CSAT Solved Papers/ 2025/Q69

2025 CSAT — Q69

Quant Data sufficiency 2.5 marks Hard

A question is given followed by two Statements I and II. Consider the Question and the Statements and mark the correct option.

Question: Let $P, Q, R, S$ be distinct non-zero digits. If $PP \times PQ = RRSS$ , where $P \le 3$ and $Q \le 4$ , then what is $Q$ equal to?

Statement I: $R = 1$ .

Statement II: $S = 2$ .

Which one of the following is correct in respect of the above Question and the Statements?

A The Question can be answered by using one of the Statements alone, but cannot be answered using the other statement alone
B The Question can be answered by using either Statement alone
C The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone
D The Question can be answered even without using any of the Statements Answer

Worked rationale

Answer the question on its own first. $PP = 11P$ and $RRSS = 11(100R + S)$ , so

PP \times PQ = 11P(10P + Q) = 11(100R + S) \;\Longrightarrow\; P(10P + Q) = 100R + S.

The right side is $\ge 100$ (as $R \ge 1$ ). Test $P \le 3$ :

$P=1$ : $10 + Q \le 14 < 100$ — impossible.
$P=2$ : $2(20+Q) \le 48 < 100$ — impossible.
$P=3$ : $3(30+Q) = 90 + 3Q$ . To reach $\ge 100$ needs $3Q \ge 10$ , i.e. $Q \ge 4$ . With $Q \le 4$ , this forces $Q = 4$ : $3 \times 34 = 102 = 100(1) + 2$ , so $R=1, S=2$ .

Check: $33 \times 34 = 1122 = RRSS$ with $R=1, S=2$ ✓, all digits $\{3,4,1,2\}$ distinct and non-zero. The solution is unique and $Q = 4$ is fixed by the question alone ( $P\le3,\ Q\le4$ already pin it). Statements I ( $R=1$ ) and II ( $S=2$ ) are true of the solution but add nothing.

Answer: (d) The Question can be answered even without using any of the Statements.

Why the other options miss

A

thought a statement was needed when the question stands alone: thinks one statement is needed to break ties, not seeing the constraints $P\le3, Q\le4$ already force $Q=4$ .
B

called a statement “sufficient” when none is even needed: notes each statement is consistent with the unique solution and labels each “sufficient,” overlooking that neither is needed.
C

combined statements the question never required: the engineered trap — combines $R=1$ and $S=2$ to “reconstruct” $1122$ and pick (c), never testing whether the question was self-contained.

Specialist insight

The gold DS reflex is solve the question before reading the statements. Here the cryptarithm $PP \times PQ = RRSS$ with the bounds $P\le3, Q\le4$ has a unique solution ( $33\times34=1122$ ), so $Q=4$ is determined unaided — making the statements pure decoys. The key structural step is factoring out $11$ from both $PP$ and $RRSS$ , which exposes $P(10P+Q)=100R+S$ and the $\ge 100$ bound that kills $P=1,2$ . Train to test self-sufficiency first: if the question pins the value alone, neither “alone,” “either,” nor “both” can be the answer — it is (d). The $-1/3$ marking pays for catching that the statements never did any work.

The trap, in one line

The bounds $P\le3, Q\le4$ already force the unique $33\times34=1122$ , so $Q=4$ before any statement — the answer is (d), not the "both together" (c) decoy.

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