The average of 20 numbers is zero. Of them, How many of them may be greater than zero , at the most?

Options
1) 1
2) 20
3) 0
4) 19

• At the most 19 numbers can be greater than zero.
  Since the 20th number can be equal to the negation of the sum of all 19 numbers .
  consider n1, n2, n3....,n19 are all positive numbers and n20 = - (n1+ n2 + ...... n19)
  then average = {(n1+ n2 + ..... n19) + (- (n1 + n2 +... n19)) }/ 20
  = 0
  
• 9 years agoHelpfull: Yes(7) No(2)
• Ans : 3)0
  Sol: let no is X1,X2,X3------------------------------------------X20;
  average of those no is
  =>(x1+x2+x3+----------------------------------+X20)/20=0
  so X1+x2+x3+------------------------------------+x20=0
  it possible only when at lest 1 digit should be negative and at most 19 digit should be negative
  we put, put X1=19 and X2=X3=X4=-----------------------------------X20=-1
  therefore (19-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1)/2-
  =(19-19)/20
  =0
  
• 9 years agoHelpfull: Yes(2) No(0)
• 4) 19 bcoz sum of 19 positive number must be counter balanced by negative number and that number could be equal to sum of 19 positive numbers
• 8 years agoHelpfull: Yes(2) No(0)
• atmost 19 because...
  
  take any 19 elements +ve add them take 20th element -ve of above sum....
  hence avg=0...
• 9 years agoHelpfull: Yes(1) No(0)
• option 4) 19
• 9 years agoHelpfull: Yes(0) No(2)
• please tell me elitmus 16th result is declare or not?
  
• 9 years agoHelpfull: Yes(0) No(0)
• the ans is 19
• 8 years agoHelpfull: Yes(0) No(0)
• 19 is the correct answer
  
• 8 years agoHelpfull: Yes(0) No(0)