'B' + A'B + AB' +AB) . which simply means on any combination of A and B the o/p generated is C.

Now the counter statements for the options:

A) (ABC)' + A'BC' + ABC

ABC + (ABC)' = 1.  Therefore , 1 + A'BC' = 1. That means this function will o/p 1 every time regardless of                  value of C which makes it wrong.

B) A + B + C

if A = 1 and B = 1 and C=0, then the equation becomes 1 + 1 + 0 = 1 . which is again wrong because                      value  of C = 0 but the o/p generated is 1.

C) A ⊕ B ⊕ C

if A = 1 and B = 0 and C = 0, then the equation becomes 1 XOR 0 XOR 0 = 1 , which is again wrong                           because of C = 0 but the o/p generated is 1.

D) AB + BC + CA

if A = 0 and B = 0 and C = 1, then the equation becomes 0.0 + 0.1 +1.0 = 0, which is again wrong because               of C = 1 but the o/p generated is 0.

any help will be appreciated

Ans is "C". Not option C.

the Equation made from MuX = A'B'C + A'BC + AB'C + ABC

By solving this equation using K Map it will give ans as C

A'B'C + A'BC + AB'C + ABC = C.
1 vote
This ques was asked in GATE 1996....for 2,3 it is C' not C
f=A'B'C+A'BC+AB'C+ABC

=1

OPTION 1.   (ABC)'+A'BC'+ABC

=(A'+B'+C')+A'BC'+ABC

=A'(1+BC')+B'+C'+ABC

=A'+B'+C'+ABC

=(ABC)'+ABC

=1
0

here i think you are missing something....

f=A'B'C+A'BC+AB'C+ABC !=1 it's is going to be..C(A'B'+A'B+AB'+AB)-> C(1)->C

0
sorry...... i have missed.....

f=C.1=C

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