Question

If $\displaystyle \sum _{n = 1}^{k}\left[\frac{1}{3} + \frac{n}{90}\right]=21,$ where $\left[\right.x\left]\right.$ denotes the integral part of $x,$ then $k$ is equal to

• A. $84$
• B. $80$
• C. $85$
• D. $86$

Answer 1

Answer: (B)

If $\left[\frac{1}{3} + \frac{n}{90}\right]=0\Rightarrow 0\leq \frac{1}{3}+\frac{n}{90} < 1$
$\Rightarrow 0\leq \frac{\left(\right. 30 + n \left.\right)}{90} < 1\Rightarrow 0\leq \left(\right.n+30\left.\right) < 90$
$\Rightarrow 1\leq n < 60$
and If $\left[\frac{1}{3} + \frac{n}{90}\right]=1\Rightarrow 1\leq \frac{1}{3}+\frac{n}{90} < 2$
$\Rightarrow 90\leq \left(\right.n+30\left.\right) < 180\Rightarrow 60\leq n < 150$
$\therefore \displaystyle \sum _{n = 1}^{59}\left[\frac{1}{3} + \frac{n}{90}\right]+\displaystyle \sum _{n = 60}^{k}\left[\frac{1}{3} + \frac{n}{90}\right]=21$
$\Rightarrow 0+\displaystyle \sum _{n = 60}^{k}\left[\frac{1}{3} + \frac{n}{90}\right]=21\Rightarrow \displaystyle \sum _{n = 60}^{k}\left[\frac{1}{3} + \frac{n}{90}\right]=21$
which is possible only when $k=80.$ $\left[\because \left[\frac{1}{3} + \frac{n}{90}\right] = 1\right]$