MCQOPTIONS
Saved Bookmarks
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.
| 601. |
Karnaugh map is used for the purpose of |
| A. | reducing the electronic circuits used. |
| B. | to map the given boolean logic function. |
| C. | to minimize the terms in a boolean expression. |
| D. | to maximize the terms of a given a boolean expression . |
| Answer» D. to maximize the terms of a given a boolean expression . | |
| 602. |
A universal logic gate is one, which can be used to generate any logic function. Which of the following is a universal logic gate? |
| A. | or |
| B. | and |
| C. | xor |
| D. | nand |
| Answer» E. | |
| 603. |
The result of adding hexadecimal number A6 to 3A is |
| A. | dd |
| B. | e0 |
| C. | f0 |
| D. | ef |
| Answer» C. f0 | |
| 604. |
When an input signal A=11001 is applied to a NOT gate serially, its output signal is |
| A. | 111 |
| B. | 110 |
| C. | 10101 |
| D. | 11001 |
| Answer» C. 10101 | |
| 605. |
The excess-3 code of decimal 7 is represented by |
| A. | 1100 |
| B. | 1001 |
| C. | 1011 |
| D. | 1010 |
| Answer» E. | |
| 606. |
The binary equivalent of (FA)16 is |
| A. | 1010 1111 |
| B. | 1111 1010 |
| C. | 10110011 |
| D. | none of these |
| Answer» C. 10110011 | |
| 607. |
How many two-input AND and OR gates are required to realize Y=CD+EF+G |
| A. | 22 |
| B. | 23 |
| C. | 33 |
| D. | none of these |
| Answer» B. 23 | |
| 608. |
The decimal equivalent of ( 1100)2 is |
| A. | 12 |
| B. | 16 |
| C. | 18 |
| D. | 20 |
| Answer» B. 16 | |
| 609. |
Convert decimal 153 to octal. Equivalent in octal will be |
| A. | (231)8 |
| B. | ( 331) 8 |
| C. | ( 431) 8 . |
| D. | none of these. |
| Answer» B. ( 331) 8 | |
| 610. |
The NOR gate output will be low if the two inputs are |
| A. | 11 |
| B. | 1 |
| C. | 10 |
| D. | all |
| Answer» E. | |
| 611. |
The number 140 in octal is equivalent to |
| A. | (96)10 . |
| B. | ( 86) 10 |
| C. | (90) 10 . |
| D. | none of these. |
| Answer» B. ( 86) 10 | |
| 612. |
The octal equivalent of (247) 10 is |
| A. | ( 252) 8 |
| B. | (350) 8 |
| C. | ( 367) 8 |
| D. | ( 400) 8 |
| Answer» D. ( 400) 8 | |
| 613. |
The hexadecimal number for (95.5)10 is |
| A. | (5f.8) 16 |
| B. | (9a.b) 16 |
| C. | ( 2e.f) 16 |
| D. | ( 5a.4) 16 |
| Answer» B. (9a.b) 16 | |
| 614. |
How many AND gates are required to realize Y = CD+EF+G |
| A. | 4 |
| B. | 5 |
| C. | 3 |
| D. | 2 |
| Answer» E. | |
| 615. |
The excess 3 code of decimal number 26 is |
| A. | 0100 1001 |
| B. | 1011001 |
| C. | 1000 1001 |
| D. | 1001101 |
| Answer» C. 1000 1001 | |
| 616. |
1’s complement representation of decimal number of -17 by using 8 bit |
| A. | 1110 1110 |
| B. | 1101 1101 |
| C. | 1100 1100 |
| D. | 0001 0001 |
| Answer» B. 1101 1101 | |
| 617. |
The decimal equivalent of Binary number 11010 is |
| A. | 26 |
| B. | 36 |
| C. | 16 |
| D. | 23 |
| Answer» B. 36 | |
| 618. |
When signed numbers are used in binary arithmetic, then which one of the following notations would have unique representation for zero. |
| A. | sign- magnitude. |
| B. | 1’s complement. |
| C. | 2’s complement. |
| D. | 9’s compleme nt. |
| Answer» B. 1’s complement. | |
| 619. |
DeMorgan’s first theorem shows the equivalence of |
| A. | or gate and exclusive or gate. |
| B. | nor gate and bubbled and gate. |
| C. | nor gate and nand gate. |
| D. | nand gate and not gate |
| Answer» C. nor gate and nand gate. | |
| 620. |
-8 is equal to signed binary number |
| A. | 10001000 |
| B. | oooo1000 |
| C. | 10000000 |
| D. | 11000000 |
| Answer» B. oooo1000 | |
| 621. |
The code where all successive numbers differ from their preceding number by single bit is |
| A. | binary code. |
| B. | bcd. |
| C. | excess – 3. |
| D. | gray. |
| Answer» E. | |
| 622. |
When simplified with Boolean Algebra (x + y)(x + z) simplifies to |
| A. | x |
| B. | x + x(y + z) |
| C. | x(1 + yz) |
| D. | x + yz |
| Answer» E. | |
| 623. |
The 2’s complement of the number 1101101 is |
| A. | 101110 |
| B. | 111110 |
| C. | 110010 |
| D. | 10011 |
| Answer» E. | |
| 624. |
The Boolean expression A.B+ A.B+ A.B is equivalent to |
| A. | a + b |
| B. | a\.b |
| C. | (a + b)\ |
| D. | a.b |
| Answer» B. a\.b | |
| 625. |
The hexadecimal number ‘A0’ has the decimal value equivalent to |
| A. | 80 |
| B. | 256 |
| C. | 100 |
| D. | 160 |
| Answer» E. | |
| 626. |
The simplification of the Boolean expression (A'BC')'+ (AB'C)' is |
| A. | 0 |
| B. | 1 |
| C. | a |
| D. | bc |
| Answer» C. a | |
| 627. |
( 734)8 =( )16 |
| A. | c 1 d |
| B. | d c 1 |
| C. | 1 c d |
| D. | 1 d c |
| Answer» E. | |
| 628. |
The decimal equivalent of hex number 1A53 is |
| A. | 6793 |
| B. | 6739 |
| C. | 6973 |
| D. | 6379 |
| Answer» C. 6973 | |
| 629. |
What is the binary equivalent of the decimal number 1011 |
| A. | 1111110111 |
| B. | 1111000111 |
| C. | 1111110011 |
| D. | 1.11e+09 |
| Answer» D. 1.11e+09 | |
| 630. |
The gray code equivalent of (1011)2 is |
| A. | 1101 |
| B. | 1010 |
| C. | 1111 |
| D. | 1110 |
| Answer» E. | |
| 631. |
What is the binary equivalent of the Octal number 367 |
| A. | 11110111 |
| B. | 11100111 |
| C. | 110001101 |
| D. | 10111011 1 |
| Answer» B. 11100111 | |
| 632. |
What is the binary equivalent of the Hexadecimal number 368 |
| A. | 1111101000 |
| B. | 1101101000 |
| C. | 1101111000 |
| D. | 1.11e+09 |
| Answer» C. 1101111000 | |
| 633. |
What is the binary equivalent of the decimal number 368 |
| A. | 101110000 |
| B. | 110110000 |
| C. | 111010000 |
| D. | 1.11e+08 |
| Answer» B. 110110000 | |
| 634. |
A binary digit is called a |
| A. | bit |
| B. | character |
| C. | number |
| D. | byte |
| Answer» B. character | |
| 635. |
The NAND gate output will be low if the two inputs are |
| A. | 0 |
| B. | 1 |
| C. | 10 |
| D. | 11 |
| Answer» E. | |
| 636. |
The Hexadecimal number equivalent of (4057.06)8 is |
| A. | 82f.027 |
| B. | 82f.014 |
| C. | 82f.937 |
| D. | 83f.014 |
| Answer» C. 82f.937 | |
| 637. |
The answer of the operation (10111)2*(1110)2 in hex equivalence is |
| A. | 150 |
| B. | 241 |
| C. | 142 |
| D. | 1.01e+08 |
| Answer» D. 1.01e+08 | |
| 638. |
Negative numbers cannot be represented in |
| A. | signed magnitude form |
| B. | 1’s complement form |
| C. | 2’s complement form |
| D. | none of the above |
| Answer» E. | |
| 639. |
The negative numbers in the binary system can be represented by |
| A. | sign magnitude |
| B. | 2\s complement |
| C. | 1\s complement |
| D. | all of the above |
| Answer» B. 2\s complement | |
| 640. |
What is decimal equivalent of (11011.1000)2 ? |
| A. | 22 |
| B. | 22.2 |
| C. | 20.2 |
| D. | 27.5 |
| Answer» E. | |
| 641. |
Logic X-OR operation of (4ACO)H & (B53F)H results |
| A. | aacb |
| B. | 0 |
| C. | abcd |
| D. | ffff |
| Answer» E. | |
| 642. |
A three input NOR gate gives logic high output only when |
| A. | one input is high |
| B. | one input is low |
| C. | two input are low |
| D. | all input are low |
| Answer» E. | |
| 643. |
The absorption law in Boolean algebra say that |
| A. | x + x = x |
| B. | x . y = x |
| C. | x + x . y = x |
| D. | none of the above |
| Answer» D. none of the above | |
| 644. |
The hexadecimal number equivalent to (1762.46)8 is |
| A. | 3f2.89 |
| B. | 3f2.98 |
| C. | 2f3.89 |
| D. | 2f3.98 |
| Answer» C. 2f3.89 | |
| 645. |
AB+(A+B)’ is equivalent to |
| A. | a ex-nor b |
| B. | a ex or b |
| C. | (a+b)a |
| D. | (a+b)b |
| Answer» B. a ex or b | |
| 646. |
The 2s compliment form (Use 6 bit word) of the number 1010 is |
| A. | 111100 |
| B. | 110110 |
| C. | 110111 |
| D. | 1011 |
| Answer» C. 110111 | |