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: For two distinct real numbers x and y, which of them is bigger?

Statement I: $x^{2} < y < 1$
Statement II: $y < \sqrt{x} < 1$

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

Statement I alone: $x^{2} < y < 1$
This statement tells us that y is greater than $x^{2}$ and both are less than 1. However, it does not provide a definitive relationship between x and y because x can be positive or negative.
Case 1: Let $x = 0.1$ . Then $x^{2} = 0.01$ . If $y = 0.5$ , then $0.01 < 0.5 < 1$ is satisfied. Here, $y > x$ .
Case 2: Let $x = - 0.9$ . Then $x^{2} = 0.81$ . If $y = 0.85$ , then $0.81 < 0.85 < 1$ is satisfied. Here, $x < y$ (since $- 0.9 < 0.85$ ).
Case 3: Let $x = 0.9$ . Then $x^{2} = 0.81$ . If $y = 0.85$ , then $0.81 < 0.85 < 1$ is satisfied. Here, $x > y$ (since $0.9 > 0.85$ ).
Since we can get both $x < y$ and $x > y$ , Statement I is not sufficient.

Statement II alone: $y < \sqrt{x} < 1$
For $\sqrt{x}$ to be a real number and less than 1, x must be in the range $0 <= x < 1$ . The statement says y < $\sqrt{x}$ .
Case 1: Let $x = 0.25$ . Then $\sqrt{x} = 0.5$ . If $y = 0.1$ , then $0.1 < 0.5 < 1$ is satisfied. Here, $x > y$ ( $0.25 > 0.1$ ).
Case 2: Let $x = 0.01$ . Then $\sqrt{x} = 0.1$ . If $y = 0.05$ , then $0.05 < 0.1 < 1$ is satisfied. Here, $x < y$ ( $0.01 < 0.05$ ).
Since we can get both $x < y$ and $x > y$ , Statement II is not sufficient.

Both statements together:
From Statement II, we know $0 <= x < 1$ . In this range, $x^{2} < x < \sqrt{x}$ .
Combining the inequalities: $x^{2} < y < \sqrt{x} < 1$ .
Even with this combined constraint, the relationship between x and y is not fixed. For any x in (0, 1), y is simply trapped between $x^{2}$ and $\sqrt{x}$ . Since x also lies between $x^{2}$ and $\sqrt{x}$ , y could be smaller than x (if it's near $x^{2}$ ) or larger than x (if it's near $\sqrt{x}$ ).
Example: If $x = 0.25$ , then $x^{2}$ = 0.0625 and $\sqrt{x} = 0.5$ . y can be 0.1 (making $x > y$ ) or y can be 0.4 (making $x < y$ ). Both values of y satisfy $x^{2} < y < \sqrt{x}$ .
Thus, even together, the statements are insufficient.

Why the other options are incorrect

Option (a) - one statement alone sufficient: As shown above, Statement I fails because x can be negative or positive, and Statement II fails because the relative positions of x and y within the (0, 1) interval are not fixed.
Option (b) - either statement alone sufficient: This is incorrect because neither statement provides a unique comparison between x and y.
Option (c) - both statements together sufficient: This is incorrect because even when combined, y is restricted to the interval $(x^{2}, \sqrt{x})$ , which contains x itself, allowing y to be either side of x.

Key Concept

In the interval (0, 1), the relative order of powers and roots is $x^{2} < x < \sqrt{x}$ ; any variable y constrained between the extremes ( $x^{2}$ and $\sqrt{x}$ ) cannot be definitively compared to the middle value (x) without further information.