Boolean Algebra GATE PYQ Set-1 (1998-2007) – Full Options
Q1.1 [GATE 1998 : 1 Mark] Let f(x,y,z)= x’ + y’x + xz be a switching function. Which one of the following is valid?
(a) y’x is a prime implicant of f
(b) xz is a minterm of f
(c) xz is an implicant of f
(d) y is a prime implicant of f
Answer: (c)
Explanation Hinglish: xz=1 hua to x=1,z=1, to f=1 ho jayega. Isliye xz implicant hai, minterm nahi hai kyunki isme y missing hai.
Q1.2 [GATE 1999 : 1 Mark] Let * be defined as x*y = x’ + y. Let z = x*y. Then the value of z*z is?
(a) x*y
(b) x
(c) 0
(d) 1
Answer: (d) 1
Explanation Hinglish: A*A = A’ + A = 1. To z*z = 1 hi hoga.
Q1.3 [GATE 2001 : 2 Marks] Consider the circuit with 2 NAND gates. Input A and C waveforms are given in figure. How many transitions will occur at output B in 0-10ns interval?
(a) 2
(b) 3
(c) 5
(d) 4
Answer: (d) 4
Explanation Hinglish: NAND gate ka output har input edge par change hota hai. Count karne par 4 transition aate hain.
Q1.4 [GATE 2002 : 2 Marks] f1(w,x,y,z)=Σm(8,9,10), f2(w,x,y,z)=Σm(7,8,12,13,14,15), f(w,x,y,z)=Σm(8,9). If f = f1.f2 + f3, then f3 is?
(a) Σm(0,1,2)
(b) Σm(9)
(c) Σm(8,9)
(d) Σm(7,8,9)
Answer: (b) Σm(9)
Explanation Hinglish: f1.f2 = common = 8. f ko 8,9 chahiye, to 9 f3 se aana chahiye.
Q1.5 [GATE 2003 : 1 Mark] Which of the following operations is commutative but not associative?
(a) AND
(b) OR
(c) NAND
(d) EXOR
Answer: (c) NAND
Explanation Hinglish: A NAND B = B NAND A hai lekin associative nahi hai.
Q1.6 [GATE 2004 : 1 Mark] What is the value of B ⊕ (B ⊕ (B ⊕ (… n times)))?
(a) 0 when n is odd
(b) 0 when n is even
(c) B when n is odd
(d) B when n is even
Answer: (d) B when n is even
Explanation Hinglish: B XOR B = 0, to even baar B aane par B hi bachega.
Q1.7 [GATE 2004 : 1 Mark] The K-map for a Boolean function is given. The function f is? [K-map has 1s at 01,10 pattern]
(a) B
(b) C
(c) B XNOR C (BC + B’C’)
(d) B XOR C
Answer: (c) B XNOR C
Explanation Hinglish: K-map se F=BC+B’C’ aata hai jo XNOR hai.
Q1.8 [GATE 2004 : 1 Mark] Which of the following expressions is NOT equivalent to complement of x i.e. x’?
(a) x NAND x
(b) x NOR x
(c) x XOR 1
(d) x NOR 1
Answer: (d) x NOR 1
Explanation Hinglish: x NOR 1 = 0 hamesha, x’ nahi.
Q1.9 [GATE 2005 : 2 Marks] The K-map with don’t cares (X) is given. The minimal SOP form is? [K-map image]
(a) D'(C+A)
(b) D(C+A)
(c) D'(C’+A’)
(d) D(C’+A)
Answer: (b) D(C+A)
Explanation Hinglish: Don’t care ko 1 maano to D(C+A) minimal aayega.
Q1.10 [GATE 2005 : 2 Marks] Given x+y+z=1, xy=0, xz+w=1, xy+zw’=0, then w,x,y,z are?
(a) 0101
(b) 0110
(c) 1011
(d) 1101
Answer: (c) 1011
Explanation Hinglish: Option test karo, 1011 hi satisfy karta hai.
Q1.11 [GATE 2006 : 1 Mark] Simplify: x’y’ + xy + x’y = ?
(a) x’ + y’
(b) x’ + y
(c) x + y’
(d) x + y
Answer: (b) x’ + y
Explanation Hinglish: x’y’+x’y = x’, fir x’+xy = x’+y.
Q1.12 [GATE 2007 : 1 Mark] Minimum number of 2-input NAND gates required to realize a 2-input EXOR gate is?
(a) 3
(b) 4
(c) 5
(d) 6
Answer: (b) 4
Explanation Hinglish: EXOR ka NAND-only design 4 gate me banta hai.
Logic Function & Minimization GATE PYQ Set-2 (2008-2015)
Q1.13 GATE 2008 1 Mark What is the minimal SOP form of f(w,x,y,z)=Σ(0,1,2,4,5,6,8,9,12,13,14)?
(a) y’ + w’ + z’
(b) y’ + w’z’
(c) y’ + wz’
(d) y’ + w’z
Answer: (a) y’ (simplified) Actually minimal is y’
Explanation Hinglish: K-map me y’ sab jagah common hai. Saare minterms me y=0 hai, to function sirf y’ hai. Simplest hai bhai.
Q1.14 GATE 2008 2 Marks For f(w,x,y,z)=Σ(0,2,5,7,8,10,13,15), min number of 2-input NAND gates required when complement available is?
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (b) 3
Explanation Hinglish: Ye function w’ + x’ type ka XNOR hai. Complement available hai to 3 NAND gate me ho jayega.
Q1.15 GATE 2009 1 Mark Minimal POS form of f(x,y,z)=Σ(0,2,3,5) is?
(a) (x’+y’)(y+z’)
(b) (x+y’)(y’+z)
(c) (x’+y)(y’+z’)
(d) (x+y)(y’+z’)
Answer: (c) (x’+y)(y’+z’)
Explanation Hinglish: POS ke liye 0 ko group karo. 1,4,6,7 zero hain, unse ye POS banega.
Q1.16 GATE 2009 2 Marks f(w,x,y,z)=Σ(0,1,2,4,5,6,8,9,12,13,14) has how many prime implicants?
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (c) 3
Explanation Hinglish: K-map me 3 groups prime hain, 2 essential + 1 extra.
Q1.17 GATE 2010 1 Mark (P+Q’)(P’Q+QR+PR’)(P’Q’+R’) simplifies to?
(a) P’Q’ + QR’
(b) P’Q + Q’R
(c) PQ’ + QR
(d) PQ + Q’R’
Answer: (a) P’Q’ + QR’
Explanation Hinglish: Multiply karo, absorption lagao, PQ’Q cancel ho jayega, last me ye bachega.
Q1.18 GATE 2010 2 Marks Min number of 2-input NOR gates to realize f=Σ(0,2,5,7,8,10,13,15) with complement available?
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (b) 3
Explanation Hinglish: Same XNOR function hai, NOR se bhi 3 gate me banega jab complement diya ho.
Q1.19 GATE 2011 1 Mark Let f = xy + yz + zx (3-variable majority). Which is equivalent?
(a) x⊕y⊕z
(b) (x⊕y)’z + xy
(c) x’y + y’z + z’x
(d) x’y’z’ + xyz
Answer: (b) (x⊕y)’z + xy
Explanation Hinglish: Majority ka matlab 2 ya 3 inputs 1 hone chahiye. Ye formula wahi deta hai.
Q1.20 GATE 2011 2 Marks f(w,x,y,z)=Σ(0,2,4,6,8,10,12,14). Min literals in minimal SOP is?
(a) 1 (z’)
(b) 2
(c) 3
(d) 4
Answer: (a) 1
Explanation Hinglish: Saare even numbers matlab z’=0, to pura function sirf z’ hai. 1 literal.
Q1.21 GATE 2012 1 Mark f(w,x,y,z)=Σ(0,1,2,3,7,8,10)+d(5,6,11,15) minimal POS is?
(a) f=(w’+z’)(x’+z)
(b) f=(w’+z)(x’+z’)
(c) f=(w+z’)(x’+z)
(d) f=(w+z)(x+z’)
Answer: (a) f=(w’+z’)(x’+z)
Explanation Hinglish: POS ke liye 0 ko group karo, don’t care use karke ye minimal POS aayega.
Q1.22 GATE 2012 2 Marks Which function has maximum number of minterms? f = x’+y’+z’ vs others
(a) f=x’y’z’
(b) f=x’+y’+z’
(c) f=x’y + yz
(d) f=xy + yz + zx
Answer: (b) f=x’+y’+z’
Explanation Hinglish: x’+y’+z’ = (xyz)’ hai, isme 7 minterm hote hain, sabse zyada.
Q1.23 GATE 2013 1 Mark Let x1⊕x2⊕x3⊕x4=0. Which is always TRUE?
(a) x1⊕x2⊕x3 = x4
(b) x1⊕x3 = x2⊕x4
(c) x1’⊕x3′ = x2’⊕x4′
(d) x1⊕x2 = x3⊕x4
Answer: (c)
Explanation Hinglish: XOR sum 0 matlab even number of 1s. Isliye complemented wala bhi equal rahega.
Q1.24 GATE 2013 2 Marks f(P,Q,R,S)=Σ(0,2,5,7,9,11)+d(3,8,10,12,14). Essential prime implicants?
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (b) 2
Explanation Hinglish: K-map me 2 groups essential hain jo unique minterm cover karte hain.
Q1.25 GATE 2014 2 Marks Which is NOT a valid Boolean identity?
(a) x⊕y = (xy + x’y’)’
(b) x⊕y = x+y if xy=0
(c) (x⊕y)⊕z = x⊕(y⊕z)
(d) (x+y)⊕z = x⊕(y+z)
Answer: (d) (x+y)⊕z = x⊕(y+z)
Explanation Hinglish: XOR associative hai lekin OR ke saath mix karne par associative fail ho jata hai, to (d) galat hai.