Question

For $x\in R$, let $[x]$ denotes the greatest integer $\le x$, then the value of $\left[-\frac{1}{3}\right]+\left[-\frac{1}{3}-\frac{1}{100}\right]+\left[-\frac{1}{3}-\frac{2}{100}\right]+,\dots ,+\left[-\frac{1}{3}-\frac{99}{100}\right]$ is

• A. -100
• B. -123
• C. -135
• D. -153

Answer 1

Answer: (C)

For $0\le r\le 66,0\le \frac{r}{100}<\frac{2}{3}$
$\Rightarrow -\frac{1}{3}-\frac{2}{3}<-\frac{1}{3}-\frac{r}{100}\le -\frac{1}{3}$
$\therefore \left(-\frac{1}{3}-\frac{r}{100}\right)=-1$
for $0\le r\le 66$
Also, for $67\le r\le 100,\frac{67}{100}\le \frac{r}{100}\le 1$
$\Rightarrow -1\le -\frac{r}{100}\le -\frac{67}{100}$
$\Rightarrow -\frac{1}{3}-1\le -\frac{1}{3}-\frac{r}{100}\le -\frac{1}{3}-\frac{67}{100}$
$\therefore \left(-\frac{1}{3}-\frac{r}{100}\right)=-2$ for $67\le r\le 100$
Hence, $\sum _{r=0}^{100}\left(-\frac{1}{3}-\frac{r}{100}\right)$
$=67(-1)+2(-34)=-135$