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This section includes 315 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. |
In a multiple-choice question paper of 15 questions, the answers can be A, B, C or D. The number of different ways of answering the question paper are |
| A. | 65536 x 47 |
| B. | 194536 x 45 |
| C. | 23650 x 49 |
| D. | 11287435 |
| Answer» B. 194536 x 45 | |
| 202. |
A polygon with 25 sides can be triangulated into |
| A. | 23 |
| B. | 20 |
| C. | 22 |
| D. | 21 |
| Answer» B. 20 | |
| 203. |
Suppose that P(n) is a propositional function. Determine for which positive integers n the statement P(n) must be true if: P(1) and P(2) is true; for all positive integers n, if P(n) and P(n+1) is true then P(n+2) is true. |
| A. | p(1) |
| B. | p(2) |
| C. | p(4) |
| D. | p(n) |
| Answer» E. | |
| 204. |
Suppose that P(n) is a propositional function. Determine for which positive integers n the statement P(n) must be true if: P(1) is true; for all positive integers n, if P(n) is true then P(n+2) is true. |
| A. | p(3) |
| B. | p(2) |
| C. | p(4) |
| D. | p(6) |
| Answer» B. p(2) | |
| 205. |
Which amount of postage can be formed using just 4-cent and 11-cent stamps? |
| A. | 2 |
| B. | 5 |
| C. | 30 |
| D. | 10 |
| Answer» E. | |
| 206. |
A polygon with 12 sides can be triangulated into |
| A. | 7 |
| B. | 10 |
| C. | 5 |
| D. | 12 |
| Answer» C. 5 | |
| 207. |
A polygon with 7 sides can be triangulated into |
| A. | 7 |
| B. | 14 |
| C. | 5 |
| D. | 10 |
| Answer» D. 10 | |
| 208. |
What is the induction hypothesis assumption for the inequality m ! > 2m where m>=4? |
| A. | for m=k, k+1!>2k holds |
| B. | for m=k, k!>2k holds |
| C. | for m=k, k!>3k holds |
| D. | for m=k, k!>2k+1 holds |
| Answer» C. for m=k, k!>3k holds | |
| 209. |
For any positive integer m              is divisible by 4. |
| A. | 5m2 + 2 |
| B. | 3m + 1 |
| C. | m2 + 3 |
| D. | m3 + 3m |
| Answer» E. | |
| 210. |
For every natural number k, which of the following is true? |
| A. | (mn)k = mknk |
| B. | m*k = n + 1 |
| C. | (m+n)k = k + 1 |
| D. | mkn = mnk |
| Answer» B. m*k = n + 1 | |
| 211. |
For any integer m>=3, the series 2+4+6+… +(4m) can be equivalent to |
| A. | m2+3 |
| B. | m+1 |
| C. | mm |
| D. | 3m2+4 |
| Answer» B. m+1 | |
| 212. |
For m = 1, 2, …, 4m+2 is a multiple of is known as |
| A. | lemma |
| B. | corollary |
| C. | conjecture |
| D. | none of the mentioned |
| Answer» B. corollary | |
| 213. |
In the principle of mathematical induction, which of the following steps is mandatory? |
| A. | induction hypothesis |
| B. | inductive reference |
| C. | induction set assumption |
| D. | minimal set representation |
| Answer» B. inductive reference | |
| 214. |
A proof that p → q is true based on the fact that q is true, such proofs are known as |
| A. | direct proof |
| B. | contrapositive proofs |
| C. | trivial proof |
| D. | proof by cases |
| Answer» D. proof by cases | |
| 215. |
A proof broken into distinct cases, where these cases cover all prospects, such proofs are known as |
| A. | direct proof |
| B. | contrapositive proofs |
| C. | vacuous proof |
| D. | proof by cases |
| Answer» D. proof by cases | |
| 216. |
Which of the arguments is not valid in proving sum of two odd number is not odd. |
| A. | 3 + 3 = 6, hence true for all |
| B. | 2n +1 + 2m +1 = 2(n+m+1) hence true for all |
| C. | all of the mentioned |
| D. | none of the mentioned |
| Answer» B. 2n +1 + 2m +1 = 2(n+m+1) hence true for all | |
| 217. |
A proof covering all the possible cases, such type of proofs are known as |
| A. | direct proof |
| B. | proof by contradiction |
| C. | vacuous proof |
| D. | exhaustive proof |
| Answer» E. | |
| 218. |
In proving √5 as irrational, we begin with assumption √5 is rational in which type of proof? |
| A. | direct proof |
| B. | proof by contradiction |
| C. | vacuous proof |
| D. | mathematical induction |
| Answer» C. vacuous proof | |
| 219. |
When to proof P→Q true, we proof P false, that type of proof is known as |
| A. | direct proof |
| B. | contrapositive proofs |
| C. | vacuous proof |
| D. | mathematical induction |
| Answer» D. mathematical induction | |
| 220. |
Which of the following can only be used in disproving the statements? |
| A. | direct proof |
| B. | contrapositive proofs |
| C. | counter example |
| D. | mathematical induction |
| Answer» D. mathematical induction | |
| 221. |
Let the statement be “If n is not an odd integer then square of n is not odd.â€, then if P(n) is “n is an not an odd integer†and Q(n) is “(square of n) is not odd.†For direct proof we should prove |
| A. | ∀np ((n) → q(n)) |
| B. | ∃ np ((n) → q(n)) |
| C. | ∀n~(p ((n)) → q(n)) |
| D. | ∀np ((n) → ~(q(n))) |
| Answer» B. ∃ np ((n) → q(n)) | |
| 222. |
“Parul is out for a trip or it is not snowing†and “It is snowing or Raju is playing chess†imply that |
| A. | parul is out for trip |
| B. | raju is playing chess |
| C. | parul is out for a trip and raju is playing chess |
| D. | parul is out for a trip or raju is playing chess |
| Answer» E. | |
| 223. |
What rules of inference are used in this argument? “All students in this science class has taken a course in physics†and “Marry is a student in this class†imply the conclusion “Marry has taken a course in physics.†|
| A. | universal instantiation |
| B. | universal generalization |
| C. | existential instantiation |
| D. | existential generalization |
| Answer» B. universal generalization | |
| 224. |
The premises (p ∧ q) ∨ r and r → s imply which of the conclusion? |
| A. | p ∨ r |
| B. | p ∨ s |
| C. | p ∨ q |
| D. | q ∨ r |
| Answer» C. p ∨ q | |
| 225. |
What rules of inference are used in this argument? “Jay is an awesome student. Jay is also a good dancer. Therefore, Jay is an awesome student and a good dancer.†|
| A. | conjunction |
| B. | modus ponens |
| C. | disjunctive syllogism |
| D. | simplification |
| Answer» B. modus ponens | |
| 226. |
What rules of inference are used in this argument? “It is either colder than Himalaya today or the pollution is harmful. It is hotter than Himalaya today. Therefore, the pollution is harmful.†|
| A. | conjunction |
| B. | modus ponens |
| C. | disjunctive syllogism |
| D. | hypothetical syllogism |
| Answer» D. hypothetical syllogism | |
| 227. |
Which rule of inference is used, â€Bhavika will work in an enterprise this summer. Therefore, this summer Bhavika will work in an enterprise or he will go to beach.†|
| A. | simplification |
| B. | conjunction |
| C. | addition |
| D. | disjunctive syllogism |
| Answer» D. disjunctive syllogism | |
| 228. |
Which rule of inference is used in each of these arguments, “If it hailstoday, the local office will be closed. The local office is not closed today. Thus, it did not hailed today.†|
| A. | modus tollens |
| B. | conjunction |
| C. | hypothetical syllogism |
| D. | simplification |
| Answer» B. conjunction | |
| 229. |
Which rule of inference is used in each of these arguments, “If it is Wednesday, then the Smartmart will be crowded. It is Wednesday. Thus, the Smartmart is crowded.†|
| A. | modus tollens |
| B. | modus ponens |
| C. | disjunctive syllogism |
| D. | simplification |
| Answer» C. disjunctive syllogism | |
| 230. |
Find a counterexample of ∀x∀y(xy > y), where the domain for all variables consists of all integers. |
| A. | x = -1, y = 17 |
| B. | x = -2 y = 8 |
| C. | both x = -1, y = 17 and x = -2 y = 8 |
| D. | does not have any counter example |
| Answer» D. does not have any counter example | |
| 231. |
Use quantifiers and predicates with more than one variable to express, “There is a pupil in this lecture who has taken at least one course in Discrete Maths.†|
| A. | ∃x∃yp (x, y), where p (x, y) is “x has taken y,†the domain for x consists of all pupil in this class, and the domain for y consists of all discrete maths lectures |
| B. | ∃x∃yp (x, y), where p (x, y) is “x has taken y,†the domain for x consists of all discrete maths lectures, and the domain for y consists of all pupil in this class |
| C. | ∀x∀yp(x, y), where p (x, y) is “x has taken y,†the domain for x consists of all pupil in this class, and the domain for y consists of all discrete maths lectures |
| D. | ∃x∀yp(x, y), where p (x, y) is “x has taken y,†the domain for x consists of all pupil in this class, and the domain for y consists of all discrete maths lectures |
| Answer» B. ∃x∃yp (x, y), where p (x, y) is “x has taken y,†the domain for x consists of all discrete maths lectures, and the domain for y consists of all pupil in this class | |
| 232. |
Let T (x, y) mean that student x likes dish y, where the domain for x consists of all students at your school and the domain for y consists of all dishes. Express ¬T (Amit, South Indian) by a simple English sentence. |
| A. | all students does not like south indian dishes. |
| B. | amit does not like south indian people. |
| C. | amit does not like south indian dishes. |
| D. | amit does not like some dishes. |
| Answer» E. | |
| 233. |
Let L(x, y) be the statement “x loves y,†where the domain for both x and y consists of all people in the world. Use quantifiers to express, “Joy is loved by everyone.†|
| A. | ∀x l(x, joy) |
| B. | ∀y l(joy,y) |
| C. | ∃y∀x l(x, y) |
| D. | ∃x ¬l(joy, x) |
| Answer» B. ∀y l(joy,y) | |
| 234. |
“The product of two negative real numbers is not negative.†Is given by? |
| A. | ∃x ∀y ((x < 0) ∧ (y < 0) → (xy > 0)) |
| B. | ∃x ∃y ((x < 0) ∧ (y < 0) ∧ (xy > 0)) |
| C. | ∀x ∃y ((x < 0) ∧ (y < 0) ∧ (xy > 0)) |
| D. | ∀x ∀y ((x < 0) ∧ (y < 0) → (xy > 0)) |
| Answer» E. | |
| 235. |
Translate ∀x∃y(x < y) in English, considering domain as a real number for both the variable. |
| A. | for all real number x there exists a real number y such that x is less than y |
| B. | for every real number y there exists a real number x such that x is less than y |
| C. | for some real number x there exists a real number y such that x is less than y |
| D. | for each and every real number x and y such that x is less than y |
| Answer» B. for every real number y there exists a real number x such that x is less than y | |
| 236. |
Let domain of m includes all students, P (m) be the statement “m spends more than 2 hours in playing poloâ€. Express ∀m ¬P (m) quantification in English. |
| A. | a student is there who spends more than 2 hours in playing polo |
| B. | there is a student who does not spend more than 2 hours in playing polo |
| C. | all students spends more than 2 hours in playing polo |
| D. | no student spends more than 2 hours in playing polo |
| Answer» E. | |
| 237. |
â€Everyone wants to learn cosmology.†This argument may be true for which domains? |
| A. | all students in your cosmology class |
| B. | all the cosmology learning students in the world |
| C. | both of the mentioned |
| D. | none of the mentioned |
| Answer» D. none of the mentioned | |
| 238. |
The statement, “At least one of your friends is perfectâ€. Let P (x) be “x is perfect†and let F (x) be “x is your friend†and let the domain be all people. |
| A. | ∀x (f (x) → p (x)) |
| B. | ∀x (f (x) ∧ p (x)) |
| C. | ∃x (f (x) ∧ p (x)) |
| D. | ∃x (f (x) → p (x)) |
| Answer» D. ∃x (f (x) → p (x)) | |
| 239. |
The statement,†Every comedian is funny†where C(x) is “x is a comedian†and F (x) is “x is funny†and the domain consists of all people. |
| A. | ∃x(c(x) ∧ f (x)) |
| B. | ∀x(c(x) ∧ f (x)) |
| C. | ∃x(c(x) → f (x)) |
| D. | ∀x(c(x) → f (x)) |
| Answer» E. | |
| 240. |
Let P (x) denote the statement “x >7.†Which of these have truth value true? |
| A. | p (0) |
| B. | p (4) |
| C. | p (6) |
| D. | p (9) |
| Answer» E. | |
| 241. |
(p → r) ∨ (q → r) is logically equivalent to |
| A. | (p ∧ q) ∨ r |
| B. | (p ∨ q) → r |
| C. | (p ∧ q) → r |
| D. | (p → q) → r |
| Answer» D. (p → q) → r | |
| 242. |
(p → q) ∧ (p → r) is logically equivalent to |
| A. | p → (q ∧ r) |
| B. | p → (q ∨ r) |
| C. | p ∧ (q ∨ r) |
| D. | p ∨ (q ∧ r) |
| Answer» B. p → (q ∨ r) | |
| 243. |
p ↔ q is logically equivalent to |
| A. | (p → q) → (q → p) |
| B. | (p → q) ∨ (q → p) |
| C. | (p → q) ∧ (q → p) |
| D. | (p ∧ q) → (q ∧ p) |
| Answer» D. (p ∧ q) → (q ∧ p) | |
| 244. |
¬ (p ↔ q) is logically equivalent to |
| A. | q↔p |
| B. | p↔¬q |
| C. | ¬p↔¬q |
| D. | ¬q↔¬p |
| Answer» C. ¬p↔¬q | |
| 245. |
p ∨ q is logically equivalent to |
| A. | ¬q → ¬p |
| B. | q → p |
| C. | ¬p → ¬q |
| D. | ¬p → q |
| Answer» E. | |
| 246. |
p → q is logically equivalent to |
| A. | ¬p ∨ ¬q |
| B. | p ∨ ¬q |
| C. | ¬p ∨ q |
| D. | ¬p ∧ q |
| Answer» D. ¬p ∧ q | |
| 247. |
The compound propositions p and q are called logically equivalent if                  is a tautology. |
| A. | p ↔ q |
| B. | p → q |
| C. | ¬ (p ∨ q) |
| D. | ¬p ∨ ¬q |
| Answer» B. p → q | |
| 248. |
Let P: We should be honest., Q: We should be dedicated., R: We should be overconfident. Then ‘We should be honest or dedicated but not overconfident.’ is best represented by? |
| A. | ~p v ~q v r |
| B. | p ∧ ~q ∧ r |
| C. | p v q ∧ r |
| D. | p v q ∧ ~r |
| Answer» E. | |
| 249. |
Let P: This is a great website, Q: You should not come back here. Then ‘This is a great website and you should come back here.’ is best represented by? |
| A. | ~p v ~q |
| B. | p ∧ ~q |
| C. | p v q |
| D. | p ∧ q |
| Answer» C. p v q | |
| 250. |
Let P: I am in Delhi.; Q: Delhi is clean.; then q ^ p(q and p) is? |
| A. | delhi is clean and i am in delhi |
| B. | delhi is not clean or i am in delhi |
| C. | i am in delhi and delhi is not clean |
| D. | delhi is clean but i am in mumbai |
| Answer» B. delhi is not clean or i am in delhi | |