CSAT Solved Papers/ 2023/Q56

2023 CSAT — Q56

Quant Data sufficiency 2.5 marks Hard

Question: Is $p$ greater than $q$ ?

Statement-1: $p \times q$ is greater than zero.

Statement-2: $p^{2}$ is greater than $q^{2}$ .

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 cannot be answered even by using both the Statements together Answer

Worked rationale

DS yes/no: hunt for one “yes” and one “no” case under each statement (and under both).

Statement-1 alone ( $pq > 0$ , same sign): $p=3, q=1 \Rightarrow p > q$ (yes); $p=1, q=3 \Rightarrow p < q$ (no). Both keep $pq > 0$ . Insufficient.

Statement-2 alone ( $p^2 > q^2$ , i.e. $|p| > |q|$ ): $p=3, q=1 \Rightarrow$ yes; $p=-3, q=1 \Rightarrow p < q$ , no. Insufficient.

Both together ( $pq > 0$ and $|p| > |q|$ ):

Both positive: $p=3, q=1$ — $pq>0$ , $|p|>|q|$ , and $p > q$ (yes).
Both negative: $p=-3, q=-1$ — $pq = 3 > 0$ , $p^2 = 9 > 1 = q^2$ , but $p = -3 < -1 = q$ (no).

A “yes” and a “no” survive both statements, so even together the answer is undecidable.

Answer: (d) The Question cannot be answered even by using both the Statements together.

Why the other options miss

A

over-trusts a single statement: thinks $p^2 > q^2$ alone settles $p > q$ , ignoring the sign ambiguity ( $p$ could be the large negative).
B

claims each statement is enough on its own: treats “same sign” or “bigger square” as fixing the order, missing both counterexamples.
C

misses a case: combines to “both positive, $|p|>|q| \Rightarrow p>q$ ” but forgets the both-negative branch, where the order flips.

Specialist insight

The decisive case is both negative: $pq > 0$ and $|p| > |q|$ then force $p$ to be more negative, so $p < q$ — the exact opposite of the both-positive conclusion. DS discipline is to always test the negative branch when only signs and squares are given; the “yes/no” pair ( $p=3,q=1$ vs $p=-3,q=-1$ ) breaks sufficiency even with both statements, giving (d).

The trap, in one line

Both-negative case ( $p=-3,q=-1$ ) satisfies $pq>0$ and $p^2>q^2$ yet $p<q$ , so even together it's undecidable $=$ (d).

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