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Chapter 13 Boolean Algebra Solutions

Question - 21 : - Verify the following using Boolean Laws X + Z = X + X’. Z + Y. Z

Answer - 21 : -

Taking RHS
X + X’Z + YZ
= (X + X’). (X + Z) + YZ (Distribution Law)
= 1. (X + Z) + YZ    (A + A’ = 1)
= X + Z + YZ
= X + Z (1 + Y)
= X + Z    (1 + A = 1; 1. A = A)
= Hence verified

Question - 22 : - Verify the following using Boolean Laws : A + C = A + A. C + B.C

Answer - 22 : -

A + C = A + A’.C + BC
Solve RHS
A + A’C + BC
(A + A). (A + C) + BC [Using distributive law]
1. (A + C) + BC
= A + C + BC
= A + C(1 + B)
= A + C.1
= A + C
= LHS (Hence, verified)

Question - 23 : -
Verify the following using truth table :
(i) X . X’ = 0
(ii) X + 1 = 1

Answer - 23 : -


Question - 24 : - Write the equivalent boolean expression for the following logic circuit :

Answer - 24 : -

Y = U  +  

Question - 25 : - Write the equivalent boolean expression for the following logic circuit :

Answer - 25 : -

Y = (U +  ). (U+  )

Question - 26 : -
Verify the following using truth table :
(i) X + 0 = X
(ii) X + X’ = 1

Answer - 26 : -


Question - 27 : - Derive a Canonical SOP expression for a Boolean function F, represented by the following truth table :

Answer - 27 : -

F(A, B, C) = A’B’C +A’BC + AB’C + ABC
OR
F(A,B,C) =∑(0, 3,4,7)

Question - 28 : -
Obtain a simplified form for a Boolean expression :
F(U, V, W, Z) = II (0,1,3,5, 6, 7,15)

Answer - 28 : -


(u + v +w).(u+z’).(v’+w’).(u’+w’+z)

Question - 29 : -
Reduce the following Boolean Expression to its simplest form using K-Map :
F (X, X Z, W) = X (0,1, 6, 8, 9,10,11,12,15)

Answer - 29 : -


Simplified Expression : XY + Y’Z’ + XZ’W’ + XZW + X’YZW’

Question - 30 : -
Reduce the following Boolean Expression to its simplest form using K-Map :
F(X, Y, Z, W) = X (0,1,4, 5,6, 7,8, 9,11,15)

Answer - 30 : -


Simplified Expression: Y’Z’ + XY + XZW

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