Question

If [x] denotes the greatest integer x, then $\left[\frac{2}{3}\right]+[\frac{2}{3}+\frac{1}{99}]+[\frac{2}{3}+\frac{2}{99}]+\dots ..+[\frac{2}{3}+\frac{98}{99}]$is equal to

• (A) 99
• (B) 98
• (C) 66
• (D) 65

Answer 1

The Correct Option is C

Explanation:
$\begin{array}{c}[\frac{2}{3}+\frac{r}{99}]=\{\begin{array}{ll}0& x<33\\ 1& r\ge 33\end{array}\\ \therefore \sum _{r=0}^{28}[\frac{2}{3}+\frac{r}{99}]=\sum _{r=0}^{32}[\frac{2}{3}+\frac{r}{99}]+\sum _{r=33}^{98}[\frac{2}{3}+\frac{r}{99}]\\ =& 0+66=66\end{array}$