CAT 2024 Question Paper | Quant Slot 1

CAT Questions

CAT 2024 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 2 :If \((a+b \sqrt{n})\) is the positive square root of \((29-12 \sqrt{5})\), where \(a\) and \(b\) are integers, and \(n\) is a natural number, then the maximum possible value of \((a+b+n)\) is

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Video Explanation

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Explanatory Answer

Given the square root of \(29-12 \sqrt{5}\) is \(a+b \sqrt{n}\). \[ \begin{array}{l} \Rightarrow(a+b \sqrt{n})^2=29-12 \sqrt{5} \\ \Rightarrow a^2+n b^2=29 \\ 2 a b \sqrt{n}=-12 \sqrt{5} \end{array} \] Equating on both sides, we get \(n=5\) and \(a b=-6\) \[ \begin{array}{l} a^2+5 b^2=29 \\ 9+5(4)=29 \end{array} \] The values that will satisfy the above equation, ( \(a=3\) and \(b=-2\) )or \((a=-3\) and \(b=2)\)
Positive root is \(-3+2 \sqrt{5}\)
Negative root is \(3-2 \sqrt{5}\)
Now maximum possible value of \((a+b+n)\) in \(-3+2 \sqrt{5}=-3+1 \sqrt{20}\) (modified to get maximum value of \(a+b+n\) ) \[ \begin{array}{l} a=-3 \\ b=1 \\ n=20 \end{array} \] Now the maximum possible value of \((a+b+n)=-3+1+20=18\)

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The question is " If \((a+b \sqrt{n})\) is the positive square root of \((29-12 \sqrt{5})\), where \(a\) and \(b\) are integers, and \(n\) is a natural number, then the maximum possible value of \((a+b+n)\) is "

Hence, the answer is '18'

Choice C is the correct answer.

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