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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