Question

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

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

Answer 1

Answer: (B)

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