Question

Let $\left[k\right]$ denotes the greatest integer less than or equal to $k.$ Then the number of positive integral solutions of the equation $\left[\frac{x}{\left[{\pi }^2\right]}\right]=\left[\frac{x}{\left[11\frac{1}{2}\right]}\right]$ is

• A. $29$
• B. $24$
• C. $21$
• D. $34$

Answer 1

The correct option is B $24$

$\left[\frac{x}{\left[{\pi }^2\right]}\right]=\left[\frac{x}{\left[11\frac{1}{2}\right]}\right]$
$\Rightarrow \left[\frac{x}{\left[9.87\right]}\right]=\left[\frac{x}{\left[11.5\right]}\right]$
$\Rightarrow \left[\frac{x}{9}\right]=\left[\frac{x}{11}\right]$

$\text{Case I}:0\le \frac{x}{9}<1$ and $0\le \frac{x}{11}<1$
$\Rightarrow 0\le x<9$ and $0\le x<11$
Common value of $x$ is $\{1,2,3,\dots ,8\}$

$\text{Case II}:1\le \frac{x}{9}<2$ and $1\le \frac{x}{11}<2$
$\Rightarrow 9\le x<18$ and $11\le x<22$
$\Rightarrow x\in \{11,12,\dots ,17\}$

$\text{Case III}:2\le \frac{x}{9}<3$ and $2\le \frac{x}{11}<3$$\Rightarrow 18\le x<27$ and $22\le x<33$
$\Rightarrow x\in \{22,23,\dots ,26\}$

$\text{Case IV}:3\le \frac{x}{9}<4$ and $3\le \frac{x}{11}<4$$\Rightarrow 27\le x<36$ and $33\le x<44$
$\Rightarrow x\in \{33,34,35\}$

$\text{Case V}:4\le \frac{x}{9}<5$ and $4\le \frac{x}{11}<5$
$\Rightarrow 36\le x<45$ and $44\le x<55$
$\Rightarrow x\in \left\{44\right\}$

$\therefore$ Total number of positive integers $=8+7+5+3+1=24$