Digital Electronics And Logic Design Set 3 MCQs

Q1 | Convert the octal number 7401 to Binary.

1.111e+11
1.1111e+11
1.111e+11
1.11e+11

Answer: 1.111e+11

Q2 | Find the hex sum of (93)16 + (DE)16 .

(171)16
(271)16
(179)16
(181)16

Answer: (171)16

Q3 | Perform 2’s complement subtraction of (7)10 − (11)10 .

1100 (or -4)
1101 (or -5)
1011 (or -3)
1110 (or-6)

Answer: 1100 (or -4)

Q4 | What is the Gray equivalent of (25)10

1101
110101
10110
10101

Answer: 10101

Q5 | Simplify the Boolean expression F = C(B + C)(A + B + C).

c
bc
abc
a+bc

Answer: c

Q6 | Simplify the following expression into sum of products using Karnaugh map F(A,B,C,D) = (1,3,4,5,6,7,9,12,13)

a\b+c\ d+ a\d+bc\
a\b\+c\ d\+a\d\+b\c\
a\b\+c\ d+ a\d+bc\
a\b+c\ d\+a\d\+bc\

Answer: a\b+c\ d+ a\d+bc\

Q7 | Simplify F = (ABC)'+( AB)'C+ A'BC'+ A(BC)'+ AB'C.

( a\ + b\ +c\ )
( a\ + b +c )
( a + b +c )
( a + b\+c\ )

Answer: ( a\ + b\ +c\ )

Q8 | Determine the binary numbers represented by 25.5

11001.1
11011.101
10101.11
11001.0101

Answer: 11001.1

Q9 | Conversion of decimal number 10.625 into binary number:

1010.101
1110.101
1001.11
1001.101

Answer: 1010.101

Q10 | Conversion of fractional number 0.6875 into its equivalent binarynumber:

0.1011
0.1111
0.10111
0.0101

Answer: 0.1011

Q11 | Perform the following subtractions using 2’s complement method. 01000 – 01001

1
10
11
11110

Answer: 1

Q12 | Subtraction of 01100-00011 using 2’s complement method. :

1001
1000
1010
110

Answer: 1001

Q13 | Minimize the logic functionY(A,B,C,D) = IZm(0,1,2,3,5,7,8,9,11,14) . UsingKarnaugh map.

abc d\ + a\ b\+ b\ c\ + b\ d+ a\d
abc d + a b+ b\ c\ + b\ d+ a\d
a\ b\ + b\ c\ + b\ d+ a\d
abc d\ + a\ b\ + b\ c\ + b\ d

Answer: abc d\ + a\ b\+ b\ c\ + b\ d+ a\d

Q14 | Simplify the given expression to its Sum of Products (SOP) form Y = (A + B)(A + (AB)')C + A'(B+C')+ A'B+ ABC

ac+ bc+ a\b + a\ c\
ac+ bc+ a\b
bc+ a\b + a\ c\
ac+ a\b + a\ c\

Answer: ac+ bc+ a\b + a\ c\

Q15 | Convert the decimal number 82.67 to its binary, hexadecimal and octal equivalents

(1010010.10101011)2; (52.ab)16 ;
(1010010.10101011)2; (52.ab)16 ;
(1010010.10101011)2; (52.ab)16 ;
(1010010.

Answer: (1010010.10101011)2; (52.ab)16 ;

Q16 | Add 20 and (-15) using 2’s complement.

(100100 )2 or(+4)10
(000100 )2or (-4)10
both (a) and (b)
none of the above

Answer: (100100 )2 or(+4)10

Q17 | Add 648 and 487 in BCD code.

1135
1136
1235
1138

Answer: 1135

Q18 | (23.6)10 = (X)2 FIND X

(10111.1001100)2
(10101.1001100)2
(10001.1001100)2
(10111.1000011)2

Answer: (10111.1001100)2

Q19 | (65.535)10 =(X)16 FIND X

(41.88f5c28)16.
(42.88f5c28)16.
(41.88f5c)16.
(42.88f5c)16.

Answer: (41.88f5c28)16.

Q20 | Convert the decimal number 430 to Excess-3 code:

110110001
110110000
110110011
110100001

Answer: 110110001

Q21 | Minimize the following logic function using K-maps F(A,B,C,D) = m(1,3,5,8,9,11,15) + d(2,13)

a b\ c\ + c\ d + b\d + ad
c\ d + b\d + ad
a b\ c\ + b\d+ ad
a b\ c\ + c\ d + b\d

Answer: a b\ c\ + c\ d + b\d + ad

Q22 | Convert (2222)10 in Hexadecimal number.

8ae
8be
93c
fff

Answer: 8ae

Q23 | Divide ( 101110) 2 by ( 101)2.

quotient -1001 remainder - 001
quotient - 1000remainder - 001
quotient - 1001remainder - 011
quotient - 1001remainder-000

Answer: quotient -1001 remainder - 001

Q24 | Minimise the logic function (POS Form) F A,B,C,D) = PI M (1, 2, 3, 8, 9,10, 11,14)× d (7, 15)

f=[(b+d’)+(b+ c’)’(a’+c’)+(a’+b)]’
f=[(b+d’)+(b+c’)’(‘a’+c’)+ (a’+b)]’
f=[(b+d’)+(b+c’)’(‘a’+c’)+ (a’+b)]’
f=[(b+d’)+ (b+c’)’(‘a’+c’)+(a’+b)]’

Answer: f=[(b+d’)+(b+ c’)’(a’+c’)+(a’+b)]’

Q25 | Perform following subtraction (i) 11001-10110 using 1’s complement

11
111
10
10011

Answer: 11