crazy new logical puzzles

Replies

• 

  bharathpb

  4 trials..is It??
• 

  vineetdiit

  no.....
  want to hear some different answer...
• 

  bharathpb

  differenT answer??
  
  To which are you exactly refering as the ODD BALL....CONFIRM PLZ..
• 

  vineetdiit

  the ball which is different in weight when compared to others
• 

  vineetdiit

  the minimum possibility checks in the weighing balance required?....
  odd ball : the ball which is different from others (weight defective 1)
• 

  Shruthi.S.D

  first u weigh 1 ball then 2nd ball with first then 1st and 2nd ball with 3rd and 4th and so on ... by 6 trials /weighing u wil come to the odd ball . Hope ur looking for this ans.
• 

  vineetdiit

  1st check 10 vs 10...if they are equal...then the remaining 1 is the odd ball
  if its not then....
  take the (10) set which has more weight....and again check for 5 vs 5......take the set(5) which has more weight...
  check 2 vs 2 in that (5)set ...if equal then the remaining is the odd ball.....if not...
  then check the 2 balls set which has more weight....
  
  
  so to get correct always.....the answer is 4 checks
  
  
  
  
  
  but the minimum possibility check is always ...................1
• 

  mohan_185

  maximum probability is 4 and minimum is 1
• 

  rushi4444

  vineetdiit

  1st check 10 vs 10...if they are equal...then the remaining 1 is the odd ball
  if its not then....
  take the (10) set which has more weight....and again check for 5 vs 5......take the set(5) which has more weight...
  check 2 vs 2 in that (5)set ...if equal then the remaining is the odd ball.....if not...
  then check the 2 balls set which has more weight....
  
  so to get correct always.....the answer is 4 checks
  
  but the minimum possibility check is always ...................1

  in question it is not mention that odd ball is heavier than others or lighter than others......so u cant proceed with havier group.......u have to decide first that faulty ball is havier than others or lighter than others....
  
  second thing is.....in this kind of problems never partition the given number into group of half.....by grouping them into half of the elements ,every time u r eleminating only 50% elements.........better approach is partition them into n/3, n/3, n/3..........so every time u can eliminate 66% of remaining elements.....
  
  for example...in this question partition the given no 21 into p1=7,p2=7,p3=7.....now compare p1 & p2 ..if equal tehn p3 contains odd ball but still u dont know with more weight or less weight than others...To know that compare p3 with either p1 or p2......now partition p3 into p3/3 ,p3/3 ,p3/3...i.e. g1=2, g2=2, g3=3......compare g1 & g2.....if equal then g3 contains faulty ball otherwise either g1 or g2 contains faulty ball........
  
  if u know the comaritive weight of faulty piece than....maximum attempt is ceil(log3(n)) (i.e. ceiling value of log n to the base 3)......for 21 it is 3....for 27 it is 3 but for 28 it is 4....
• 

  Jayesh289jd

  maximum is 3times we get answer.
• 

  Farjand

  Way 2 balls one by one.
  If you get different ball then OK.
  next you half the balls that is 20 on one side and 21 on other. weigh them. Now you know the weight of ball and number of balls. So you can easily tell the odd ball. So total 4 chances at the maximum.