A can contains a mixture of two liquids A and B in proportion 7:5. When 9 litres of mixture was drawn off and the can was filled with B, the proportion of A and B becomes 1:2. How many litres of mixture was contained by the can initially?

• let initially mixer=x
  A=7x/12 and b=5x/12
  and 9 liter drown
  
  A=7(x-9)/12 b=5(x-9)/12
  filled with b
  b=5(x-9)/12+9
  A=7(x-9)/12
  
  now ratio of A and B is 1:2
  means b=2A
  5(x-9)/12+9=2*7(x-9)/12
  9=9(x-9)/12
  x=21
  
  ans=21 liter mixer
  thank you
• 10 years agoHelpfull: Yes(43) No(3)
• Let a:b=7:5 =7x:5x (let assume x)
  then total 12x
  if 9 litre of mixture is drown off then same then remaining parts of a and b in the mixture
  a= 7x-7x*9/12x =7x-21/4 and thus b also 5x-15/4
  now 9 liters of b is added then the amount of b became (7x-15/4)+9
  now the ratio of a and b is given as 1:2
  so 7x-21/4:(7x-15/4)+9=1:2
  after solving value of x is 7/4
  so the total amount of mixture initially was 12x= 12*7/4= 21 liters
  ans. 21 liters .
• 10 years agoHelpfull: Yes(6) No(0)
• (7x-9)/(5x+9)=1/2
  by solving x=3
  
  so can intially contains 7x+3x=21+15=36
• 10 years agoHelpfull: Yes(2) No(5)
• Suppose the can initially contains 7x and 5x of mixtures A and B respectively.
  
  Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
  12 4
  Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
  12 4
  
  7x - 21
  4
  = 7
  5x - 15 + 9
  4
  9
  28x - 21 = 7
  20x + 21 9
  252x - 189 = 140x + 147
  
  112x = 336
  
  x = 3.
  
  So, the can contained 21 litres of A.
• 10 years agoHelpfull: Yes(2) No(1)
• 48 liters was contained initially in the can....
  
  
• 10 years agoHelpfull: Yes(1) No(4)
• ans is = 21 litres
• 10 years agoHelpfull: Yes(0) No(0)