Mixture Question

[engg.manik]

A can contains a mixture of two liquids A & B in ratio 7:5.A 9litre of mix is drawn out and can is filled with B. Te new ratio becomes 7:9. How many litres of A was contained by can initially

[a_forest]

Nasty one...

Let's say the can contains x liters. So at first the a content is 7/12*x and the b content is 5/12*x.

After 9 liters are drawn, the ratio is maintained. So the a content is 7/12*(x-9) and b content is 5/12*(x-9).

After 9 liters of b are added, the b content is 5/12*(x-9)+9. The ratio then is 7:9 so you can also say the b content is 9/16*x (the can is full again so we use x).

So: 5/12*(x-9)+9 = 9/16x
...
x=36

The original a content was 7/12*x=7/12*36= 21 liters.

Am I right?

[aa2kash]

Let x be the total vol.
B is 5x/12
if we take out 9 litres. then amount of B taken out is
5*9/12=15/4
and we are adding 9 litres of B.
but the final volumes remains the same i.e x.
And in final volume B is 9*x/16.
so the equation will be
5x/12 - 15/4 + 9 = 9*x/16
this gives x= 36.
And initial amount of A is 7*36/12 = 21 litres