There are four teams A, B, C, D playing game. If any one team loses, it will play twice the money to all other teams. They play 3 games. B, C, D loses one game each in the order. Finally A & B has Rs. 40 each & C has Rs. 80 & D has Rs. 16.
Which team has started with minimum money? i) A ii) B iii) C iv) D

• A will have minimum money.
  starting from end....
  A have 40 rs, B - 40, C - 80, D - 16
  at last D loses the game that mean D pays the twice money to all hence before D pays A, B and C would have half of the money,
  hence before D loses
  A have 20 rs, B - 20, C - 40, D =16+ 20+20+40= 96 rs
  
  Now before D, C looses the game. Hence before C looses the game all would be having half of money i.e. and C would be having money he has now plus money he pay to all.
  Now A have 10 rs, B have 10 rs, D have 48 rs, C have =40+ 10+10+48=108
  
  Similarly before B looses everyone would be having
  A - 5 rs, C - 54, D - 24, B have = 10+ 5+54+24=93
  
  Hence A started with minimum money is 5 and B started with max is 93
  
  
• 11 years agoHelpfull: Yes(13) No(3)