Question

lf $x,y\in (0,30)$ such that $\left[\frac{x}{3}\right]+\left[\frac{3x}{2}\right]+\left[\frac{y}{2}\right]+\left[\frac{3y}{4}\right]=\frac{11x}{6}+\frac{5y}{4}$ (where [x] denote greatest integer $\le x$) then the number of ordered pairs $(x,y)$ is

• A. 10
• B. 20
• C. 24
• D. 28

Answer 1

The correct option is D 28

Let $\left\{x\right\}=x-\left[x\right]$ denote fractional part of $xi.e0\le$ $\left\{x\right\}<1$
Now given expression is
$\frac{x}{3}-\left\{\frac{x}{3}\right\}+\frac{3x}{2}-\left\{\frac{3x}{2}\right\}+\frac{y}{2}-\left\{\frac{y}{2}\right\}+\frac{3y}{4}-\left\{\frac{3y}{4}\right\}=\frac{11x}{6}+\frac{5y}{4}$
$\Rightarrow \left\{\frac{x}{3}\right\}+\left\{\frac{3x}{2}\right\}+\left\{\frac{y}{2}\right\}+\left\{\frac{3y}{4}\right\}=0$
$\therefore \left\{\frac{x}{3}\right\}=\left\{\frac{3x}{2}\right\}=\left\{\frac{y}{2}\right\}=\left\{\frac{3y}{4}\right\}=0$
$\Rightarrow x=6,12,18,24$;
and $y=4,8,12,16,20,24,28$
Hence no. of ordered pairs $=4\times 7=28$