Question : If x and y are integers, then is x even? Statement I : x²y² is even. Statement II : 1 + x² + y² is odd.
- (a) Select this option if the question can be answered using one of these statements alone, but cannot be answered using other statement
- (b) Select this option if the question can be answered using either statement alone
- (c) Select this option if the question can be answered using both the statements together, but cannot be answered using either statement alone ✓ UPSC's answer
- (d) Select this option if the question cannot be answered even using any of the statements
Why the answer is (c)
• Statement I states that $x^2y^2$ is even, which implies that the product $xy$ is even, meaning at least one of $x$ or $y$ is even, but it does not specify which one, so $x$ could be odd or even.
• Statement II states that $1 + x^2 + y^2$ is odd, which implies that $x^2 + y^2$ is even, meaning $x$ and $y$ must have the same parity (both even or both odd), but it does not determine if they are specifically even or odd.
• Combining both statements: Since $x$ and $y$ have the same parity (from Statement II) and their product is even (from Statement I), they cannot both be odd (as the product of two odd numbers is odd).
• Therefore, $x$ and $y$ must both be even.
• Since $x$ is even, the question 'is x even?' can be answered affirmatively only when both statements are used together.
• Thus, the correct option is (c), as the question can be answered using both statements together but not either alone.
Why the other options are wrong
- (a) Select this option if the question can be answered using one of these statements alone…
- Statement I alone is insufficient because $x$ could be odd if $y$ is even, and Statement II alone is insufficient because $x$ could be odd if $y$ is also odd.
- (b) Select this option if the question can be answered using either statement alone
- Neither statement alone provides enough information to determine the parity of $x$ definitively; Statement I allows $x$ to be odd, and Statement II allows $x$ to be odd.
- (d) Select this option if the question cannot be answered even using any of the statements
- The question can be answered by combining the information from both statements, which forces $x$ and $y$ to be even, so it is not unanswerable.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2026, held on 24 May 2026. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.