GATE IT 2004 | Question: 44

Question

The function $A \bar B C + \bar A B C + AB \bar C+ \bar A \bar B C+ A \bar B \bar C$ is equivalent to

• A. $A \bar C + AB+ \bar A C$
• B. $A \bar B+ A \bar C+ \bar A C$
• C. $\bar A B+ A \bar C+ A \bar B$
• D. $\bar AB+ AC+ A \bar B$

Answer 1

So, the equivalent expression will be $\bar AC + A\bar C + A\bar B$ 

(B) option

Comments

$AB'C + AB'C' \leftrightarrow AB'$ 
$A'BC + A'B'C \leftrightarrow A'C$
$ABC' + AB'C' \leftrightarrow AC'$

Note -
$i) AB'C'$ can be used multiple times as $X+X =X.$

$ii)K- Map$ method is more reliable as we can get "minimal expressions" not "minimum expression" i.e multiple minimum expressions are possible that's why we call it "minimal" not "minimum".

$iii)$The result of Elimination method depends on the order in which terms are combined or eliminated so the final expression obtained is not necessarily minimal or we may not get desired minimal expression. 

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thanks this is going to save minutes...

Answer 2

Let us solve the given function

$F = AB’C + A’BC + ABC’ + A’B’C +  AB’C'$

$F = AB'(C + C') + A'C(B + B') + ABC'$

$F = AB'.1 + A'C.1 + ABC'   [C + C' = 1,  B + B '= 1 ]$

$F = AB' + A'C + ABC'$

$F = A(B' + BC')+ A'C$

$F=A( ( B'+B ).( B' + C' ) ) + A'C \:\:\:\:\:[\because \text{Using Distributive law}: A + B.C = (A + B).(A + C)]$

$F = A.1.( B' + C' ) + A'C  \:\:\:\:\:\:\:\:[\because B' + B = B + B' = 1  ]$

$F = A.(B' + C' ) + A'C$

$F = AB' + AC' + A'C$
So, the correct answer is $(B)$.

Answer 3

Answer is "B"

Answer 4

AB’C + A’BC + ABC’ + A’B’C + AB’C’

AB’(C + C') + A'C ( B+ B') +  AC’ ( B + B' )

AB' + A'C + AC'

Hence option b ..!

Comments

both are right just do it by making K-map.2 minimum expression will be possible but here in given options (A'C + AC' + B'C) is not given.so now there should not be any confusion

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I am getting  AC'+ A'C + B'C

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@Sandeep Verma sometimes there are some time more than one minimal boolean expression are possible.this is one of the example