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MCQOPTIONS
This section includes 646 Mcqs, each offering curated multiple-choice questions to sharpen your Computer Science Engineering (CSE) knowledge and support exam preparation. Choose a topic below to get started.
| 551. |
which of the following rules states that if one input of an AND gate is always 1, the output is equal to the other input? |
| A. | a +1 =1 |
| B. | a +a =a |
| C. | a.a = a |
| D. | a.1= a |
| Answer» D. a.1= a | |
| 552. |
The Unsigned Binary representation can only represent positive binary numbers |
| A. | true |
| B. | false |
| C. | both (a) and (b) |
| D. | none of above |
| Answer» B. false | |
| 553. |
The difference of 111 - 001 equals |
| A. | 100 |
| B. | 111 |
| C. | 1 |
| D. | 110 |
| Answer» E. | |
| 554. |
The output of a logic gate is 1 when all its inputs are at logic 0. the gate is either |
| A. | nand or an xo |
| B. | or or an xn |
| C. | o an and or xor |
| D. | a nor or an xnor |
| Answer» E. | |
| 555. |
2's complement of any binary number can be calculated by |
| A. | adding 1\s complement twice |
| B. | adding 1 to 1\s complement |
| C. | subtracting 1 from 1\s complement. |
| D. | calculating 1\s compleme nt and inverting most significant bit |
| Answer» C. subtracting 1 from 1\s complement. | |
| 556. |
The complement of a variable is always |
| A. | 1 |
| B. | 0 |
| C. | inverse |
| D. | none |
| Answer» D. none | |
| 557. |
Sum-of-Weights method is used |
| A. | to convert from one number system to other |
| B. | to encode data |
| C. | to decode data |
| D. | to convert from serial to parralel data |
| Answer» B. to encode data | |
| 558. |
The simplified form of the Boolean expression (X+Y+XY)(X+Z) is |
| A. | x + y + zx + y |
| B. | xy – yz |
| C. | x + yz |
| D. | xz + y |
| Answer» D. xz + y | |
| 559. |
The octal equivalent of hexadecimal (A.B)16 is |
| A. | 47.21 |
| B. | 12.74 |
| C. | 12.71 |
| D. | 17.21 |
| Answer» C. 12.71 | |
| 560. |
(2FAOC)16 is equivalent to |
| A. | (195 084)10 |
| B. | (001011111 010 0000 1100)2 |
| C. | both (a) and (b) |
| D. | none of these |
| Answer» C. both (a) and (b) | |
| 561. |
What is decimal equivalent of BCD 11011.1100 ? |
| A. | 22 |
| B. | 22.2 |
| C. | 20.2 |
| D. | 21.2 |
| Answer» C. 20.2 | |
| 562. |
12-bit 2’s complement of –73.75 is |
| A. | 01001001.110 0 |
| B. | 11001001.11 00 |
| C. | 10110110.01 00 |
| D. | 10110110. 1100 |
| Answer» D. 10110110. 1100 | |
| 563. |
The decimal number equivalent of (4057.06)8 is |
| A. | 2095.75 |
| B. | 2095.075 |
| C. | 2095.937 |
| D. | 2095.094 |
| Answer» E. | |
| 564. |
Reduce the following equation using k-map Y = B C' D'+ A' B C' D+ A B C' D+ A' B C D+ A B C D |
| A. | bc’ + bd |
| B. | bc’ + bd+a |
| C. | bc’ + bd + ac |
| D. | bc’ + bd + ad |
| Answer» B. bc’ + bd+a | |
| 565. |
8-bit 1’s complement form of –77.25 is |
| A. | 1001101.01 |
| B. | 10110010.10 11 |
| C. | 01001101.00 10 |
| D. | 10110010 |
| Answer» C. 01001101.00 10 | |
| 566. |
In computers, subtraction is generally carried out by |
| A. | 9’s complement |
| B. | 10’s complement |
| C. | 1’s complement |
| D. | 2’s compleme nt |
| Answer» E. | |
| 567. |
Convert (177.25)10 to octal. |
| A. | (261.2)8 |
| B. | (260.2)8 |
| C. | (361.2)8 |
| D. | (251.2)8 |
| Answer» B. (260.2)8 | |
| 568. |
Write the expression for Boolean function F (A, B, C) = m (1,4,5,6,7) in standard POS form. |
| A. | = (a+b+c)(a+b\ +c)(a+b\ +c\ ) |
| B. | = (a+b\ +c)(a+b\ +c\ ) |
| C. | = (a+b+c)(a+b \ +c) |
| D. | = (a+b+c)(a +b\ +c\ ) |
| Answer» B. = (a+b\ +c)(a+b\ +c\ ) | |
| 569. |
Reduce the following equation using k-map Y = (ABC)'+ A(CD)'+ AB'+ ABCD'+ (AB)'C |
| A. | b\+(ad)\ |
| B. | b\ |
| C. | (ad)\ |
| D. | b\+ad |
| Answer» B. b\ | |
| 570. |
Convert the decimal number 45678 to its hexadecimal equivalent number. |
| A. | (b26e)16 |
| B. | (a26e)16 |
| C. | (b26b)16 |
| D. | (b32e)16 |
| Answer» B. (a26e)16 | |
| 571. |
Perform following subtraction(ii) 11011-11001 using 2’s complement |
| A. | 10 |
| B. | 111 |
| C. | 11 |
| D. | 10011 |
| Answer» B. 111 | |
| 572. |
Perform following subtraction (i) 11001-10110 using 1’s complement |
| A. | 11 |
| B. | 111 |
| C. | 10 |
| D. | 10011 |
| Answer» B. 111 | |
| 573. |
Minimise the logic function (POS Form) F A,B,C,D) = PI M (1, 2, 3, 8, 9, 10, 11,14)× d (7, 15) |
| A. | f=[(b+d’)+(b+ c’)’(a’+c’)+(a’ +b)]’ |
| B. | f=[(b+d’)+(b +c’)’(‘a’+c’)+ (a’+b)]’ |
| C. | f=[(b+d’)+(b +c’)’(‘a’+c’)+ (a’+b)]’ |
| D. | f=[(b+d’)+ (b+c’)’(‘a’ +c’)+(a’+b )]’ |
| Answer» B. f=[(b+d’)+(b +c’)’(‘a’+c’)+ (a’+b)]’ | |
| 574. |
Divide ( 101110) 2 by ( 101)2. |
| A. | quotient -1001 remainder - 001 |
| B. | quotient - 1000 remainder - 001 |
| C. | quotient - 1001 remainder - 011 |
| D. | quotient - 1001 remainder -000 |
| Answer» B. quotient - 1000 remainder - 001 | |
| 575. |
Convert (2222)10 in Hexadecimal number. |
| A. | 8ae |
| B. | 8be |
| C. | 93c |
| D. | fff |
| Answer» B. 8be | |
| 576. |
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. | a b\ c\ + c\ d + b\d + ad |
| B. | c\ d + b\d + ad |
| C. | a b\ c\ + b\d + ad |
| D. | a b\ c\ + c\ d + b\d |
| Answer» B. c\ d + b\d + ad | |
| 577. |
Convert the decimal number 430 to Excess-3 code: |
| A. | 110110001 |
| B. | 110110000 |
| C. | 110110011 |
| D. | 11010000 1 |
| Answer» B. 110110000 | |
| 578. |
(65.535)10 =(X)16 FIND X |
| A. | (41.88f5c28)16 . |
| B. | (42.88f5c28 )16. |
| C. | (41.88f5c)16. |
| D. | (42.88f5c )16. |
| Answer» B. (42.88f5c28 )16. | |
| 579. |
Add 648 and 487 in BCD code. |
| A. | 1135 |
| B. | 1136 |
| C. | 1235 |
| D. | 1138 |
| Answer» B. 1136 | |
| 580. |
Add 20 and (-15) using 2’s complement. |
| A. | (100100 )2 or (+4)10 |
| B. | (000100 )2 or (-4)10 |
| C. | both (a) and (b) |
| D. | none of the above |
| Answer» B. (000100 )2 or (-4)10 | |
| 581. |
Convert the decimal number 82.67 to its binary, hexadecimal and octal equivalents |
| A. | (1010010.1010 1011)2; (52.ab)16 ; |
| B. | (1010010.10 101011)2; (52.ab)16 ; |
| C. | (1010010.10 101011)2; (52.ab)16 ; |
| D. | (1010010. |
| Answer» B. (1010010.10 101011)2; (52.ab)16 ; | |
| 582. |
(23.6)10 = (X)2 FIND X |
| A. | (10111.100110 0)2 |
| B. | (10101.1001 100)2 |
| C. | (10001.1001 100)2 |
| D. | (10111.10 00011)2 |
| Answer» B. (10101.1001 100)2 | |
| 583. |
Simplify the given expression to its Sum of Products (SOP) form Y = (A + B)(A + (AB)')C + A'(B+C')+ A'B+ ABC |
| A. | ac+ bc+ a\b + a\ c\ |
| B. | ac+ bc+ a\b |
| C. | bc+ a\b + a\ c\ |
| D. | ac+ a\b + a\ c\ |
| Answer» B. ac+ bc+ a\b | |
| 584. |
Subtraction of 01100-00011 using 2’s complement method. : |
| A. | 1001 |
| B. | 1000 |
| C. | 1010 |
| D. | 110 |
| Answer» B. 1000 | |
| 585. |
Minimize the logic functionY(A,B,C,D) = IZm(0,1,2,3,5,7,8,9,11,14) . Using Karnaugh map. |
| A. | abc d\ + a\ b\ + b\ c\ + b\ d+ a\d |
| B. | abc d + a b + b\ c\ + b\ d+ a\d |
| C. | a\ b\ + b\ c\ + b\ d+ a\d |
| D. | abc d\ + a\ b\ + b\ c\ + b\ d |
| Answer» B. abc d + a b + b\ c\ + b\ d+ a\d | |
| 586. |
Conversion of fractional number 0.6875 into its equivalent binary number: |
| A. | 0.1011 |
| B. | 0.1111 |
| C. | 0.10111 |
| D. | 0.0101 |
| Answer» B. 0.1111 | |
| 587. |
Perform the following subtractions using 2’s complement method. 01000 – 01001 |
| A. | 1 |
| B. | 10 |
| C. | 11 |
| D. | 11110 |
| Answer» B. 10 | |
| 588. |
Simplify F = (ABC)'+( AB)'C+ A'BC'+ A(BC)'+ AB'C. |
| A. | ( a\ + b\ +c\ ) |
| B. | ( a\ + b +c ) |
| C. | ( a + b +c ) |
| D. | ( a + b\ +c\ ) |
| Answer» B. ( a\ + b +c ) | |
| 589. |
Conversion of decimal number 10.625 into binary number: |
| A. | 1010.101 |
| B. | 1110.101 |
| C. | 1001.11 |
| D. | 1001.101 |
| Answer» B. 1110.101 | |
| 590. |
Determine the binary numbers represented by 25.5 |
| A. | 11001.1 |
| B. | 11011.101 |
| C. | 10101.11 |
| D. | 11001.010 1 |
| Answer» B. 11011.101 | |
| 591. |
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. | a\b+c\ d+ a\d+bc\ |
| B. | a\b\+c\ d\+ a\d\+b\c\ |
| C. | a\b\+c\ d+ a\d+bc\ |
| D. | a\b+c\ d\+ a\d\+bc\ |
| Answer» B. a\b\+c\ d\+ a\d\+b\c\ | |
| 592. |
Find the hex sum of (93)16 + (DE)16 . |
| A. | (171)16 |
| B. | (271)16 |
| C. | (179)16 |
| D. | (181)16 |
| Answer» B. (271)16 | |
| 593. |
Simplify the Boolean expression F = C(B + C)(A + B + C). |
| A. | c |
| B. | bc |
| C. | abc |
| D. | a+bc |
| Answer» B. bc | |
| 594. |
What is the Gray equivalent of (25)10 |
| A. | 1101 |
| B. | 110101 |
| C. | 10110 |
| D. | 10101 |
| Answer» E. | |
| 595. |
Perform 2’s complement subtraction of (7)10 − (11)10 . |
| A. | 1100 (or -4) |
| B. | 1101 (or -5) |
| C. | 1011 (or -3) |
| D. | 1110 (or -6) |
| Answer» B. 1101 (or -5) | |
| 596. |
Which of following are known as universal gates |
| A. | nand & nor |
| B. | and & or. |
| C. | xor & or. |
| D. | none. |
| Answer» B. and & or. | |
| 597. |
Convert the octal number 7401 to Binary. |
| A. | 1.111e+11 |
| B. | 1.1111e+11 |
| C. | 1.111e+11 |
| D. | 1.11e+11 |
| Answer» B. 1.1111e+11 | |
| 598. |
How many two input AND gates and two input OR gates are required to realize Y = BD+CE+AB |
| A. | 11 |
| B. | 42 |
| C. | 32 |
| D. | 23 |
| Answer» B. 42 | |
| 599. |
The decimal equivalent of Binary number 10101 is |
| A. | 21 |
| B. | 31 |
| C. | 26 |
| D. | 28 |
| Answer» B. 31 | |
| 600. |
The 2’s complement of the number 1101110 is |
| A. | 10001 |
| B. | 10001 |
| C. | 10010 |
| D. | none |
| Answer» D. none | |