A container contains two liquids A and B in the ratio 8 : 5 .When 13 liters of mixture is drawn off and is completely replaced with liquid B, then the ratio of A and B in the container becomes 1 : 1. How many liter of liquid A was in the container initially.

The correct option is $\frac{128}{3}$ liter.

Let us solve this problem step-by-step to find the initial quantity of liquid A in the container.

Step 1: Express initial quantities in terms of a variable
Given that the initial ratio of liquid A to liquid B in the container is 8 : 5.
Let the initial quantity of liquid A be $8x$ liters and liquid B be $5x$ liters.
Thus, the total initial volume of the mixture is:

$8x+5x=13x$ liters.

Step 2: Determine the amounts of liquid A and B removed
When 13 liters of the mixture is drawn off, liquid A and liquid B are removed in the same ratio of 8 : 5.
Amount of liquid A removed =

$13\times \frac{8}{8+5}=13\times \frac{8}{13}=8$ liters.

$13\times \frac{5}{8+5}=13\times \frac{5}{13}=5$ liters.

Step 3: Calculate remaining quantities after replacement
After removing 13 liters of mixture, 13 liters of pure liquid B is added to the container.
Remaining quantity of liquid A =

$8x-8$ liters.

$(5x-5)+13=5x+8$ liters.

Step 4: Set up the equation using the new ratio
The new ratio of liquid A to liquid B is given as 1 : 1.
Therefore, we can set up the proportion:

$\frac{8x-8}{5x+8}=\frac{1}{1}$

$8x-8=5x+8$

$8x-5x=8+8$

$3x=16$

$x=\frac{16}{3}$

Step 5: Calculate the initial quantity of liquid A
The initial quantity of liquid A was $8x$ liters.
Substituting the value of $x$:

$\text{Initial A}=8\times \frac{16}{3}=\frac{128}{3}$ liters.

Hence, the initial quantity of liquid A in the container was $\frac{128}{3}$ liters.