IAS 2026 — CSAT Reasoning

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

Analysis

Statement I alone: $x^{2} < y < 1$x2<y<1
This statement tells us that y is greater than $x^{2}$x2 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 $x = 0.1$x=0.1. Then $x^{2} = 0.01$x2=0.01. If $y = 0.5$y=0.5, then $0.01 < 0.5 < 1$0.01<0.5<1 is satisfied. Here, $y > x$y>x.
Case 2: Let $x = -0.9$x=−0.9. Then $x^{2} = 0.81$x2=0.81. If $y = 0.85$y=0.85, then $0.81 < 0.85 < 1$0.81<0.85<1 is satisfied. Here, $x < y$x<y (since $-0.9 < 0.85$−0.9<0.85).
Case 3: Let $x = 0.9$x=0.9. Then $x^{2} = 0.81$x2=0.81. If $y = 0.85$y=0.85, then $0.81 < 0.85 < 1$0.81<0.85<1 is satisfied. Here, $x > y$x>y (since $0.9 > 0.85$0.9>0.85).
Since we can get both $x < y$x<y and $x > y$x>y, Statement I is not sufficient.

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

Both statements together:
From Statement II, we know $0 <= x < 1$0<=x<1. In this range, $x^{2} < x < \sqrt{x}$x2<x<x​.
Combining the inequalities: $x^{2} < y < \sqrt{x} < 1$x2<y<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 $x^{2}$x2 and $\sqrt{x}$x​. Since x also lies between $x^{2}$x2 and $\sqrt{x}$x​, y could be smaller than x (if it's near $x^{2}$x2) or larger than x (if it's near $\sqrt{x}$x​).
Example: If $x = 0.25$x=0.25, then $x^{2}$x2 = 0.0625 and $\sqrt{x} = 0.5$x​=0.5. y can be 0.1 (making $x > y$x>y) or y can be 0.4 (making $x < y$x<y). Both values of y satisfy $x^{2} < y < \sqrt{x}$x2<y<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 $(x^{2}, \sqrt{x})$(x2,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 $x^{2} < x < \sqrt{x}$x2<x<x​; any variable y constrained between the extremes ($x^{2}$x2 and $\sqrt{x}$x​) cannot be definitively compared to the middle value (x) without further information.