JEE MAIN - Mathematics (2023 - 25th January Morning Shift - No. 6)
Let M be the maximum value of the product of two positive integers when their sum is 66. Let the sample space $$S = \left\{ {x \in \mathbb{Z}:x(66 - x) \ge {5 \over 9}M} \right\}$$ and the event $$\mathrm{A = \{ x \in S:x\,is\,a\,multiple\,of\,3\}}$$. Then P(A) is equal to :
$$\frac{1}{3}$$
$$\frac{1}{5}$$
$$\frac{7}{22}$$
$$\frac{15}{44}$$
Explanation
$x+y=66$
$$ \begin{aligned} & \frac{x+y}{2} \geq \sqrt{x y} \\\\ \Rightarrow & 33 \geq \sqrt{x y} \\\\ \Rightarrow & x y \leq 1089 \\\\ \therefore & M=1089 \\\\ S: & x(66-x) \geq \frac{5}{9} \cdot 1089 \\\\ & 66 x-x^{2} \geq 605 \\\\ \Rightarrow & x^{2}-66 x+605 \leq 0 \end{aligned} $$
$\Rightarrow(x-55)(x-11) \leq 0 ; 11 \leq x \leq 55$
Therefore $S=\{11,12,13 \ldots 55\} $
$\Rightarrow n(S)=45$
Elements of $S$ which are multiple of 3 are
$$ \begin{aligned} & 12+(n-1) 3=54 \Rightarrow 3(n-1)=42 \Rightarrow n=15 \\\\ & n(A)=15 \end{aligned} $$
$\Rightarrow P(A)=\frac{15}{45}=\frac{1}{3}$
$$ \begin{aligned} & \frac{x+y}{2} \geq \sqrt{x y} \\\\ \Rightarrow & 33 \geq \sqrt{x y} \\\\ \Rightarrow & x y \leq 1089 \\\\ \therefore & M=1089 \\\\ S: & x(66-x) \geq \frac{5}{9} \cdot 1089 \\\\ & 66 x-x^{2} \geq 605 \\\\ \Rightarrow & x^{2}-66 x+605 \leq 0 \end{aligned} $$
$\Rightarrow(x-55)(x-11) \leq 0 ; 11 \leq x \leq 55$
Therefore $S=\{11,12,13 \ldots 55\} $
$\Rightarrow n(S)=45$
Elements of $S$ which are multiple of 3 are
$$ \begin{aligned} & 12+(n-1) 3=54 \Rightarrow 3(n-1)=42 \Rightarrow n=15 \\\\ & n(A)=15 \end{aligned} $$
$\Rightarrow P(A)=\frac{15}{45}=\frac{1}{3}$
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