A bucket contains a mixture of two liquids a and b in the proportion 5: 7 . if 6 litres of the mixture is replaced by 6 litres of liquid b, then the ratio of the two liquids becomes 3 : 5. how much of the liquid b was there in the bucket initially?
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Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively.
When 9 litres of mixture are drawn off, quantity of A in mixture left:
[7x–(7/12)×9] =[7x - (21/4)] litres.
Similarly quantity of B in mixture left,
[5x−(5/12)×9] = [5x - (15/4)] litres
Therefore ratio becomes,
[7x −(21/4)]/ [5x - (15/4)] = 7/9
=>(28x -21)/(20x+21)=7/9
=>(252x -189)=140x+147
=>112x=336
=> x=3
So the can contained,
7*x=7*3= 21 litres of A initially.
i think so..
When 9 litres of mixture are drawn off, quantity of A in mixture left:
[7x–(7/12)×9] =[7x - (21/4)] litres.
Similarly quantity of B in mixture left,
[5x−(5/12)×9] = [5x - (15/4)] litres
Therefore ratio becomes,
[7x −(21/4)]/ [5x - (15/4)] = 7/9
=>(28x -21)/(20x+21)=7/9
=>(252x -189)=140x+147
=>112x=336
=> x=3
So the can contained,
7*x=7*3= 21 litres of A initially.
i think so..
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