There is csk full of milk.'E' litres are drawn from the cask and replaced by water.this process is repeated.now the ration of milk and water in cask is 16:9.what is capacity of cask in litres

• capacity 5E
  milk finally +nt= 16/25
  x(4/5)*(4/5) implies (1/5)*x was drawn in each turn...
  x/5=E thus x=5E
• 11 years agoHelpfull: Yes(10) No(1)
• total milk=xE liters
  after first replacement milk=(x-1)E and water= E
  after second replacement milk=(x-2)E and water=2E
  ratio= (x-2)E/2E=16/9
  X=50/9E liters
• 11 years agoHelpfull: Yes(9) No(4)
• cant be determined
  bcz....no. of repetation is nt given
  
• 11 years agoHelpfull: Yes(6) No(3)
• x=5E.
  let cask capacity be x
  (x*(x-E)/x*(x-E)/x)/((x-(x*(x-E)/x*(x-E)/x))=16/9
• 11 years agoHelpfull: Yes(2) No(0)
• The total capacity is 5l and E=1.
  By Assuming Total value as X and replaced value as E and putting in [X(1-(Y/X)^n]
  we get
  [X(1-(E/X)^2] would be value or remaining milk
  and so Water added would be X-[X(1-(Y/X)2] (Total capacity- Milk present=Water)
  
  And putting against the given ratio of 16:9
  X-[X(1-(Y/X)2]/X-[X(1-(Y/X)2]=16/9
  Solving would give the result of X=5 and E=1
  :)
• 11 years agoHelpfull: Yes(2) No(1)
• E=(9/25)X
  
  Solve for any values of E and X
• 11 years agoHelpfull: Yes(1) No(0)
• answer is 7 ltr.
  let capacity of cask is y ltr,when the 1st time E ltr milk is replaced by water,then the ratio becomes (y-E)/E,next time ratio becomes (y-E)^2/E^2 & so on as we know that 9 is the square of 3,and 9 has no other power of any no,so e=3 and y-3=4 hence y=7 ltr.
• 11 years agoHelpfull: Yes(1) No(1)
• capacity = 25*E/9
• 11 years agoHelpfull: Yes(1) No(1)