A cask full of milk,'E' liters is removed and water is mixed. This process is repeated and now the ratio in milk to water is 16:9. Then what is the amount of milk in liter.
E+9 9E 5E cant be determined

• Ans is 5E...
  EXPLANATION-------------
  
  After 'n' operations ,the amount of pure liquid remaining is = [x { ( 1-E/x )^n } ]...where x is original amount of liquid and E is amount of liquid removed..This is standard formula in RS aggarwal...
  
  Now ,in our case the operations is performed twice...
  
  Let the original milk be x liters..
  
  E liters removed twice n = 2,
  
  milk remaining (M) = [ x { (1-E/x)^2 } ]....water is (W) = x - [x { (1-E/x)^2 } ]
  
  M/W = 16/9 Take (1-E/x)^2 = z
  
  now solving we get z = 16/25 that is (1 - E/x)^2 = (4/5)^2
  
  1 - E/x = 4/5
  
  x = 5E (Answer)
  
• 9 years agoHelpfull: Yes(22) No(0)
• 5E
  ......................
  suppose initially milk was X
  now after process
  milk=k(1-E/k)^2...........eq
  water_remained=k-milk......eq
  eq/eq=16/9
  k=5E
  
• 9 years agoHelpfull: Yes(1) No(1)
• i think cant be determined is ans.
  
• 9 years agoHelpfull: Yes(0) No(7)
• can't be determine..
  
• 9 years agoHelpfull: Yes(0) No(2)