CAT 2022 Slot 1 QA Question 14: For any real number $x$, let $[x]$ be the largest integer less than or equal to $x$. If $\sum_{n=1}^{N} \left[\dfrac{1}{5} + \dfrac{n}{25}\right] = 25$ then $N$ is

Question

For any real number $x$, let $[x]$ be the largest integer less than or equal to $x$. If $\sum_{n=1}^{N} \left[\dfrac{1}{5} + \dfrac{n}{25}\right] = 25$ then $N$ is

Accepted Answer

$\left[\dfrac{1}{5} + \dfrac{n}{25}\right] = \left[\dfrac{n+5}{25}\right]$
Now check when the floor value changes.
For $n = 1$ to $19$, $\dfrac{n+5}{25} 
For $n = 20$ to $44$, $1 \leq \dfrac{n+5}{25} 
There are $25$ values from $20$ to $44$ inclusive, and their contribution to the sum is $25$.
$N =

For any real number xxx, let [x][x][x] be the largest integer less than or equal to xxx. If ∑n=1N[15+n25]=25\sum_{n=1}^{N} \left[\dfrac{1}{5} + \dfrac{n}{25}\right] = 25∑n=1N​[51​+25n​]=25 then NNN is

For any real number $x$ , let $[x]$ be the largest integer less than or equal to $x$ . If $\sum_{n=1}^{N} \left[\dfrac{1}{5} + \dfrac{n}{25}\right] = 25$ then $N$ is