2022 CSAT — Q19

DS yes/no: for each statement, hunt one “yes” (integer) and one “no” (non-integer) case.

Statement-1 alone ($x/3 \notin \mathbb{Z}$): $x = 1 \Rightarrow 1/3 \notin \mathbb{Z}$, and $x=1$ is an integer (yes); $x = 1/2 \Rightarrow 1/6 \notin \mathbb{Z}$, and $x=1/2$ is not an integer (no). Insufficient.

Statement-2 alone ($3x \in \mathbb{Z}$, i.e. $x = k/3$): $x = 1 \Rightarrow 3x = 3$, integer (yes); $x = 1/3 \Rightarrow 3x = 1$, and $x = 1/3$ is not an integer (no). Insufficient.

Both together ($x/3 \notin \mathbb{Z}$ and $3x \in \mathbb{Z}$): write $x = m/3$.

• $m = 3 \Rightarrow x = 1$ (integer); $x/3 = 1/3 \notin \mathbb{Z}$ ✓ — yes.
• $m = 1 \Rightarrow x = 1/3$ (not integer); $x/3 = 1/9 \notin \mathbb{Z}$ ✓ — no.

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

Answer: (d) Both Statement-1 and Statement-2 are not sufficient.

The decisive pair is $x = 1$ (integer) versus $x = 1/3$ (non-integer): both satisfy ”$3x$ integer” and ”$x/3$ not integer” simultaneously. Writing $x = m/3$ exposes that the two statements only constrain $m$ to not be a multiple of $9$ — they never pin whether $3 \mid m$. The counterexample pair breaks joint sufficiency, giving (d).