GATE CSE 2011 | Question: 14

Question

The simplified SOP (Sum of Product) from the Boolean expression

$$(P + \bar{Q} + \bar{R}) . (P + \bar{Q} + R) . (P + Q +\bar{R})$$ is 

• A. $(\bar{P}.Q+\bar{R})$
• B. $(P+\bar{Q}.\bar{R})$
• C. $(\bar{P}.Q+R)$
• D. $(P.Q+R)$

Comments

we can simply by case method in very less time

---

By Case method is so under rated in Boolean algebra that no one is using
let me share
any variable can have either 0 or 1 in our boolean algebra
take two case of our given boolean function F
take Q=0  F becomes P+R"

Take Q=1 F becomes P
now put same q=0,1 and yes only B is satisfying

use this method whenever u asked a boolean expression and tell us that which are equal

---

Watch the Detailed Video Solution by clicking the button below..!

Watch Detailed Video Solution   

Answer 1

K-map

Answer is B

Comments

@Pardhyuman

In question, given expression is POS, so we can make K-map using them.

---

I have observed one thing to convert SOP to POS or vice versa :

1. Take Complement of the expression
2. Appy Demorgans law 
3. Take complement again             
   
   I got the answer through this

---

Caption

Answer 2

First Write POS max term then write absent term as min term and Solve using K- Map 

Hope this image clears everything.

Answer is B

Comments

Detailed explanation. Great work!

Answer 3

(P+Q'+R').(P+Q'+R).(P+Q+R')

=(P+R`)(Q+Q`)(P+Q`+R)

=P+PR+PQ`+R`P+R`Q`

=P(1+Q`)+R`P+R`Q`

=P(1+R`)+R`Q`

=P+R`Q`

Comments

Yo can also solve this using Kmap.

---

depends on practice watever u prefer ....

---

q+q'=1

1+q'=1

qq'=0

Answer 4

Given Boolean Expression:

(P+Q'+R').(P+Q'+R).(P+Q+R')

we know that dual of a boolean expression is equivalent. Take dual of it:

PQ'R'+PQ'R+PQR'

Now simplify it.(I am taking dual because solving the given expression will be difficult hence, take dual and solve it)

=PQ'(R'+R)+PQR'

=PQ'+PQR'

=P(Q'+QR')

=P(Q'+R').

Now if we see option b) and take dual of it then we will be getting the above simplified expression.

Answer 5

(P+Q'+R') (P+Q'+R) (P+Q+R')

complement the whole equation

P'QR+P'QR'+P'Q'R

taking P'R common from 1st and 3rd minterm

P'R(Q+Q') + P'QR'

we know A+A' = 1

P'R+P'QR'

P'(R+QR')

P'((R+Q)(R+R'))

P'(R+Q)

Complement it back

P+R'Q'

P+Q'R'

Answer 6

By solving it we get option b. So answer is b.