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 $=4\times17=68$, with the result that $35\frac{7}{17}$ will be close to the median of the 68 mostly consecutive integers. ∴ Sum $=68\times35\frac{7}{17}=2408$ 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 $=\frac{n(n+1)}{2}$ ∴ Sum $=69\frac{70}{2}=2415$ Removed integer = original sum - sum after one integer is removed $=2415-2408=7$.