You have eight balls all of the same size. 7 of them weigh the same, and one of them weighs slightly more. How can you fine the ball that is heavier by using a balance and only two weighings?

• we can split the balls as 3,3,2(say groups as a,b,c)
  Here are the possible cases
  Case 1:
  If the balance a and b is unbalance then use one more time by putting one ball each side . If you find same weight then the 3rd is heavier
  Case 2:
  If the balance a and b is unbalance then use one more time by putting one ball each side . If you find same weight then the 3rd is heavier
  Case 3:
  If the a and b are of equal weights then you can choose the next group
• 12 years agoHelpfull: Yes(18) No(5)
• To the above answers : We can only use the balance 2 times.Only 2 weighings can be done.
  We should split the balls into 3 groups as 3 balls, 3 balls and 2 balls.
  
  First keep the 2 3-balls group on each side of the balance.
  Note the side of fall or whether the balance is in equilibrium.
  Remove the balls from the balance.
  If one side falls, go to step1. else go to step 2.
  
  step 1
  If one side falls,then it is understood that the heavier ball is in that 3-ball group.Choose that 3-ball group;Keep any 2 of those 3 balls on oppossite sides of the balance and the third one aside.Note which side of the balance falls.The ball in that side which falls is the heaviest ball.Or if the balance is in equilibrium,then the third ball kept aside is the heaviest.
  
  step 2
  If the balance containing the 3-3 balls groups are in equilibrium,then al those 6 balls are of equal weight.So the third group of 2 balls has the heaviest one.
  Weigh the 2 balls on opposite sides;the side which falls has the heaviest.
• 11 years agoHelpfull: Yes(11) No(0)
• first time 3-3 balls
  if more weightage ball is not there
  then again 1-1 ball can weight
• 12 years agoHelpfull: Yes(1) No(11)