Question

the sum of five distinct positive integers is 90. what can be the second largest number of the five at most ?

Answer 1

   a + b + c + d + e  = 90

   let a,b,c,d,e  be in increasing order.  If d and e are to be the largest, then the others have to be the least.  This is because the sum is always the same.

   least positive integer is 1.  so  a = 1.  b = 2..  c = 3.

                d + e = 90-1-2-3 = 84
        
     so d can be from 4 (as it is more than c)  to 41 (less than 84/2).
      As  e is more than d,  so it is from 80 to 43.

     second largest number can at most be 41.

Answer 2

When five distinct no are joined the smaller three number would be 1,2 and 3
Sum of other two numbers would be= 90-1-2-3= 84
As the mean is 42
One number is 42+1 and other no is 42-1.
The second largest no is 41