Simple question but big one on average weight. a,b,c weighted separately 1st like a, b, c , then a & b, then b & c ,then c & a at last a & b & c, the last weight was 167,then what will be the average weight of the 7 reading?
a) 95 b) 95.428 c) 95.45 d) 94

• The sum of 7 weights = weights of[a+b+c+(a+b)+(b+c)+(c+a)+(a+b+c)]= weights of[4*(a+b+c)=4*167=668
  Therefore The average weight of 7 readings is 668/7=95.428
  so, option is 'b'
• 14 years agoHelpfull: Yes(95) No(3)
• (167+3(a+b+c))/7;(as a+b+c=167)
  hence=95.428
• 14 years agoHelpfull: Yes(6) No(5)
• [a+b+c+(a+b)+(b+c)+(c+a)+(a+b+c)]/7;
  [3a+3b+3c+(a+b+c)]/7;
  [3(a+b+c)+(a+b+c)]/7;
  [3(182)+(182)]/7;
  95.428 is the ans...........
• 9 years agoHelpfull: Yes(2) No(0)
• Last weight abc is 167i.e three persons weight is 167 .in first 6 combinations a,b,c,ab,bc,ac i.e a checked weight for 3 times totally like that band c also so total weight in all 7 combinations is (4*167)
  Average is (668/7)=95.42
• 6 years agoHelpfull: Yes(0) No(0)