cat 2022 Question 12

Question

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

Accepted Answer

It is given,

$\sum_{n=1}^{N} \left[ \frac{1}{5} + \frac{n}{25} \right] = 25$

$\sum_{n=1}^{N} \left[ \frac{5 + n}{25} \right] = 25$

For n = 1 to n = 19, value of function is zero.

For n = 20 to n = 44, value of function will be 1.

$44 = 20 + n - 1$

n = 25 which is equal to given value.

This implies N = 44

Question 12.