2024 CSAT — Q68

In data sufficiency you must decide, not compute — and to break a statement you hunt for two allowed cases that give opposite answers (one “yes,” one “no”).

Statement-I alone ($2x+y \in \mathbb{Z}$): try $x=0,\,y=1 \Rightarrow 2x+y=1$ ✓ and $x+y=1$ (integer, “yes”). Now try $x=\tfrac12,\,y=0 \Rightarrow 2x+y=1$ ✓ but $x+y=\tfrac12$ (“no”). Two answers $\Rightarrow$ I alone insufficient.

Statement-II alone ($x+2y \in \mathbb{Z}$): symmetric — insufficient by the same pair with $x,y$ swapped.

Both together. Add them: $(2x+y)+(x+2y) = 3(x+y) \in \mathbb{Z}$, so $x+y = \tfrac{m}{3}$ for some integer $m$. That permits $x+y$ to be a non-integer. Concretely:

• $x=y=\tfrac13$: $2x+y = 1$ ✓, $x+2y = 1$ ✓, but $x+y = \tfrac23$ — not an integer (“no”).
• $x=y=1$: both statements hold and $x+y = 2$ — is an integer (“yes”).

Both statements hold in each case, yet $x+y$ is an integer in one and not the other. So even together they cannot settle the question.

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

The structural read: from the two statements you can only manufacture integer linear combinations of them. Subtracting gives $(2x+y)-(x+2y) = x-y \in \mathbb{Z}$ and adding gives $3(x+y)\in\mathbb Z$ — but $x-y$ and $3(x+y)$ being integers does not force $x+y$ to be one (take $x=y=\tfrac13$: $x-y=0\in\mathbb Z$, $3(x+y)=2\in\mathbb Z$, yet $x+y=\tfrac23$). The trap (c) is engineered for the student who knows the “add the equations” move but forgets that a multiple of a number being an integer says nothing about the number itself unless the multiplier is $\pm1$. The gold habit on every DS item: before answering (c), explicitly produce one “yes” case and one “no” case under both statements; if you can, the answer is (d). Spending $20$ seconds on that counterexample search is exactly the discipline the negative-marking scheme rewards.