A vessel contains a solution of acid and water in the ratio 5 : 7. When 9 litres of the solution are taken out and the vessel is filled with equal quantity of acid, the ratio of acid and water in the vessel becomes 9 : 7. How many litres of solution was there in the vessel, initially?

Correct option is A

Given:
Initial ratio of acid : water = 5 : 7
9 litres of solution are removed.
Same 9 litres of acid are added.
New ratio of acid : water = 9 : 7
Formula Used:
Final Acid Quantity = Initial Acid - Acid Removed + Acid Added
Final Water Quantity = Initial Water - Water Removed
Solution:
Let the initial quantity = x litres.
Initial acid = $$\frac{5}{12}x$$​​

Initial water = $$\frac{7}{12}x$$​

When 9 litres are removed:

Acid removed = $$\frac{5}{12} \times 9 = \frac{15}{4} = 3.75$$ litres​

Water removed = $$\frac{7}{12} \times 9 = \frac{21}{4} = 5.25$$​ litres

Now,
Acid left = $$\frac{5}{12}x - 3.75$$​​

Water left = $$\frac{7}{12}x - 5.25$$​

After adding 9 litres of acid:

New Acid = $$\frac{5}{12}x - 3.75 + 9 = \frac{5}{12}x + 5.25$$​

Water remains = $$\frac{7}{12}x - 5.25$$​

New ratio = 9 : 7 Thus,

​$$\frac{\frac{5}{12}x + 5.25}{\frac{7}{12}x - 5.25} = \frac{9}{7}$$​

​$$7\left( \frac{5}{12}x + 5.25 \right) = 9\left( \frac{7}{12}x - 5.25 \right)$$​

​$$\frac{35}{12}x + 36.75 = \frac{63}{12}x - 47.25$$​

35x + 441 = 63x - 567

63x - 35x = 441 + 567

28x = 1008

x = 36

Thus, the correct option is (a) ₹36 litres

Alternate Method:

Acid  : Water

5       :      7      = 12 units

After removing 9 litre solution and adding 9 litre acid:

9       :       7

Change in acid Ratio = 9 - 5 = 4 units

Initial amount = $$\frac{9 \space\text{litre}}{4\space\text{units}} \times 12\space \text{units}$$ + 9 litres = 27 litres + 9 litres = ₹36 litres​