anceer wrote:

A can contains a mixture of liquids A and B is the ratio 7:5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7:9. How many liter of liquid A was contained by the can initially?

A. 10

B. 20

C. 21

D. 25

E. 27

Suppose the can initially contain 7x and 5x litres of mixtures A and B respectively. The quantity of A in the mixture left = [7x – (7/12)×9] litres = [7x – 21/4] litres.

The quantity of B in mixture left = [5x – (5/12)×9] litres = [5x – 15/4] litres.

Therefore we can write: the ratio of the two quantities as [7x – 21/4]/ [5x – 15/4] = 7/9

In other words, we may say that 252x – 189 = 140x + 147

Hence, x = 3. Therefore the can had 7(3) = 21 litres of the quantity A. Thus the correct option is C.

Or

WAY 2 :

let A = 7x & B = 5x

total mixture = 7x+5x=12x

after 9 litre of mixture are drawn off remaining misture = 12x-9

after filling can with B:

A = (12x-9) * 7/12= 28x-21

B = {(12x-9)* 5/12}+9= 20x+21

So, (28x-21)/(20x+21)= 7/9

by solving above equation, we get

x=3

So liquid A was contained by the can initially = 7x = 7*3 = 21