CSAT Solved Papers/ 2025/Q74

2025 CSAT — Q74

Quant Data sufficiency 2.5 marks Medium

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

Question: Is (p+q)2−4pq(p + q)^2 - 4pq, where p,qp, q are natural numbers, positive?

Statement I: p<qp < q.

Statement II: p>qp > q.

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

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

Worked rationale

Simplify the expression first — it is a perfect square:

(p+q)2−4pq=p2+2pq+q2−4pq=p2−2pq+q2=(p−q)2.(p+q)^2 - 4pq = p^2 + 2pq + q^2 - 4pq = p^2 - 2pq + q^2 = (p - q)^2.

So "(p+q)2−4pq>0(p+q)^2 - 4pq > 0" is exactly "(p−q)2>0(p-q)^2 > 0", i.e. "p≠qp \ne q".

  • Statement I (p<qp < q): then p≠qp \ne q, so (p−q)2>0(p-q)^2 > 0 — YES. Sufficient alone.
  • Statement II (p>qp > q): then p≠qp \ne q, so (p−q)2>0(p-q)^2 > 0 — YES. Sufficient alone.

Each statement alone settles the question (both give “positive”). So either alone works.

Answer: (b) The Question can be answered by using either Statement alone.

Why the other options miss

  • A
    saw only one direction as decisive: thinks only the "p>qp>q" direction makes the square positive, not seeing (p−q)2>0(p-q)^2 > 0 holds for p<qp<q equally.
  • C
    insisted on combining when each statement alone suffices: misses that the simplified (p−q)2(p-q)^2 needs only p≠qp \ne q, which each statement already gives.
  • D
    claimed self-contained when a statement is genuinely needed: claims it’s always positive even without a statement — false, since p=qp = q (allowed a priori) gives 00, not positive. A statement is needed to rule out p=qp=q.

Specialist insight

The whole item turns on the algebraic identity (p+q)2−4pq=(p−q)2(p+q)^2 - 4pq = (p-q)^2. Once simplified, the question is merely ”p≠qp \ne q?” — and both statements assert non-equality (one via <<, one via >>), so either alone suffices. The subtle wrong turn is (d): without any statement, p=qp = q is permitted and the value is 00 (not positive), so the question is not self-contained here — contrast this with self-determined items. Simplify before classifying sufficiency; the identity is what converts a messy-looking inequality into a one-line decidability check.

The trap, in one line

(p+q)2−4pq=(p−q)2(p+q)^2 - 4pq = (p-q)^2, positive iff p≠qp \ne q — and each statement (p<qp<q or p>qp>q) gives p≠qp\ne q, so either alone suffices: (b).

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