Digital Electronics And Logic Design Set 3
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This set of Digital Electronics and Logic Design Multiple Choice Questions & Answers (MCQs) focuses on Digital Electronics And Logic Design Set 3
Q1 | Convert the octal number 7401 to Binary.
- 1.111e+11
- 1.1111e+11
- 1.111e+11
- 1.11e+11
Q2 | Find the hex sum of (93)16 + (DE)16 .
- (171)16
- (271)16
- (179)16
- (181)16
Q3 | Perform 2’s complement subtraction of (7)10 − (11)10 .
- 1100 (or -4)
- 1101 (or -5)
- 1011 (or -3)
- 1110 (or-6)
Q4 | What is the Gray equivalent of (25)10
- 1101
- 110101
- 10110
- 10101
Q5 | Simplify the Boolean expression F = C(B + C)(A + B + C).
- c
- bc
- abc
- a+bc
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\
Q7 | Simplify F = (ABC)'+( AB)'C+ A'BC'+ A(BC)'+ AB'C.
- ( a\ + b\ +c\ )
- ( a\ + b +c )
- ( a + b +c )
- ( a + b\+c\ )
Q8 | Determine the binary numbers represented by 25.5
- 11001.1
- 11011.101
- 10101.11
- 11001.0101
Q9 | Conversion of decimal number 10.625 into binary number:
- 1010.101
- 1110.101
- 1001.11
- 1001.101
Q10 | Conversion of fractional number 0.6875 into its equivalent binarynumber:
- 0.1011
- 0.1111
- 0.10111
- 0.0101
Q11 | Perform the following subtractions using 2’s complement method. 01000 – 01001
- 1
- 10
- 11
- 11110
Q12 | Subtraction of 01100-00011 using 2’s complement method. :
- 1001
- 1000
- 1010
- 110
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
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\
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.
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
Q17 | Add 648 and 487 in BCD code.
- 1135
- 1136
- 1235
- 1138
Q18 | (23.6)10 = (X)2 FIND X
- (10111.1001100)2
- (10101.1001100)2
- (10001.1001100)2
- (10111.1000011)2
Q19 | (65.535)10 =(X)16 FIND X
- (41.88f5c28)16.
- (42.88f5c28)16.
- (41.88f5c)16.
- (42.88f5c)16.
Q20 | Convert the decimal number 430 to Excess-3 code:
- 110110001
- 110110000
- 110110011
- 110100001
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
Q22 | Convert (2222)10 in Hexadecimal number.
- 8ae
- 8be
- 93c
- fff
Q23 | Divide ( 101110) 2 by ( 101)2.
- quotient -1001 remainder - 001
- quotient - 1000remainder - 001
- quotient - 1001remainder - 011
- quotient - 1001remainder-000
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)]’
Q25 | Perform following subtraction (i) 11001-10110 using 1’s complement
- 11
- 111
- 10
- 10011