GATE CSE 2015 | Set 3 | Question: 44

Question

Given the function $F = P' +QR$, where $F$ is a function in three Boolean variables $P, Q$ and $R$ and $P'=!P$, consider the following statements.

$(S1) F = \sum(4, 5, 6)$

$(S2) F = \sum(0, 1, 2, 3, 7)$

$(S3) F = \Pi (4, 5, 6)$

$(S4) F = \Pi (0, 1, 2, 3, 7)$

Which of the following is true?

• A. (S1)-False, (S2)-True, (S3)-True, (S4)-False
• B. (S1)-True, (S2)-False, (S3)-False, (S4)-True
• C. (S1)-False, (S2)-False, (S3)-True, (S4)-True
• D. (S1)-True, (S2)-True, (S3)-False, (S4)-False

Comments

$f=\sum(0,1,2,3,7)$

$f=\Pi(4.5.6)$

---

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

Watch Detailed Video Solution   

---

option A sahi H.

f = p' + qr

just p' = 000,001,010,011  = (0,1,2,3)

       qr = 011,111 = (3,7)

sop = singma(0,1,2,3,7)

pos = pi(4,5,6)

Answer 1

$F=P'+QR$,  draw the Kmap for this

We can find the minterm $\sum (0,1,2,3,7)$ and maxterm $\Pi (4,5,6) $

So, option A is correct: $(S1)$-False, $(S2)$-True, $(S3)$-True, $(S4)$-False

Answer 2

$F = P{}' + QR$

for SOP we have :

$F = P{}.1.1 + 1.QR'=P{}'(Q+Q{}')(R+R{}') + (P+P{}')QR$

$P{}'QR + P{}'QR{}' +P{}'Q{}'R+P{}'Q{}'R{}'+PQR+P{}'QR$

$P{}'QR + P{}'QR{}' +P{}'Q{}'R+P{}'Q{}'R{}'+PQR$

$F=\sum(0, 1, 2, 3, 7)$ (considering barred terms as 0 and unbarred as 1 and converting them to binary and then to decimal).

now for POS we have :

$F = P{}' + QR = (P{}'+Q)(P{}'+R) = (P{}'+Q+0)(P{}'+R+0)$

$(P{}'+Q+R.R{}')(P{}'+R+Q.Q{}') $

$(P{}'+Q+R)(P{}'+Q+R)(P{}'+Q+R{}')(P{}'+Q{}'+R)$

$(P{}'+Q+R)(P{}'+Q+R{}')(P{}'+Q{}'+R)$

$F=\prod (4, 5, 6)$(considering barred terms as 1 and unbarred as 0 and converting them to binary and then to decimal).

http://mcs.uwsuper.edu/sb/461/PDF/sop.html

Comments

But K-map is better time saving approach rt?

---

may be but that also kind of varies from individual to individual don't you think ?

---

Yes. Sure :)
But you needn't solve for both min terms and max terms rt? Solve one and take complement for other.

---

i was thinking of doing that, but just did it to show the general method to solve it both ways( POS ans SOP). Feel free to edit.

Answer 3

Answer = A

Answer 4

S2 and S3 are true

Answer 5

Option A is correct

P'QR + P'QR' +P'Q'R + P'Q'R' + PQR = ⋿(0,1,2,3,7) = ∏(4,5,6)

Answer 6

option A is correct.

F=~P+QR

=011+000+010+001+011+111=∑(0,1,2,3,7)=Π(4,5,6)