UNIT 4 GATE EC BY RK Kanodia Digital Electronics
A - B = A + B,
SOLUTIONS
A 010 10110
B + 00010011
0110 1001
1. (D) 100110 2 = 2 5 + 2 2 + 21 = 3810
- A - B = A + B, A 1010 1001
2616 = 2 ´ 16 + 6 = 3810
B + 00010011
468 = 4 ´ 8 + 6 = 3810
10111100
212 4 = 2 ´ 4 2 + 41 = 3810
So 3610 is not equivalent. 7. (B) Here A , B are 2’s complement
A + B, A 0100 0110
2. (C) 2 x 2 + x + 1 = 64 + 5 ´ 8 + 2 Þ x =7 B + 1101 0011
1 0001 1001
3. (C) All are 2’s complement of 7
11001 Þ 00110 Discard the carry 1
+ 1
A - B = A + B, A 010 0 0110
00111 = 710 B + 0010 1101
1001 Þ 0110
0111 0011
+ 1
0111 = 710 B - A, B 1101 0011
111001 Þ 000110 A + 1011 1010
+ 1 1 1000 1101
000111 = 710
Discard the carry 1
4. (C) See a example - A - B = A + B, A 1011 1010
B + 0010 1101
42 in a byte 00101010
1110 0111
42in a word 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0
-42 in a byte 11010110
8. (B) 1110 = 10112
-42 in a word 1 1 1 1 1 1 1 1 1 1 0 1 0 1 1 0
0.3 2Fi-1 Bi Fi
Therefore (C) is correct.
0.6 0 0.6
5. (C) 4810 = 00110000 2
1.2 1 0.2
-4810 = 1100 1111
+ 1 0.4 0 0.4
11010000 0.8 0 0.8
1.6 1 0.6
6. (D) Here A , B are 1’s complement
A + B, A 01010110 Repeat from the second line 0.310 = 0.01001 2
B + 1110 1100
10100 0010 , 9. (C)
+ 1 b4 b3 b2 p3 b1 p2 p1
0100 0011 Received 1 1 0 1 1 0 0
B - A = B + A, B 1110 1100
C1* = b4 Å b2 Å b1 Å p1 = 0
A + 1010 1001
C2* = b4 Å b3 Å b1 Å p2 = 1
110010101
C3* = b4 Å b3 Å b2 Å p3 = 1
+ 1
10010110
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Number Systems & Boolean Algebra GATE EC BY RK Kanodia Chap 4.1
C3* C2* C1* = 110 which indicate position 6 in error = ( AA + AB)( A + B + C) = A( A + B + C) = A
Transmitted code 1001100. Therefore No gate is required to implement this
function.
10. (D) X = MNQ + M NQ + M NQ
= MQ + M NQ = Q( M + M N ) = Q( M + N ) 23. (A)
A B C ( A + BC) ( A + B)( A + C)
11. (A) The logic circuit can be modified as shown in fig.
S. 4.1.11 0 0 0 0 0
A 0 0 1 0 0
B
Z 0 1 0 0 0
C+D
E 0 1 1 1 1
Fig. S4.1.11
1 0 0 1 1
Now Z = AB + ( C + D) E 1 0 1 1 1
Fig. S 4.1.23
12. (D) You can see that input to last XNOR gate is
same. So output will be HIGH. 24. (B) X = ABC + ABC + ABC = BC + ABC
13. (D) Z = A + ( AB + BC) + C
25. (B) ( A + B)( B + C ) = ( AB)( BC) = ABC
= A + ( A + B + B + C) + C = A + B + C
( A + B)( B + C ) = ( A + B) + ( B + C) = A + B + C
ABC = A + B + C
( A + B)( B + C) = ( A + B) + ( B + C)
AB + BC + AC = A + B + B + C + A + C = A + B + C
= AB + B + C = A + B + C
14. (C) ( X + Y )( X + Y ) = XY + X Y From truth table Z = A + B + C
Thus (B) is correct.
( X + Y )( X + Y )( X + Y ) = ( X + Y )( XY + X Y )
= XY + XY = XY 26. (D) AC + BC = AC( B + B) + ( A + A) BC
= ABC + ABC + ABC + ABC
15. (B) Using duality
( A + B)( A + C)( B + C) = ( A + B)( A + C) 27. (D) F = A + AB + A BC + A B C( D + DE)
Thus (B) is correct option. = A + AB + A B( C + C( D + E))
= A + A( B + B( C + D + E)) = A + B + C + D + E
16. (B) Z = ( AB)( CD)( EF ) = AB + CD + EF
28. (B) A( B + C ( AB + AC)) = AB + AC ( AB × AC)
17. (A) X = ( A B + AB)( A + B) = ( AB + A B)( AB) = AB
= AB + AC[( A + B)( A + C)]
= AB + AC ( A + AC + AB + B C) = AB
18. (B) Y = ( A Å B) × C = ( AB + AC) × C
= ( AB + AB) + C = A B + AB + C
29. (C) ( AB ) × ( AB) = AB + AB = AB + AB
19. (C) Z = A( A + A) BC = ABC
30. (B) X Z + XZ = X ( XY + XY ) + X ( X Y + XY )
20. (A) Z = AB( B + C) = ABC = X ( XY + X Y ) + XY = XY + XY = Y
31. (A) X Å Y = X Y + XY = ( XY + XY ) = ( XY ) = X + Y
21. (A) Z = ( A + B ) × BC = ( AB) × BC = ABC
22. (A) A( A + B)( A + B + C)
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