A storage container C initially holds 50 liters of solvent A and the remaining volume as solvent B. When 20% of the solution is drawn out from container C, and 14 liters of solvent A together with 17 liters of solvent B are poured into container C, the ratio of solvent A to solvent B becomes 2:3. Calculate the initial volume (in liters) of solvent B in the mixture.

The correct answer is 80.

Step-by-Step Solution:

Step 1: Define variables for the initial setup

Let the initial volume of solvent B in container C be $x$ liters.

The initial volume of solvent A is given as 50 liters.

Thus, the initial total volume of the solution is $(50+x)$ liters.

Step 2: Calculate the remaining solvent volumes after drawing out 20%

When 20% of the mixture is drawn out, 80% (or 0.8) of each solvent remains in the container.

Remaining volume of solvent A = $50\times 0.8=40$ liters.

Remaining volume of solvent B = $0.8x$ liters.

Step 3: Calculate the new solvent volumes after additions

Next, 14 liters of solvent A and 17 liters of solvent B are poured into container C.

New volume of solvent A = $40+14=54$ liters.

New volume of solvent B = $0.8x+17$ liters.

Step 4: Set up the equation using the final ratio and solve for $x$

We are given that the final ratio of solvent A to solvent B is 2 : 3.

$\frac{54}{0.8x+17}=\frac{2}{3}$

Cross-multiplying both sides gives:

$54\times 3=2\times (0.8x+17)$

$162=1.6x+34$

Subtract 34 from both sides:

$1.6x=162-34$

$1.6x=128$

Divide both sides by 1.6:

$x=\frac{128}{1.6}=80$

Therefore, the initial volume of solvent B in the mixture was 80 liters.