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MCQOPTIONS
This section includes 242 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.
| 201. |
Product-of-Sums expressions can be implemented using |
| A. | 2-level or-and logic circuits |
| B. | 2-level nor logic circuits |
| C. | 2-level xor logic circuits |
| D. | both 2-level or-and and nor logic circuits |
| Answer» E. | |
| 202. |
The prime implicant which has at least one element that is not present in any other implicant is known as |
| A. | essential prime implicant |
| B. | implicant |
| C. | complement |
| D. | prime complement |
| Answer» B. implicant | |
| 203. |
Each product term of a group, w’.x.y’ and w.y, represents the                         in that group. |
| A. | input |
| B. | pos |
| C. | sum-of-minterms |
| D. | sum of maxterms |
| Answer» D. sum of maxterms | |
| 204. |
There are              cells in a 4-variable K-map. |
| A. | 12 |
| B. | 16 |
| C. | 18 |
| D. | 8 |
| Answer» C. 18 | |
| 205. |
                           expressions can be implemented using either (1) 2-level AND-OR logic circuits or (2) 2-level NAND logic circuits. |
| A. | pos |
| B. | literals |
| C. | sop |
| D. | pos |
| Answer» D. pos | |
| 206. |
There are                            Minterms for 3 variables (a, b, c). |
| A. | 0 |
| B. | 2 |
| C. | 8 |
| D. | 1 |
| Answer» D. 1 | |
| 207. |
The K-map based Boolean reduction is based on the following Unifying Theorem: A + A’ = 1. |
| A. | impact |
| B. | non impact |
| C. | force |
| D. | complementarity |
| Answer» C. force | |
| 208. |
Maxterm is the sum of                      of the corresponding Minterm with its literal complemented. |
| A. | terms |
| B. | words |
| C. | numbers |
| D. | nibble |
| Answer» B. words | |
| 209. |
A product term containing all K variables of the function in either complemented or uncomplemented form is called a |
| A. | minterm |
| B. | maxterm |
| C. | midterm |
| D. | ∑ term |
| Answer» B. maxterm | |
| 210. |
The max term when X=Y=Z=1 is |
| A. | x’+y’+z’ |
| B. | xyz |
| C. | x’y’z’ |
| D. | x+y+z |
| Answer» B. xyz | |
| 211. |
The min term when X=Y=Z=0 is |
| A. | x’+y’+z’ |
| B. | xyz |
| C. | x’y’z’ |
| D. | x+y+z |
| Answer» D. x+y+z | |
| 212. |
The minterm of any expression is denoted by |
| A. | mt |
| B. | m |
| C. | m |
| D. | min |
| Answer» C. m | |
| 213. |
What is the complement of X’Y’Z? |
| A. | x+yz |
| B. | x’+y+’z’ |
| C. | x+y+z’ |
| D. | xyz’ |
| Answer» D. xyz’ | |
| 214. |
The complement term for X’.Y’.Z + X.Y will be |
| A. | xyz’+x’y’ |
| B. | (x+y+z’)(x’+y’) |
| C. | (x+y+z’)(x’+y) |
| D. | (x+y+z’)(x’+y) |
| Answer» C. (x+y+z’)(x’+y) | |
| 215. |
The number of literals in the expression F=X.Y’ + Z are |
| A. | 4 |
| B. | 3 |
| C. | 2 |
| D. | 1 |
| Answer» C. 2 | |
| 216. |
The general form for calculating the number of rows in a truth table is |
| A. | 2n |
| B. | 2n+1 |
| C. | 2n |
| D. | 2n+1 |
| Answer» D. 2n+1 | |
| 217. |
In the boolean function w=f(X,Y,Z), what is the RHS referred to as |
| A. | right hand side |
| B. | expression |
| C. | literals |
| D. | boolean |
| Answer» C. literals | |
| 218. |
The result of X+X.Y is X. |
| A. | true |
| B. | false |
| Answer» B. false | |
| 219. |
Boolean Function is of the form of |
| A. | truth values |
| B. | k=f(x,y,x) |
| C. | algebraic expression |
| D. | truth table |
| Answer» B. k=f(x,y,x) | |
| 220. |
How many full adders are required to construct an m-bit parallel adder? |
| A. | m/2 |
| B. | m |
| C. | m-1 |
| D. | m+1 |
| Answer» D. m+1 | |
| 221. |
Which of the following will give the sum of full adders as output? |
| A. | three point major circuit |
| B. | three bit parity checker |
| C. | three bit comparator |
| D. | three bit counter |
| Answer» E. | |
| 222. |
The number of full and half adders are required to add 16-bit number is |
| A. | 8 half adders, 8 full adders |
| B. | 1 half adders, 15 full adders |
| C. | 16 half adders, 0 full adders |
| D. | 4 half adders, 12 full adders |
| Answer» C. 16 half adders, 0 full adders | |
| 223. |
Odd parity of word can be conveniently tested by |
| A. | or gate |
| B. | and gate |
| C. | nand gate |
| D. | xor gate |
| Answer» E. | |
| 224. |
What is the minimum number of two input NAND gates used to perform the function of two input OR gates? |
| A. | one |
| B. | two |
| C. | three |
| D. | four |
| Answer» D. four | |
| 225. |
In which of the following gates the output is 1 if and only if at least one input is 1? |
| A. | and |
| B. | nor |
| C. | nand |
| D. | or |
| Answer» E. | |
| 226. |
The boolean function A + BC is a reduced form of |
| A. | ab + bc |
| B. | (a + b)(a + c) |
| C. | a’b + ab’c |
| D. | (a + c)b |
| Answer» C. a’b + ab’c | |
| 227. |
Simplify Y = AB’ + (A’ + B)C. |
| A. | ab’ + c |
| B. | ab + ac |
| C. | a’b + ac’ |
| D. | ab + a |
| Answer» B. ab + ac | |
| 228. |
The involution of A is equal to |
| A. | a |
| B. | a’ |
| C. | 1 |
| Answer» B. a’ | |
| 229. |
According to boolean law: A + 1 = ? |
| A. | 1 |
| B. | a |
| C. | a’ |
| Answer» B. a | |
| 230. |
(A + B)(A’ * B’) = ? |
| A. | 1 |
| B. | 0 |
| C. | ab |
| D. | ab’ |
| Answer» C. ab | |
| 231. |
The expression for Absorption law is given by |
| A. | a + ab = a |
| B. | a + ab = b |
| C. | ab + aa’ = a |
| D. | a + b = b + a |
| Answer» B. a + ab = b | |
| 232. |
DeMorgan’s theorem states that |
| A. | (ab)’ = a’ + b’ |
| B. | (a + b)’ = a’ * b |
| C. | a’ + b’ = a’b’ |
| D. | (ab)’ = a’ + b |
| Answer» B. (a + b)’ = a’ * b | |
| 233. |
In boolean algebra, the OR operation is performed by which properties? |
| A. | associative properties |
| B. | commutative properties |
| C. | distributive properties |
| D. | all of the mentioned |
| Answer» E. | |
| 234. |
The decimal equivalent of the excess-3 number 110010100011.01110101 is |
| A. | 970.42 |
| B. | 1253.75 |
| C. | 861.75 |
| D. | 1132.87 |
| Answer» B. 1253.75 | |
| 235. |
The excess-3 code for 597 is given by |
| A. | 100011001010 |
| B. | 100010100111 |
| C. | 010110010111 |
| D. | 010110101101 |
| Answer» B. 100010100111 | |
| 236. |
How many bits would be required to encode decimal numbers 0 to 9999 in straight binary codes? |
| A. | 12 |
| B. | 14 |
| C. | 16 |
| D. | 18 |
| Answer» C. 16 | |
| 237. |
Carry out BCD subtraction for (68) – (61) using 10’s complement method. |
| A. | 00000111 |
| B. | 01110000 |
| C. | 100000111 |
| D. | 011111000 |
| Answer» B. 01110000 | |
| 238. |
A three digit decimal number requires                  for representation in the conventional BCD format. |
| A. | 3 bits |
| B. | 6 bits |
| C. | 12 bits |
| D. | 24 bits |
| Answer» D. 24 bits | |
| 239. |
The decimal number 10 is represented in its BCD form as |
| A. | 10100000 |
| B. | 01010111 |
| C. | 00010000 |
| D. | 00101011 |
| Answer» D. 00101011 | |
| 240. |
When numbers, letters or words are represented by a special group of symbols, this process is called |
| A. | decoding |
| B. | encoding |
| C. | digitizing |
| D. | inverting |
| Answer» C. digitizing | |
| 241. |
Binary coded decimal is a combination of |
| A. | two binary digits |
| B. | three binary digits |
| C. | four binary digits |
| D. | five binary digits |
| Answer» D. five binary digits | |
| 242. |
If the decimal number is a fraction then its binary equivalent is obtained by                  the number continuously by 2. |
| A. | dividing |
| B. | multiplying |
| C. | adding |
| D. | subtracting |
| Answer» C. adding | |