Question

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

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

Answer 1

Answer: (C)

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