3 friends A, B, C went for week end party to McDonald’s restaurant and there they measure there weights in some order IN 7 rounds. A B C AB BC AC ABC. Final round measure is 155 kg then find the average weight of all the 7 rounds?

• suppose the wait for A is x,wait for B is y and the wait for c is z...then for
  A-> x, b-> y, c->z, ab-> x+y, bc->y+z , ac-> x+z, abc -> x+y+z
  x+y+z=155;;; now the total wait will be x+y+z+x+y+y+z+x+z+x+y+z=(x+y+z)+2*(x+y+z)+(x+y+z)=155+2*155+155=620
  so the average will be 620/7=88.56..
• 9 years agoHelpfull: Yes(42) No(0)
• let a=x,b=y,c=z
  x+y+z=155
  x+y+z+x+y+y+z+z+x+x+y+z
  4(x+y+z)
  4(155)
  620
  620/7
  =88.57
• 9 years agoHelpfull: Yes(15) No(0)
• A=X;
  B=Y;
  C=Z;
  X+Y+Z=155
  IN 7 ROUNDS=X+Y+Z+[X+Y]+[Y+Z]+[Z+A]+[X+Y+Z]
  ASSUME VALUE
  =155+2(155)+155=310+210=620
  AVERAGE WEIGHT OF ALL THE 7 ROUNDS=620/7=88.571...
• 9 years agoHelpfull: Yes(3) No(0)
• given A+B+C=155
  so (A+B+C+A+B+B+C+C+A+A+B+C)/7
  =4(A+B+C)/7
  =4*155/7
  =88.5
• 9 years agoHelpfull: Yes(2) No(0)
• A+B+C=155
  THEN, A+B+C+(A+B)+(B+C)+(A+C)+(A+B+C)=4(155){total weight of 7 round}
  average weight=620/7=88.57
• 9 years agoHelpfull: Yes(2) No(0)
• think ans is 155/7=22.7
• 9 years agoHelpfull: Yes(1) No(10)
• (4*abc)=620..because total weight of abc is 155 kg..from 7 rounds a present in 4 rounds...b and c also present in 4 rounds.
  average is=620/7=88.57
• 9 years agoHelpfull: Yes(1) No(0)
• the total weight of ABC is 155. so the average weight of one person is 155/3=51.6
  weight of two ppl is 2(51.6)=103.2 A+B+C+AB+BC+AC+ABC= 51.6+51.6+51.6+103.2+103.2+103.2+155=619.4 Average weight of 7 rounds=619.4/7= 88.48
• 9 years agoHelpfull: Yes(1) No(0)
• A+B+AB+BC+CA+ABC=4(ABC).
  4(ABC_=4(155)=620.
  Average weight = 620/4 since we are grouping it in 4 terms.
  Average weight= 155
  
• 9 years agoHelpfull: Yes(1) No(2)
• in the last round total weight=155
  i.e.
  A+B+C=155
  now total weight in 7 round=A+B+C+(A+B)+(B+C)+(C+A)+(A+B+C)
  =3(A+B+C)
  =3*155
  =465
  now average=465/7=66.43kg
• 9 years agoHelpfull: Yes(0) No(9)
• A+B+C +A+B +B+C +C+A+ A+B+C
  4*(A+B+C)
  4*155=620
  620/7=88.57
  
• 9 years agoHelpfull: Yes(0) No(0)
• a,b,c,d weighted for 4 times so (4*155)/7=88.4
• 9 years agoHelpfull: Yes(0) No(0)
• a+b+c=155 avg=4*155/7
• 9 years agoHelpfull: Yes(0) No(0)
• TOTAL AGE OF THREE BOY=15*3=45Yrs
  LET THE RATIO OF THEIR AGE BE 3x+5x+7x=15x
  15x=45 so x=3
  THEN THE AVG AGE OF YOUNG BOY=3x THAT IS 3*3=9YRS
• 9 years agoHelpfull: Yes(0) No(0)
• a+b+c+ab+bc+ca+abc
  i.e a+b+c+a+b+b+c+a+c+a+c+a+b+c
  i.e (a+b+c)
  a+b+c=155;
  hence 4*155
  620
  avg 620/7=88.57
  ans=88.57
• 9 years agoHelpfull: Yes(0) No(0)
• A+B+C+(A+B)+(B+C)+(A+C)+(A+B+C)/7
  4A+4B+4C/7
  4(A+B+C)/7
  4*155/7=88.57
  
• 8 years agoHelpfull: Yes(0) No(0)