

McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Computer Science Engineering (CSE) .
251. |
A Poset in which every pair of elements has both a least upper bound and a greatest lower bound is termed as |
A. | sublattice |
B. | lattice |
C. | trail |
D. | walk |
Answer» B. lattice |
252. |
If every two elements of a poset are comparable then the poset is called |
A. | sub ordered poset |
B. | totally ordered poset |
C. | sub lattice |
D. | semigroup |
Answer» B. totally ordered poset |
253. |
A has a greatest element and a least element which satisfy 0<=a<=1 for every a in the lattice(say, L). |
A. | semilattice |
B. | join semilattice |
C. | meet semilattice |
D. | bounded lattice |
Answer» D. bounded lattice |
254. |
The graph given below is an example of |
A. | non-lattice poset |
B. | semilattice |
C. | partial lattice |
D. | bounded lattice |
Answer» A. non-lattice poset |
255. |
A sublattice(say, S) of a lattice(say, L) is a convex sublattice of L if |
A. | x>=z, where x in s implies z in s, for every element x, y in l |
B. | x=y and y<=z, where x, y in s implies z in s, for every element x, y, z in l |
C. | x<=y<=z, where x, y in s implies z in s, for every element x, y, z in l |
D. | x=y and y>=z, where x, y in s implies z in s, for every element x, y, z in l |
Answer» C. x<=y<=z, where x, y in s implies z in s, for every element x, y, z in l |
256. |
The graph is the smallest non-modular lattice N5. A lattice is if and only if it does not have a isomorphic to N5. |
A. | non-modular, complete lattice |
B. | moduler, semilattice |
C. | non-modular, sublattice |
D. | modular, sublattice |
Answer» D. modular, sublattice |
257. |
Every poset that is a complete semilattice must always be a |
A. | sublattice |
B. | complete lattice |
C. | free lattice |
D. | partial lattice |
Answer» B. complete lattice |
258. |
A free semilattice has the property. |
A. | intersection |
B. | commutative and associative |
C. | identity |
D. | universal |
Answer» D. universal |
259. |
Algebra of logic is termed as |
A. | numerical logic |
B. | boolean algebra |
C. | arithmetic logic |
D. | boolean number |
Answer» C. arithmetic logic |
260. |
What is the definition of Boolean functions? |
A. | an arithmetic function with k degrees such that f:y–>yk |
B. | a special mathematical function with n degrees such that f:yn–>y |
C. | an algebraic function with n degrees such that f:xn–>x |
D. | a polynomial function with k degrees such that f:x2–>xn |
Answer» B. a special mathematical function with n degrees such that f:yn–>y |
261. |
F(X,Y,Z,M) = X`Y`Z`M`. The degree of the function is |
A. | 2 |
B. | 5 |
C. | 4 |
D. | 1 |
Answer» C. 4 |
262. |
Which of the following is a Simplification law? |
A. | m.(~m+n) = m.n |
B. | m+(n.o) = (m+n)(m+o) |
C. | ~(m+n) = ~m.~n |
D. | m.(n.o) = (m.n.o) |
Answer» A. m.(~m+n) = m.n |
263. |
What are the canonical forms of Boolean Expressions? |
A. | or and xor |
B. | nor and xnor |
C. | max and min |
D. | som and pom |
Answer» D. som and pom |
264. |
Which of the following is/are the universal logic gates? |
A. | or and nor |
B. | and |
C. | nand and nor |
D. | not |
Answer» C. nand and nor |
265. |
The logic gate that provides high output for same inputs |
A. | not |
B. | x-nor |
C. | and |
D. | xor |
Answer» B. x-nor |
266. |
The of all the variables in direct or complemented from is a maxterm. |
A. | addition |
B. | product |
C. | moduler |
D. | subtraction |
Answer» A. addition |
267. |
What is the use of Boolean identities? |
A. | minimizing the boolean expression |
B. | maximizing the boolean expression |
C. | to evaluate a logical identity |
D. | searching of an algebraic expression |
Answer» A. minimizing the boolean expression |
268. |
Inversion of single bit input to a single bit output using |
A. | not gate |
B. | nor gate |
C. | and gate |
D. | nand gate |
Answer» A. not gate |
269. |
There are numbers of Boolean functions of degree n. |
A. | n |
B. | 2(2*n) |
C. | n3 |
D. | n(n*2) |
Answer» B. 2(2*n) |
270. |
A is a Boolean variable. |
A. | literal |
B. | string |
C. | keyword |
D. | identifier |
Answer» A. literal |
271. |
Minimization of function F(A,B,C) = A*B*(B+C) is |
A. | ac |
B. | b+c |
C. | b` |
D. | ab |
Answer» D. ab |
272. |
The set for which the Boolean function is functionally complete is |
A. | {*, %, /} |
B. | {., +, -} |
C. | {^, +, -} |
D. | {%, +, *} |
Answer» B. {., +, -} |
273. |
(X+Y`)(X+Z) can be represented by |
A. | (x+y`z) |
B. | (y+x`) |
C. | xy` |
D. | (x+z`) |
Answer» A. (x+y`z) |
274. |
is a disjunctive normal form. |
A. | product-of-sums |
B. | product-of-subtractions |
C. | sum-of-products |
D. | sum-of-subtractions |
Answer» C. sum-of-products |
275. |
a ⊕ b = |
A. | (a+b)(a`+b`) |
B. | (a+b`) |
C. | b` |
D. | a` + b` |
Answer» A. (a+b)(a`+b`) |
276. |
Find the simplified expression A’BC’+AC’. |
A. | b |
B. | a+c |
C. | (a+b)c’ |
D. | b’c |
Answer» C. (a+b)c’ |
277. |
Evaluate the expression: (X + Z)(X + XZ’) + XY + Y. |
A. | xy+z’ |
B. | y+xz’+y’z |
C. | x’z+y |
D. | x+y |
Answer» D. x+y |
278. |
Simplify the expression: A’(A + BC) + (AC + B’C). |
A. | (ab’c+bc’) |
B. | (a’b+c’) |
C. | (a+ bc) |
D. | ac |
Answer» D. ac | |
Explanation: = A’(A + BC) + (AC + B’C) = A’A + A’BC + AC + B’C = A’BC + C(A + B’) = C(A’B + A + B’) = C(A + B + B’) = C(A + 1) = AC. |
279. |
What is the simplification value of MN(M + N’) + M(N + N’)? |
A. | m |
B. | mn+m’n’ c) (1+m) |
C. | d |
D. | m+n’ |
Answer» B. mn+m’n’ c) (1+m) |
280. |
Simplify the expression XZ’ + (Y + Y’Z) + XY. TOPIC 5.5 MINIMIZATION OF BOOLEAN ALGEBRA |
A. | (1+xy’) |
B. | yz + xy’ + z’ |
C. | (x + y +z) |
D. | xy’+ z’ |
Answer» C. (x + y +z) |
281. |
Find the simplified term Y’ (X’ + Y’) (X + X’Y)? |
A. | xy’ |
B. | x’y |
C. | x + y |
D. | x’y’ |
Answer» A. xy’ |
282. |
If an expression is given that x+x’y’z=x+y’z, find the minimal expression of the function F(x,y,z) = x+x’y’z+yz? |
A. | y’ + z |
B. | xz + y’ |
C. | x + z |
D. | x’ + y |
Answer» C. x + z |
283. |
Simplify the expression: XY’ + X’ + Y’X’. |
A. | x’ + y |
B. | xy’ |
C. | (xy)’ |
D. | y’ + x |
Answer» C. (xy)’ |
284. |
Minimize the Boolean expression using Boolean identities: A′B+ABC′+BC’+AB′C′. |
A. | b(ac)’ + ac’ |
B. | ac’ + b’ |
C. | abc + b’ + c |
D. | bc’ + a’b |
Answer» A. b(ac)’ + ac’ |
285. |
Minimize the following Boolean expression using Boolean identities. F(A,B,C) = (A+BC’)(AB’+C) |
A. | a + b + c’ |
B. | ac’ + b |
C. | b + ac |
D. | a(b’ + c) |
Answer» D. a(b’ + c) |
286. |
Which of the following statement is a proposition? |
A. | Get me a glass of milkshake |
B. | God bless you! |
C. | What is the time now? |
D. | The only odd prime number is 2 |
Answer» D. The only odd prime number is 2 | |
Explanation: Only this statement has got the truth value which is false. |
287. |
The truth value of ‘4+3=7 or 5 is not prime’. |
A. | False |
B. | True |
Answer» B. True | |
Explanation: Compound statement with ‘or’ is true when either of the statement is true. Here the first part of the statement is true, hence the whole is true. |
288. |
Which of the following option is true? |
A. | If the Sun is a planet, elephants will fly |
B. | 3 +2 = 8 if 5-2 = 7 |
C. | 1 > 3 and 3 is a positive integer |
D. | -2 > 3 or 3 is a negative integer |
Answer» A. If the Sun is a planet, elephants will fly | |
Explanation: Hypothesis is false, thus the whole statement is true. |
289. |
What is the value of x after this statement, assuming the initial value of x is 5? ‘If x equals to one then x=x+2 else x=0’. |
A. | 1 |
B. | 3 |
C. | 0 |
D. | 2 |
Answer» C. 0 | |
Explanation: If condition is false so value decided according to else condition. |
290. |
Let P: I am in Bangalore.; Q: I love cricket.; then q -> p(q implies p) is? |
A. | If I love cricket then I am in Bangalore |
B. | If I am in Bangalore then I love cricket |
C. | I am not in Bangalore |
D. | I love cricket |
Answer» A. If I love cricket then I am in Bangalore | |
Explanation: Q is hypothesis and P is conclusion. So the compound statement will be if hypothesis then conclusion. |
291. |
Let P: If Sahil bowls, Saurabh hits a century.; Q: If Raju bowls, Sahil gets out on first ball. Now if P is true and Q is false then which of the following can be true? |
A. | Raju bowled and Sahil got out on first ball |
B. | Raju did not bowled |
C. | Sahil bowled and Saurabh hits a century |
D. | Sahil bowled and Saurabh got out |
Answer» C. Sahil bowled and Saurabh hits a century | |
Explanation: Either hypothesis should be false or both (hypothesis and conclusion) should be true. |
292. |
The truth value ‘9 is prime then 3 is even’. |
A. | False |
B. | TTru |
Answer» B. TTru | |
Explanation: The first part of the statement is false, hence whole is true. |
293. |
If there are n distinct components in a statement then there are _______ combinations of values in the truth table. |
A. | 2^n |
B. | n+1 |
C. | n |
D. | n+2 |
Answer» A. 2^n |
294. |
If P then Q is called _________ statement |
A. | Conjunction |
B. | disjunction |
C. | conditional |
D. | bi conditional |
Answer» C. conditional |
295. |
(P->Q)-> (^Q) is __________. |
A. | not a well formed formula |
B. | tautology |
C. | contradiction |
D. | well formed formula |
Answer» A. not a well formed formula |
296. |
A relation R in a set X is symmetric if ________. |
A. | xRy, yRz => xRz. |
B. | xRy |
C. | xRy=>yRx |
D. | xRx |
Answer» C. xRy=>yRx |
297. |
If a relation is reflexive, then all the diagonal entries in the relation matrix must be________. |
A. | 0 |
B. | 1 |
C. | 2 |
D. | -1 |
Answer» B. 1 |
298. |
If R is reflexive, symmetric and transitive then the relation is said to be ________. |
A. | Binary relation |
B. | Compatibility relation |
C. | Equivalence relation |
D. | Partial order relation |
Answer» C. Equivalence relation |
299. |
S -> aAB, AB -> bB, B -> b, A -> aB satisfies ___________ type of grammar |
A. | 0 |
B. | 1 |
C. | 0,1 |
D. | 2 |
Answer» C. 0,1 |
300. |
If there are more than 2 LMD for a string then it is said to be ___________. |
A. | Ambigious |
B. | unambigious |
C. | language |
D. | finite state automata |
Answer» A. Ambigious |
Done Studing? Take A Test.
Great job completing your study session! Now it's time to put your knowledge to the test. Challenge yourself, see how much you've learned, and identify areas for improvement. Don’t worry, this is all part of the journey to mastery. Ready for the next step? Take a quiz to solidify what you've just studied.