CSAT Solved Papers/ 2024/Q39

2024 CSAT — Q39

Quant Statement validity 2.5 marks Medium

Let pp, qq, rr and ss be distinct positive integers. Let pp, qq be odd and rr, ss be even. Consider the following statements:

  1. (p−r)2(qs)(p - r)^2(qs) is even.

  2. (q−s)q2s(q - s)q^2 s is even.

  3. (q+r)2(p+s)(q + r)^2(p + s) is odd.

Which of the statements given above are correct?

  1. A 1 and 2 only
  2. B 2 and 3 only
  3. C 1 and 3 only
  4. D 1, 2 and 3 Answer

Worked rationale

Parity rules: odd±\pmeven == odd; squaring preserves parity; a product is even iff any factor is even. Here p,qp,q odd and r,sr,s even.

Statement 1: (p−r)2(qs)(p-r)^2(qs). p−r=p-r = odd−-even == odd, so (p−r)2(p-r)^2 is odd. qs=qs = odd×\timeseven == even. odd ×\times even == even. True.

Statement 2: (q−s)q2s(q-s)q^2 s. q−s=q-s = odd−-even == odd; q2q^2 odd; ss even. odd×\timesodd×\timeseven == even. True.

Statement 3: (q+r)2(p+s)(q+r)^2(p+s). q+r=q+r = odd++even == odd, so (q+r)2(q+r)^2 is odd. p+s=p+s = odd++even == odd. odd×\timesodd == odd. True.

All three hold.

Answer: (d) 1, 2 and 3.

Why the other options miss

  • A
    mislabels qq as even: treats q+rq+r as even (treating qq itself as even), wrongly making statement 3 even and rejecting it.
  • B
    thinks (p−r)2(p-r)^2 being odd makes the whole product odd in statement 1, forgetting the even factor ss inside qsqs.
  • C
    overlooks the even factor ss in statement 2, mislabelling it odd.

Specialist insight

No values needed — this is pure parity bookkeeping. Two rules carry everything: a square never changes parity (so (odd)2(\text{odd})^2 stays odd, killing the temptation to call it even), and a product is even the moment one factor is even. In statements 1 and 2 there is always a lone even factor (ss), forcing “even”; in statement 3 every factor is odd, forcing “odd”. The trap is hunting for a hidden even factor inside a squared term — but the even-ness lives in the separate factor ss, not in the square.

The trap, in one line

Squaring preserves parity; a single even factor (ss) forces "even" in 1 and 2, while every factor in 3 is odd ⇒\Rightarrow all three claims hold.

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