8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of water is 16 : 81. How much wine did the cask hold originally?

A. 18 litres B. 24 litres

C. 32 litres D. 42 litres

• Let the quantity of the wine in the cask originally be x litres.
  
  Then, quantity of wine left in cask after 4 operations = x(1-8/x)^4 litres.
  therefore, (x(1-8/x)^4)/x=16/81
  =>(1-8/x)^4=(2/3)^4
  =>x-8/x=2/3
  =>3x-24=2x
  =>x=24 litres.
• 13 years agoHelpfull: Yes(26) No(12)
• quantity of wine=x(1-8/x)^4
  quantity of water=1-quantity of wine
  (1-8/x)^4/1-(1-8/x)^4=16/81
  (1-8/x)^4=16/81-16/81(1-8/x)^4
  97/81(1-8/x)^4=16/81
  1-8/x=2/3*(81/97)
  1-8/x=54/97
  x=18 liters
  ans (A)
• 12 years agoHelpfull: Yes(10) No(8)
• i got E here
  (1) the number of female=40 (50-10)
  insufficient
  (2)the number of black=0,4*5=20
  insufficient
  both insufficient
• 13 years agoHelpfull: Yes(2) No(22)
• Answer: Option B
  
  Explanation:
  
  Let the quantity of the wine in the cask originally be x litres.
  
  Then, quantity of wine left in cask after 4 operations = x 1 - 8 4 litres.
  x
  x(1 - (8/x))4 = 16
  x 81
  1 - 8 4 = 2 4
  x 3
  x - 8 = 2
  x 3
  3x - 24 = 2x
  
  x = 24.
• 13 years agoHelpfull: Yes(2) No(16)
• Let the quantity of the wine in the cask originally be x litres.
  Using formula:
  
  Final Amount of solute that is not replaced=
  Initial Amount×(Vol. after removalVol. after replacing)n ---(1)
  
  Or
  Final ratio of solute not replaced to total=
  Initial ratio×(Vol. after removalVol. after replacing)n ---(2)
  
  Considering 2nd formula here,
  
  Then ratio of wine to total solution in cask after 4 operations:
  ⇒1×(x−8x)4⇒x−8x⇒3x−24⇒x=24 litres.=1681=23=2x
• 6 years agoHelpfull: Yes(0) No(0)