There is a cask full of milk. 'E' litres are drawn from the cask, it is then filled with water. This process is repeated. Now the ration of milk to water in the cask is 16:9. What is the capacity of the cask in litres?

a. E+9
b. 9E
c. 5E

• 5e.....
  
  let volume be x.....
  after 1 draw... volume left===x*(1-e/x)
  after 2 draw....volume left===x*(1-e/x)^2.....eq no. 1"
  
  thus on other hand...ratio==16:9...thus amount of milk===16:25..... eq 2
  
  equating both eq...we get x=5e...
• 10 years agoHelpfull: Yes(4) No(1)
• Final conc= Initial conc* [1-(Amount replaced in each operation/total amount)]^no. of times operation is repeated
  so 16/25=1*[1-(E/Capacity)]^2
• 10 years agoHelpfull: Yes(4) No(1)
• let capacity is x.then initially x l of milk.
  after 1st removal milk remains=x-(x/x)*e=(x-e)
  now there is (x-e)l milk and e l water.
  after 2nd removal milk remains=(x-e)-(x-e)/x*e = (x-e)2/x.
  now (x-e)2/(x*x)=16/25.
  this gives x=5 & e=1.
  so capacity is C.5E
• 10 years agoHelpfull: Yes(2) No(1)
• ans= -3E becoz 1- (E/X)=4/3 so
• 11 years agoHelpfull: Yes(1) No(2)
• how do u calculate volume of left milk is x*(1-e/x)?
• 10 years agoHelpfull: Yes(1) No(0)
• Let water in the cask be x
  Milk left in the cask after the process is repeated twice=x(1-E/x)^2
  =>x(1-E/x)^2
  _____________ =16/(16+9)
  x
  =>(1-E/x)^2=16/25
  x=5
  E=1
• 11 years agoHelpfull: Yes(0) No(5)