Page 1 :
VXL TUTORIAL COMPUTER CLASSES BY VINOD KHARE PH: 9336061496, , ANSWER SHEET, BOOLEAN ALGEBRA & LOGIC GATES, Аnswer 1:, The logic circuit is given:, , Аnswer 2:, The expression F(a, b) = a.b using only NOT and OR gates is:, , Аnswer 3:, 2N input combinations can be there in the truth table of a logic system., Аnswer 4:, , Аnswer 5:

Page 2 :
VXL TUTORIAL COMPUTER CLASSES BY VINOD KHARE PH: 9336061496, , Аnswer 6:, A+C = A +, .C + B.C, R.H.S =A +, .C + B.C, =(A +, ).(A + C)+B.C, = 1. (A + C)+B.C, = A+C+B.C = A+C. (1 + B), = A + C.( 1 ) = A+C, = L.H.S, , [By using Distributive law], [By using A+ =1], [∵ 1 + B = 1], , Аnswer 7:, Obtain the boolean expression for the logic circuit shown below:, , Аnswer 8:

Page 4 :
VXL TUTORIAL COMPUTER CLASSES BY VINOD KHARE PH: 9336061496, , Аnswer 11: no answer, , Аnswer 12: Given the following circuit:, , Аnswer 13::, Complement of (A + B).(B + C).(A + C), = [(A + B).(B + C).(A + C)]', = (A + B)' + [(B + C).(A + C)]', = (A + B)' + (B + C)' + (A + C)', = A'.B' + B'.C' + A'.C', Аnswer 14:

Page 5 :
VXL TUTORIAL COMPUTER CLASSES BY VINOD KHARE PH: 9336061496, Complement of A.B + (A'.B').(B.C + B'.C'), = [A.B + (A'.B').(B.C + B'.C')]', = (A.B)'.[(A'.B').(B.C + B'.C')]', = (A' + B').[(A'.B')' + (B.C + B'.C')'], = (A' + B').[A + B + [(B.C)' . (B'.C')']], = (A' + B').[A + B + [(B' + C').(B + C)]], = (A' + B').[A + B + [BB' + B'C + BC' + CC']], = (A' + B').[A + B + [0 + B'C + BC' + 0]], = (A' + B').(A + B + B'C + BC'), = (A' + B').(A + B'C + B(1 + C')), = (A' + B').(A + B + B'C), Аnswer 15:, Complement of XY'Z + XY + YZ', = [XY'Z + XY + YZ']', = (XY'Z)'.(XY)'.(YZ')', = (X' + Y + Z').(X' + Y').(Y' + Z), Аnswer 16:, Simplified expression:, F(A, B, C) = B, Аnswer 17:, (A' + C) (A' + C') (A' + B + C'D), = (A'A' + A'C' + CA' + CC')(A' + B + C'D), = (A' + A'C' + A'C + 0)(A' + B + C'D), = [A'(1 + C' + C)](A' + B + C'D), = [A'(1 + C)](A' + B + C'D), = A'(A' + B + C'D), = (A'A' + A'B + A'C'D), = (A' + A'B + A'C'D), = [A'(1 + B) + A'C'D], = [A' + A'C'D], = [A'(1 + C'D)], = A', , [Distributive Law], [∵ CC' = 0, A'A' = A'], [∵ 1 + C' = 1], [∵ 1 + C = 1], [Distributive Law], [∵ A'A' = A'], [∵ 1 + B = 1], [∵ 1 + C'D = 1], , Аnswer 18:, Dual of (P + Q').R.1 = P.R + Q'.R is:, (P.Q') + R + 0 = (P + R).(Q' + R), Аnswer 19:, (A + B')(B + CD)', = (A + B')(B'(CD)'), = (A + B')(B'(C' + D')), = (A + B')(B'C' + B'D'), , [De-Morgan's Law], [De-Morgan's Law], [Distributive Law]

Page 6 :
VXL TUTORIAL COMPUTER CLASSES BY VINOD KHARE PH: 9336061496, =, =, =, =, , AB'C' + AB'D' + B'B'C' + B'B'D', AB'C' + AB'D' + B'C' + B'D', B'C'(1 + A) + B'D'(1 + A), B'C' + B'D', , [∵ B'B' = B'], [∵ 1 + A = 1], , Аnswer 20:, Dual of (A' B) + (C' 1) = (A' + C) (B + C) is:, (A' + B).(C' + 0) = (A'.C) + (B.C), Аnswer 21:, 1., {(CD)' + A} + A + C.D + A.B, {(CD)' + A} + A + C.D + A.B, = C' + D' + A + A + C.D + A.B, = C' + D' + A + C.D + A.B, = A(1 + B) + C' + D' + C.D, = A + C' + D' + C.D, = (A + C' + D' + C).(A + C' + D' + D), = (A + D' + 1).(A + C' + 1), = 1.1, =1, , [De-Morgan's Law], [∵ A+A=A], [∵ 1+B=1], [Distributive Law], [∵ C'+C=1, D'+D=1], [∵ A+D'+1=1, A+C'+1=1], , 2., A.{B + C (A.B + A.C)'}, =A.{B + C(A.B + A.C)'}, =A.{B + C[(A.B)'.(A.C)']}, =A.{B + C[(A' + B').(A' + C')]}, =A.{B + C[A'A' + A'C' + A'B' + B'C']}, =A.{B + C[A' + A'C' + A'B' + B'C']}, =A.{B + C[A'(1 + C' + B') + B'C']}, =A.{B + C[A'.1 + B'C']}, =A.{B + A'C + B'CC'}, =A.{B + A'C + 0}, =AB + AA'C, =AB, , [De-Morgan's Law], [De-Morgan's Law], [Distributive Law], [∵ A'A'=A'], [∵ 1 + C' + B' = 1], [Distributive Law], [∵ CC'=0], [Distributive Law], [∵ AA'C=0], , Аnswer 22:, Reducing the expression using boolean laws:, (L+M+O+P).(L+M+O'+P).(L'+M+O+P).(L'+M+O'+P), = (L+M+O+P).(L'+M+O+P).(L+M+O'+P).(L'+M+O'+P), = [(M+O+P) + (LL')].[(M+O'+P) + (LL')], = [(M+O+P) + 0].[(M+O'+P) + 0], = (M+O+P).(M+O'+P), = (M+P) + (O.O'), , [Associative Law], [Distributive Law], [Complementary Law], [∵ a+0=a], [Distributive Law]

Page 7 :
VXL TUTORIAL COMPUTER CLASSES BY VINOD KHARE PH: 9336061496, = M+P+0, = M+P, , [Complementary Law], [∵ a+0=a], , Аnswer 23:, A.B' + A'.B.C' + (A.C') + B.C, = A.B' + A'.B.C' + (A.C')(B+B') + B.C, = A.B' + A'.B.C' + A.B.C' + A.B'.C' + B.C, = A.B' + A.B'.C' + A'.B.C' + A.B.C' + B.C, = A.B' + A'.B.C' + A.B.C' + B.C, = A.B' + B.C + B.C'(A' + A), = A.B' + B.C + B.C'.1, = A.B' + B(C + C'), = A.B' + B, = (A + B).(B' + B), =A+B, Аnswer 24:, a.b + a'.c + b.c, = a.b + a'.c + b.c(a + a'), = a.b + a'.c + a.b.c + a'.b.c, = a.b + a.b.c + a'.c + a'.b.c, = a.b + a'.c, , [Complementary Law: B+B'=1], [Distributive Law], [Associative Law], [Absorbtion Law:A.B'+A.B'.C'=A.B'], [Complementary Law: A+A'=1], [Complementary Law: C+C'=1], [Distributive Law], [Complementary Law: B+B'=1], , [Complementary Law: A+A'=1], [Distributive Law], [Associative Law], [Absorbtion Law], , Аnswer 25:, (a.b + x + y + z).(a.b + x'.y'.z'), = a.b + [(x + y + z).(x'.y'.z')], [Distributive Law], = a.b + [(x + y + z).(x + y + z)'], [De-Morgan's Law], = a.b + 0, [Complementary Law: (x + y + z).(x + y + z)' = 0], = a.b, Аnswer 26:, A.[B + C.(A.B + A.C')], = A.[B + A.B.C + A.C'.C], = A.[B + A.B.C + 0], = A.[B(1 + AC)], = A.B, Аnswer 27:, [a + {(b + c).(b' + d')}]', = a'.[{(b + c).(b' + d')}]', = a'.{(b + c)' + (b' + d')'}, = a'.{(b'c') + (b''.d'')}, = a'.{(b'c') + (bd)}, = a'b'c' + a'bd, , [Distributive Law], [Complementary Law: C'.C = 0], [Distributive Law], [∵ 1+AC=1], , [De-Morgan's Law], [De-Morgan's Law], [De-Morgan's Law], [Involution Law: a''=a]

Page 8 :
VXL TUTORIAL COMPUTER CLASSES BY VINOD KHARE PH: 9336061496, , Аnswer 28:, Dual of P'QR' + PQ'R + P'Q'R:, (P'+Q+R').(P+Q'+R).(P'+Q'+R), Complement of PQ'R + Q.(P'R' + PR'):, [PQ'R+Q.(P'R'+PR')]', = (PQ'R)'.[Q.(P'R'+PR')]', = (P'+Q+R').[P'QR'+PQR']', = (P'+Q+R').(P+Q'+R).(P'+Q'+R), Hence proved., Аnswer 29:, F' = [A + (B + C).(D' + E)]', = A'.[(B + C).(D' + E)]', = A'.[(B + C)' + (D' + E)'], = A'.[(B'C') + (D''E')], = A'.(B'C' + DE'), = A'B'C' + A'DE', , [De-Morgan's Law], [De-Morgan's Law], [De-Morgan's Law], [∵ D'' = D], [Distributive Law], , Аnswer 30:, Y.(A + B').(B + CD)', = Y.(A + B').[B'.(CD)'], = Y.(A + B').[B'.(C' + D')], = (AY + B'Y).(B'C' + B'D'), = AB'C'Y + AB'D'Y + B'C'Y + B'D'Y, = B'C'Y(A + 1) + B'D'Y(A + 1), = B'C'Y + B'D'Y, , [De-Morgan's Law], [De-Morgan's Law], [Distributive Law], [Distributive Law], [Distributive Law], [∵ A+1=1], , Аnswer 31:, LHS = (Z + X)(Z + X' + Y), = Z + X'Z + YZ + XZ + XX' + XY, = Z + XY + Z(Y + X + X') + 0, = Z + XY, RHS = (Z + X)(Z + Y), = Z + YZ + XZ + XY, = Z(1 + Y + X) + XY, = Z + XY, ∴ LHS = RHS, Hence Proved., Аnswer 32: