Reasoning : Logical Reasoning & Analytical Ability

Q 4 / 325

UPSC CSE Prelims 2026

Directions: 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. (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 (d) Select this option if the question cannot be answered even using any of the statements
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?
  1. Statement I: The sum of any two numbers belonging to T is even.
  2. Statement II: If both p and q are picked from T, then (p−1)q is even.

EXPLANATION

The correct option is D - Select this option if the question cannot be answered even using any of the statements.

[as per provisional answerkey]

Analysis

The question asks if every number in set T is from set Y (i.e., is every number in T even?). We are given that X contains only odd numbers and Y contains only even numbers. T is a subset of the union of X and Y.

Statement I alone: The sum of any two numbers belonging to T is even.
For the sum of any two numbers to be even, there are two possibilities:
1. All numbers in T are even (Even + Even = Even). In this case, the answer to the question is "Yes".
2. All numbers in T are odd (Odd + Odd = Even). In this case, the answer to the question is "No".
Since we cannot determine which of these two cases is true, Statement I alone is not sufficient.

Statement II alone: If both p and q are picked from T, then (p−1)q is even.
Let's test different scenarios for p and q:
1. If all numbers in T are even: q is even, so (p−1)×Even is always even. This satisfies the statement. (Answer: Yes)
2. If all numbers in T are odd: p is odd, so (p−1) is even. Thus, Even×q is always even. This also satisfies the statement. (Answer: No)
3. If T contains both even and odd numbers: If q is even, the product is even. If q is odd, the product is even only if (p−1) is even, which means p must be odd. This implies that if T contains an odd number, every p picked must be odd for the condition to hold for an odd q. This leads back to the ambiguity between all even or all odd.
Since Statement II is satisfied by both "all even" and "all odd" collections, it is not sufficient.

Both statements together:
Even when combined, both statements are satisfied if T consists entirely of odd numbers (from X) OR if T consists entirely of even numbers (from Y).
Case A: T={3,5}. Sums (8) are even; (p−1)q (e.g., (3−1)5=10) is even. Answer: No.
Case B: T={2,4}. Sums (6) are even; (p−1)q (e.g., (2−1)4=4) is even. Answer: Yes.
Because both "Yes" and "No" are possible under both conditions, the statements together are insufficient.

Why the other options are incorrect

  • Option (a) - one statement alone sufficient: Incorrect because Statement I allows for T to be all odd or all even, and Statement II does the same; neither isolates set Y.
  • Option (b) - either statement alone sufficient: Incorrect because, as shown above, both statements are independently ambiguous regarding whether the numbers are from X or Y.
  • Option (c) - both statements together sufficient: Incorrect because even when combined, the properties of parity (Even+Even=Even and Odd+Odd=Even) prevent us from distinguishing if the collection T is exclusively odd or exclusively even.

Key Concept

Parity logic: The sum of two numbers being even only implies they share the same parity (both odd or both even), but does not identify which parity they share.