Directions for the following 5 (five) items: Each item in this section contains a question followed by two statements. Answer each item using the following instructions and mark your response on the Answer Sheet accordingly.
Question: X is a collection of certain odd numbers whereas Y is a collection of certain even numbers. T consists of the numbers all of which are either from X or from Y. Is every number of T from Y?
- Statement I: The sum of any two numbers belonging to T is even.
- Statement II: If both p and q are picked from T, then (p − 1)q is even.
Answer & explanation
Answer: (d) Select this option if the question cannot be answered even using any of the statements
Each statement only tells us that T cannot mix odd and even numbers. An all-odd T and an all-even T satisfy both statements, so whether T comes wholly from Y cannot be decided.
- Statement I: the sum of two numbers is even only when both have the same parity. So T is all odd or all even. Not sufficient.
- Statement II: if T is all odd, p − 1 is even, so (p − 1)q is even. If T is all even, q is even, so (p − 1)q is even. Both cases fit.
- A mixed T fails II: an even p and an odd q give (odd) × (odd) = odd. So II also means 'all odd or all even'. Not sufficient.
- Together: an all-odd T (from X) and an all-even T (from Y) satisfy both statements. Still not answered.
- ✗ Statement I Not sufficient alone: T = {1, 3} and T = {2, 4} both have even pairwise sums.
- ✗ Statement II Not sufficient alone: for T = {1, 3}, p − 1 is always even; for T = {2, 4}, q is always even. Both give (p − 1)q even.
Remember · When every statement is satisfied by two opposite cases, even combined, mark 'cannot be answered'.
Question and answer: UPSC's provisional GS Paper II (2026, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·