Question

A container contains a mixture of two miscible liquids A and B in the ratio 5:7.after withdrawing 24 litres of the mixture from the container.same volume of liquid B was replaced .this makes change in the ratio of A and B in the container as 1:2.what is the initial volume of liquid A in the container?​

Answer 1

Solution :-

Let us Assume that initial in container liquid A was 5x and liquid B was 7x litres respectively.

Than,

→ Total mixture = 5x + 7x = 12x.

Now, we have given that, 24 litres of mixture was withdrawn from the container.

So, we can say that, in that 24 litre of mixture also the ratio of A and B was 5 : 7.

That means,

→ A withdrawn = 24 * (5/12) = 10 litre.

→ B withdrawn = 24 * (7/12) = 14 litre.

Than,

→ A left in container = initial A - withdrawn A = (5x - 10) litres.

→ B left in container = initial B - withdrawn B = (7x - 14) litres.

Now, we have given that, same volume of B was replaced and ratio of A and B becomes 1:2.

So,

→ Total A in container = (5x - 10)

→ Total B = (7x - 14) + 24 = (7x + 10)

and,

→ A : B = 1 : 2

Therefore,

→ (5x - 10) / (7x + 10) = 1/2

→ 2(5x - 10) = 7x + 10

→ 10x - 20 = 7x + 10

→ 10x - 7x = 10 + 20

→ 3x = 30

→ x = 10.

Hence,

→ initial volume of liquid A in container = 5x = 5*10 = 50 Litres. (Ans.)