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 ratio of milk to water in the cask is 16:9. what is the capacity of the cask in liters?
a)E+9
b)9E
c)5E
d)Cannot be determined

• if n litres of milk is taken out from M litres of milk and replaced by water then-
  if the process is all total repeated k times the resultant portion of milk is=
  {(M-n)/M}^k
  here k=2 M=unknown and n=E and resultant portion of milk is = 16/(16+9)=16/25
  so,
  (M-E)/M =(16/25)^0.5= 4/5 or M=5E (ans)
• 10 years agoHelpfull: Yes(12) No(4)
• Ans: 5E
  Let the quantity of the milk in the cask originally be x litres
  Then, quantity of milk left in cask after 2 operation(1 first time drawn and one for repeated)
  === x(1-E/x)^2
  so, (x(1-E/x)^2/x)=16/16+9 i.e 25;
  1-E/x=4/5
  5x-5E=4x
  x=5E;
  
  
• 10 years agoHelpfull: Yes(7) No(1)
• cannot be determine untill repitition is not given..
  It will not be given that after how many repition ratio become 16:9.
• 10 years agoHelpfull: Yes(2) No(3)
• a) E+9 is right
• 10 years agoHelpfull: Yes(2) No(6)
• Let the total Liters of Milk in cask is "T"
  Given "E" liters of milk are drawn simultaneously with repetation of N times And filled with equivalent water.
  So the first operation has a fraction => (T-E)^1/Nth times / T^Nth times = {(T-N)/T}^1/Nth times
  Again taking N=2 we get,
  =>T-E/T=4/5 =>T=5E , When Nth=2 time repetation. so the Answer is D. Cannot be determined.
• 9 years agoHelpfull: Yes(2) No(0)
• final conc=initial conc(1- amount replaced/total volume)^n
  
  n= number of times replaced
  
  conc of the material is used whose conc is decreasing
  
  16/25=(1-E/V)^2
  
  1-E/V=4/5
  
  V=5E
• 9 years agoHelpfull: Yes(1) No(0)
• ans is 5E
  Left amount = initial amount(1-replace amount / total amount)^n
  let suppose total amount = 5E
  16/25 = (1-E/5E)^n
  (4/5)^2 = (4/5)^2
  n=2..
  
  
• 10 years agoHelpfull: Yes(0) No(1)
• Shamik Maity can u plz tell me how u take .5 as power???
• 10 years agoHelpfull: Yes(0) No(0)
• Milk Left after k opeartion is : M(1-E/M)^2 [ // Formula for milk left after successive operation is M(1-E/M)^n]
  Water in mixture : M- M(1-E/M)^2
  {M(1-E/M)^2}/{M- M(1-E/M)^2}=16/9
  M=5E
• 9 years agoHelpfull: Yes(0) No(0)
• this question has 1 answer.
• 8 years agoHelpfull: Yes(0) No(0)