Logic Questions and Answers

Answer: D

The statement that fails for some \(x\) is \(fl(x) \lt ce(x)\).

When \(x\) is an integer, both the ceiling and the floor equal \(x\) itself, so \(fl(x) = ce(x)\) and the strict inequality is false. For example, \(fl(3) = ce(3) = 3\).

The other three hold for every real \(x\): by definition the ceiling is at least \(x\) and the floor is at most \(x\), and so \(ce(x) \ge x \ge fl(x)\).

Answer: D

The statement that must be true is There exists at least one mango in the basket that is not good.

'All the mangoes are good' is false exactly when at least one mango is not good; that is its logical negation, so it is necessarily true.

The other options say more than the negation allows. 'All the mangoes are not good' and 'No mango is good' claim every mango is bad, but the statement is also false when just one mango is bad. 'Some are good and some are not' requires at least one good mango, which is not guaranteed: every mango could be bad.

Answer: C

The necessarily correct option is (S1) and (S3) together imply (S2).

If the average is 25, the four numbers add to 100. If each is at least 20, any three of them add to at least 60, so the fourth is at most \(100 - 60 = 40\). That is (S2).

The others fail. (S1) and (S2) allow 40, 40, 20, 0, which breaks (S3). (S2) and (S3) allow 20, 20, 20, 20, whose average is 20, not 25. And (S1) alone allows 100, 0, 0, 0, so it does not imply (S3).

Answer: D

The answer is neither (i) nor (ii).

The passage says only that coding regions, which are read in triplets to make amino acids, are about 2% of human DNA. It does not say what the other 98% does. Non-coding DNA could still take part in protein synthesis in other ways, for example by regulating which genes are expressed, so (i), that the majority has no role in protein synthesis, does not follow.

Statement (ii) is about the state of scientific understanding, which the passage never discusses. Being non-coding is not the same as having an unknown function. So neither can be inferred with certainty from the passage alone, which rules out 'only (i)', 'only (ii)' and 'both'.

Answer: B

Only two statements are correct.

Statement 1 is correct: 'taller than' is transitive, so A > B > C gives A > C.

Statement 2 is correct: X equals Y and Y is smaller than Z, so X, being the same as Y, is smaller than Z.

Statement 3 is wrong: P > Q > R gives P > R, the opposite of what it says. So statements 1 and 2 hold and 3 reverses the conclusion.

Answer: D

The negation is Amit does not live in Delhi or Aman lives in Mumbai.

The statement has the form 'p and not q', where p is 'Amit lives in Delhi' and q is 'Aman lives in Mumbai'. By De Morgan's law the negation of 'p and not q' is 'not p or q'.

So the 'and' becomes 'or', the first part is negated, and the second part loses its 'not': Amit does not live in Delhi, or Aman lives in Mumbai.

The options keeping 'and' fail because negation swaps 'and' for 'or'. The option with 'Aman does not live in Mumbai' forgot to negate the second part.

Answer: D

Correct answer: 3, 5, 1, 4, 2

The sequence follows the natural order of cause and effect. Study comes first, then the exam that tests it, then the job the qualification obtains, then the income the job brings, and finally the independence the income makes possible. Each step is the precondition of the one after it.

Answer: D

Correct answer: Sonam

David is below Dechen and above Sonam, and Karma is above David. So Dechen and Karma are both above David and Sonam is the only one below him. The ordering is not complete - nothing settles whether Dechen or Karma is higher - but the bottom position is determined, and it is Sonam’s.

Answer: C

Correct answer: If p is true and q is false and r is false, the P1 is true

Conjunction binds more tightly than disjunction, so the formula reads \( [(p \lor \lnot q) \land (q \to r)] \lor (r \lor p) \). With p true, q false and r false: \( p \lor \lnot q \) is true, \( q \to r \) is true because the antecedent is false, so the bracket is true and the whole is true. It is not a tautology, since p false, q true and r false makes every disjunct false.

Answer: A

Correct answer: ((P → Q)^(Q → R))→ (P → R)

This is the hypothetical syllogism and it is valid: if P leads to Q and Q leads to R, then P leads to R. The second option is the fallacy of denying the antecedent, the third fails when P is false and Q false, and the fourth confuses the disjunction of two implications with the implication of a disjunction.