In a mixture of acid and water, if 1 liter of water is added, then the new mixture contains 20% acid. When 1 liter of acid is added to the new mixture, the acid in the resulting mixture is  \(33\frac{1}{33}\)%. The percentage of acid in the original mixture is

Given:

After adding 1 L water → acid becomes 20%.

Then adding 1 L acid → acid becomes 33¹/₃%.

Formula used:

Percentage = \(\dfrac{\text{Acid}}{\text{Total}} \times 100\)

Calculation:

Let acid = x, water = y

From 1st case: \(x = 0.2(x + y + 1)\)

⇒ 0.8x = 0.2y + 0.2 ... (1)

From 2nd case: \(x + 1 = \dfrac{1}{3}(x + y + 2)\)

⇒ 3x + 3 = x + y + 2 ⇒ 2x - y = -1 ... (2)

From (1): y = 4x - 1

Put in (2): 2x - (4x - 1) = -1 ⇒ -2x + 1 = -1 ⇒ x = 1

Then y = 4(1) - 1 = 3

Original % of acid = \(\frac{1}{1 + 3} \times 100 = 25\%\)

∴ The correct answer is option (3).