UPSC Prelims 2026 CSAT, question 2

Topic: Data sufficiency

Directions · Q1 to Q5

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.
(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 : $y<\sqrt{x}<1$

(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 statementsCorrect answer

Solution.

Answer: (d) Select this option if the question cannot be answered even using any of the statements

Statement I alone is not sufficient: for $x = 0.5$ ( $x^2 = 0.25$ ), $y$ can be $0.3$ (giving $y < x$ ) or $0.6$ (giving $y > x$ ).
Statement II alone is not sufficient: for $x = 0.25$ ( $\sqrt{x} = 0.5$ ), $y$ can be $0.1$ (giving $y < x$ ) or $0.4$ (giving $y > x$ ).
Both statements together give $x^2 < y < \sqrt{x} < 1$ for $0 < x < 1$ ; since $x^2 < x < \sqrt{x}$ on $(0, 1)$ , $y$ can still lie in $(x^2, x)$ where $y < x$ or in $(x, \sqrt{x})$ where $y > x$ , so both together are not sufficient.

Trap: For $0 < x < 1$ , $x$ lies strictly between $x^2$ and $\sqrt{x}$ , so bounding $y$ between $x^2$ and $\sqrt{x}$ never fixes whether $y < x$ or $y > x$ .