CSAT Solved Papers/ 2025/Q49

2025 CSAT — Q49

Verbal Reading comprehension 2.5 marks Hard Contested key

A mobile phone has been stolen. There are 3 suspects P, Q and R. They were questioned knowing that only one of them is guilty. Their responses are as follows:

P : I did not steal. Q stole it.

Q : R did not steal. I did not steal.

R : I did not steal. I do not know who did it.

Who stole the mobile phone?

  1. A P
  2. B Q UPSC official answer Our stricter read
  3. C R
  4. D Cannot be concluded

Thinking pathway

Locate. A truth-teller puzzle is forced only if the stem states a rule for how many statements are true or false. Read this stem for that rule — and notice what is missing: it says only “only one of them is guilty” and gives no rule on how many statements are true or false, or whether each person speaks uniformly. That gap is the whole story.

Test (try each suspect — and watch which convention you are using). Put each suspect in the guilty seat and count the false statements:

  • Guilty = Q → only Q’s own “I did not steal” is false; every other statement is true. One false.
  • Guilty = P → P’s “I did not steal” and “Q stole it” are both false. Two false, both from P.
  • Guilty = R → R’s “I did not steal” is false, and Q’s “R did not steal” is also false. Two false, from two different people.

Now the catch — which of these is “the answer” depends on a rule the stem never states:

  • Under “the fewest statements are false,” Q wins (one false) — this is UPSC’s key, (b).
  • Under “each person’s statements are all true or all false” (a common CSAT convention), the Q-world is forbidden (Q would be a mixed true/false speaker) and only the P-world qualifies (P uniformly false; Q and R uniformly true) → that convention forces (a) P, not Q.

Eliminate by anatomy. Because two reasonable conventions give two different answers, the stem as printed does not force a unique answer. UPSC published (b) Q, and that is what you record on the exam — but it is forced only by the unstated “fewest false statements” rule; swap in the equally common “all-or-nothing per speaker” rule and the same stem gives (a) P. A keyed answer that flips with an unstated convention is not forced by the text, which is why we show both and exclude the item from scored practice. The transferable discipline: a logic puzzle is well-posed only once its truth-rule is stated — when it is missing, name the convention you are assuming.

Evidence in the text

“only one of them is guilty” — the single load-bearing constraint. Test each suspect as the thief against the statements: if Q is the thief, exactly one statement is false (Q’s own “I did not steal”), every other statement — including P’s accusation “Q stole it” and Q’s “R did not steal” — stays true, giving a consistent single-guilty world. If P is the thief, P utters two falsehoods (“I did not steal” AND “Q stole it”); if R is the thief, two different people speak falsely (R’s “I did not steal” and Q’s “R did not steal”). Only the Q-as-thief case leaves the statements mutually consistent.

Worked rationale

The only hard fact is “exactly one of P, Q, R is guilty.” The stem gives no rule on the truth-values, so the answer depends on which standard truth-rule you supply. Counting false statements in each guilty-world:

  • Q guilty → 1 false statement (Q’s own denial); P’s “Q stole it” and Q’s “R did not steal” both true.
  • P guilty → 2 false statements, both spoken by P (“I did not steal” and “Q stole it”).
  • R guilty → 2 false statements, spoken by two people (R’s denial and Q’s “R did not steal”).

Under the “minimize false statements” rule, Q (one false) beats P and R (two each) → (b) Q, which is UPSC’s published key. Under the “each speaker is wholly true or wholly false” rule, the Q-world is disqualified (Q would be half-truthful) and only the P-world is admissible → (a) P.

The two conventions disagree, and the stem chooses neither. UPSC’s answer is (b) Q — record that for the exam. We flag the item because that key is not forced by the stem alone.

Why the other options miss

  • A
    the answer under the “each speaker is all-true-or-all-false” convention (P uniformly false; Q and R uniformly true). Not a careless trap: it is what a defensible reading of an under-specified stem yields — which is exactly why this item is contested rather than clean.
  • C
    not supported under either convention: the R-guilty world has two separate false speakers, minimises nothing, and violates the all-or-nothing rule too. A genuine elimination.
  • D
    the most honest reading of the stem as printed, since no truth-rule is stated. UPSC did not key it, and we do not penalise it; it is the answer a strict logician gives to an under-determined puzzle.

Specialist insight

The real lesson of this PYQ is about question design, and it is worth more than the answer. A truth-teller puzzle is well-posed only when it states how many statements are true or false (or that each person speaks uniformly). This stem states neither — only “one is guilty” — so the count of false statements is left to a convention the question never fixes. UPSC’s key (b) Q follows the “fewest false statements” convention; the equally standard “all-or-nothing per speaker” convention gives (a) P. On exam day, go with UPSC’s (b) — but the transferable discipline is to notice when a logic puzzle hasn’t stated its truth-rule, name the convention you are using, and not be rattled if your defensible reading differs from the key. That awareness is what separates a calm scorer from a second-guesser.

The trap, in one line

The stem never states a truth-rule, so the "answer" depends on an unstated convention: UPSC keys (b) Q (fewest false statements) while the all-or-nothing-per-speaker convention gives (a) P — record UPSC's (b), but know the item is under-determined.

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