A vessel contained a certain amount of a solution of acid and water. When 2 litres of water was added to it, the new solution had 50% acid concentration. When 15 litres of acid was further added to this new solution, the final solution had 80% acid concentra tion. The ratio of water and acid in the original solution was: Options:

The correct option is 3 : 5.

Let us denote the initial quantity of acid in the vessel as
$A$
litres and the initial quantity of water as
$W$
litres.

Step 1: Analyze the first dilution
When 2 litres of water is added to the solution, the quantity of water becomes
$W+2$
litres. The quantity of acid remains unchanged at
$A$
litres.

The new total volume of the solution is:

$\text{Total Volume}=A+W+2$

Since the new solution has an acid concentration of 50%, we can write the equation:

$\frac{A}{A+W+2}=\frac{50}{100}=\frac{1}{2}$

Cross-multiplying to simplify:

$2A=A+W+2$

$A=W+2$         — (Equation 1)

Step 2: Analyze the addition of acid
When 15 litres of acid is further added, the quantity of acid becomes
$A+15$
litres. The quantity of water remains
$W+2$
litres.

The final total volume of the solution is:

$(A+W+2)+15=A+W+17$

The final solution has an 80% acid concentration, which gives:

$\frac{A+15}{A+W+17}=\frac{80}{100}=\frac{4}{5}$

Cross-multiplying to simplify:

$5(A+15)=4(A+W+17)$

$5A+75=4A+4W+68$

$A-4W=-7$         — (Equation 2)

Step 3: Solve for the volumes of water and acid
Substitute the expression for
$A$
from Equation 1 into Equation 2:

$(W+2)-4W=-7$

$-3W+2=-7$

$-3W=-9$

$W=3$

Now substitute this value of
$W$
back into Equation 1 to find the volume of acid:

$A=3+2=5$

Step 4: Find the original ratio
The ratio of water to acid in the original solution is:

$W:A=3:5$

This matches the correct option of 3 : 5.