Question

A container holds 200 litres of a solution of acid and water, having 30% acid by volume. Atul replaces 20% of this solution with water, then replaces 10% of the resulting solution with acid, and finally replaces 15% of the solution thus obtained, with water. The percentage of acid by volume in the final solution obtained after these three replacements, is nearest to?

• A. \(25\)
• B. \(27\)
• C. \(29\)
• D. \(23\)

Hint:

In replacement problems, always track the \emph{amount} of the substance (here, acid) after each step. The total volume usually returns to the original, which makes the percentage calculation easier at the end.

Answer 1

The Correct Option is B

Step 1: Initial composition of the solution. The total volume of the solution is \(200\) L. Since acid constitutes \(30\%\) of the solution, the initial quantity of acid is \[ 0.30 \times 200 = 60 \text{ L}. \] Step 2: First replacement of solution with water. In the first operation, \(20\%\) of the solution is removed: \[ 0.20 \times 200 = 40 \text{ L}. \] As the solution is homogeneous, the amount of acid removed is \(20\%\) of 60 L: \[ 0.20 \times 60 = 12 \text{ L}. \] Thus, the remaining acid is \[ 60 - 12 = 48 \text{ L}. \] After adding 40 L of water, the total volume again becomes 200 L, with acid equal to 48 L. Step 3: Second replacement of solution with acid. Next, \(10\%\) of the solution is withdrawn: \[ 0.10 \times 200 = 20 \text{ L}. \] The acid removed at this stage is \(10\%\) of 48 L: \[ 0.10 \times 48 = 4.8 \text{ L}. \] The remaining acid is therefore \[ 48 - 4.8 = 43.2 \text{ L}. \] When 20 L of pure acid is added, the amount of acid becomes \[ 43.2 + 20 = 63.2 \text{ L}, \] while the total volume remains 200 L. Step 4: Third replacement of solution with water. Now, \(15\%\) of the solution is removed: \[ 0.15 \times 200 = 30 \text{ L}. \] The acid removed in this step is \(15\%\) of 63.2 L: \[ 0.15 \times 63.2 = 9.48 \text{ L}. \] Hence, the acid left in the solution is \[ 63.2 - 9.48 = 53.72 \text{ L}. \] After adding 30 L of water, the total volume is restored to 200 L. Step 5: Final concentration of acid. The final percentage of acid in the solution is \[ \frac{53.72}{200} \times 100 = 26.86\% \approx 27\%. \] Therefore, the percentage of acid in the final solution is closest to \(27\%\).

Answer 2

Step 1: Initial state.
Total volume of solution \(= 200\) L. Acid is \(30%\) of 200 L: \[ \text{Acid} = 0.3 \times 200 = 60 \text{ L.} \]
Step 2: First replacement (20% with water).
We remove \(20%\) of the solution: \[ 0.2 \times 200 = 40 \text{ L.} \] Since the solution is uniform, acid removed is \(20%\) of 60 L: \[ \text{Acid removed} = 0.2 \times 60 = 12 \text{ L.} \] Remaining acid: \[ 60 - 12 = 48 \text{ L.} \] We now add 40 L of water, so total volume is again 200 L. Current acid amount: \(48\) L.
Step 3: Second replacement (10% with acid).
We remove \(10%\) of the 200 L solution: \[ 0.1 \times 200 = 20 \text{ L.} \] Acid removed is \(10%\) of 48 L: \[ \text{Acid removed} = 0.1 \times 48 = 4.8 \text{ L.} \] Remaining acid: \[ 48 - 4.8 = 43.2 \text{ L.} \] Now we add 20 L of pure acid, so: \[ \text{New acid amount} = 43.2 + 20 = 63.2 \text{ L.} \] Total volume returns to 200 L.
Step 4: Third replacement (15% with water).
We remove \(15%\) of the 200 L solution: \[ 0.15 \times 200 = 30 \text{ L.} \] Acid removed is \(15%\) of 63.2 L: \[ \text{Acid removed} = 0.15 \times 63.2 = 9.48 \text{ L.} \] Remaining acid: \[ 63.2 - 9.48 = 53.72 \text{ L.} \] We add 30 L of water, so total volume is again 200 L.
Step 5: Final percentage of acid.
\[ \text{Percentage of acid} = \frac{53.72}{200} \times 100 = 26.86% \approx 27%. \] So, the percentage of acid in the final solution is nearest to \(27%\).