Tom, Dick and harry, weigh themselves in a particular order.
First Tom, Dick, and Harry weigh themselves individually and then tom and Dick, Dick and Harry, Tom and
Harry and then Tom, Dick and harry together respectively. The recorded weight for the last measure is 158 kgs.
The average of all the 7 measures is
(a)112.86 (b)52.67 (c)90.29 (d)67.71

• Add all weights like tom+dick+harry+(tom+dick)+(dick+harry)+(harry+tom)+(tom+dick+harry)=4((tom+dick+harry)weights
  so 4*158=632
  then avg of all is 632/7=90.29 hence option (c)is correct
• 14 years agoHelpfull: Yes(15) No(0)
• The sum of 7 measures = weights of[tom+dick+harry+(tom+dick)+(dick+harry)+(tom+harry)+(tom+dock+harry)]= weights of[4*(tom+dick+harry)=4*158=632
  Therefore The average of all the 7 measures is 632/7=90.29
  so, option is' c'
• 14 years agoHelpfull: Yes(3) No(0)
• 90.29
• 14 years agoHelpfull: Yes(1) No(0)
• 90.2857142857 = 90.29
• 14 years agoHelpfull: Yes(1) No(3)
• 158*4 = 632
  632/7 = 90.29
• 14 years agoHelpfull: Yes(1) No(1)
• 90.29
  
  158*4/7= 90.29
• 14 years agoHelpfull: Yes(1) No(1)