CAT 2025 Question Paper | Quant Slot 1

CAT Questions

CAT 2025 Quant was dominated by Algebra followed by Arithmetic. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 22 : Let \(3 \leq x \leq 6\) and \(\left[x^2\right]=[x]^2\), where \([x]\) is the greatest integer not exceeding \(x\). If set S represents all feasible values of \(x\), then a possible subset of S is

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Video Explanation

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Explanatory Answer

The condition \(x^2=\lfloor x\rfloor^2\) becomes:

Let \(n=\lfloor x\rfloor\).
\(x^2=n^2\Longleftrightarrow n^2\le x^2<n^2+1\)

Taking square roots:
\(\sqrt{n^2}\le x<\sqrt{n^2+1}\)
\(\Rightarrow n\le x<\sqrt{n^2+1}\)

For \(n=3\) (\(x=3\)):
\(3\le x<\sqrt{3^2+1}\)
\(x\in[3,\sqrt{10})\)

For \(n=4\) (\(x=4\)):
\(4\le x<\sqrt{4^2+1}\)
\(x\in[4,\sqrt{17})\)

For \(n=5\) (\(x=5\)):
\(5\le x<\sqrt{5^2+1}\)
\(x\in[5,\sqrt{26})\)

For \(x=6\):
\(6^2=36=6^2\Rightarrow x=6\)

Complete Feasible Solution Set:
\(S=[3,\sqrt{10})\cup[4,\sqrt{17})\cup[5,\sqrt{26})\cup\{6\}\)

Now as we have the total solution set, we can eliminate the non feasible options which lie outside our solution set.
Therefore options A, B, C are eliminated.
Therefore only D is a subset of out solution set.

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The question is " Let \(3 \leq x \leq 6\) and \(\left[x^2\right]=[x]^2\), where \([x]\) is the greatest integer not exceeding \(x\). If set S represents all feasible values of \(x\), then a possible subset of S is "

Hence, the answer is '\((3, \sqrt{10}) \cup[5, \sqrt{26}) \cup\{6\}\)'

Choice 4 is the correct answer.

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