A vessel is full of a mixture of kerosene ane petrol in which there is 18% kerosene. 8liter are drawn off and then the vessel is filled with petrol. If the kerosene is now 15% ,how much does the vessel hold?
A-40
B-32
C-36
D-48

• Let the total capacity of vessel be x litres
  Kerosene initially= 18x/100
  Kerosene finally= 15x/100
  Kerosene present in 8 liters of mixture= 18*8/100
  
  18x/100 - 18*8/100 = 15x/100
  3x = 144
  x = 48 litres
  
• 7 years agoHelpfull: Yes(7) No(1)
• Let us suppose ,vessel contains x litre of mixture Initially , kerosene=18% so petrol is (x-18)% afterwards,kerosene=15% after 8 litre is drawn off. So change is 3% or 3 litre of kerosene is drawn off . Initial ratio,18:x-18 so qty of kerosene drawn,18*8/x=3 which gives x=48litres
• 8 years agoHelpfull: Yes(2) No(0)
• Lets solve this question as a Allegatons question.
  
  Even after 8 litres are removed, the remaining quantity has 18% of kerosene, we are mixing this remaining quantity with pure petrol( 0% kerosene) to get the result of 15% kerosene
  
  These two quantities should be mixed in the ratio = 15-0/18-15 = 15/3 = 5:1 ratio
  As we know the amount of petrol is 8 litres, the amount of the other mixture is 5* 8 = 40 litres
  But 40 is not the answer, as it is the amount after 8 litres are removed
  So, the capacity of the vessel is 48 litres
  
  Answer is 48 litres
  
• 10 years agoHelpfull: Yes(1) No(2)
• even i used alligations. for petrol
  n1 values : n2 values
  8liters x liters
  85%
  100% 82%
  
  
  (85-85)/8 = 100-85/x
  here 85% is the average amount which we get after mixing and n1 are n2 are the values before mixing
  i think the answer should be 48 , 40 will be the amount to which the pure petrol is added ..
  the vessel holds 48 liters
• 8 years agoHelpfull: Yes(0) No(0)