NIMCET 2020 Question Paper with solutions

After one value is removed:
Since all of the values are integers, the sum here must be an integer.
sum=(number)×(average)
Since the average $=35\frac{7}{17}$, and the sum be an integer, the number of integers must be a multiple of 17,
For any evenly spaced set, average = median.
After one of the consecutive integers is removed , most of the remaining set will be evenly spaced.
As a result , the average of the remaining set $37\frac{7}{17}$ will still be close to the median.
Implication:
The number of integers $=4\times 17=68$, with the result that $35\frac{7}{17}$ will be close to the median of the 68 mostly consecutive integers.
∴ Sum $=68\times 35\frac{7}{17}=2408$
Original set:
Since 68 integers remain after one of the integers is removed, the original set contains 69 integers.
Sum of the first n positive integers $=\frac{\mathit{n}\left(\mathit{n}+1\right)}{2}$
∴ Sum $=69\frac{70}{2}=2415$
Removed integer = original sum - sum after one integer is removed
$=2415-2408=7$.