Question

If $0<x<1000$ and $\left[\frac{x}{10}\right]+\left[\frac{x}{20}\right]+\left[\frac{x}{30}\right]=\frac{11}{60}x$ where $[.]$ represents integral part of $x$, then the number of possible values of $x$ is

• A. $15$
• B. $12$
• C. $16$
• D. $33$

Answer 1

The correct option is C $16$

Given,
$0<x<1000$

$\left[\frac{x}{10}\right]+\left[\frac{x}{20}\right]+\left[\frac{x}{30}\right]=\frac{11}{60}x$

let,$\frac{x}{60}=t=>0<t<\frac{1000}{60}$

$\Rightarrow 0<t<\frac{50}{3}$

$\Rightarrow \left[6t\right]+\left[3t\right]+\left[2t\right]=11t$

for L.H.S, $max\left[6t\right]=6t$ when $6t\in Z$

$max\left[3t\right]=3t$ when $3t\in Z$

$max\left[2t\right]=2t$ when $2t\in Z$

but, $max\left[6t\right]+max\left[3t\right]+max\left[2t\right]=11t$

So, every part should be equal to their maximum's

$\Rightarrow 6t,3t,2t$ must be integers $3t\in Z,2t\in Z$

$\Rightarrow 3t-2t=t$, $t\in Z$

so, for $t\in Z$, this is possible

$0<t<16.66$

So, no of integer values for $t$ is $16$

hence option C is correct