Question:
A 200-litre container holds a solution that is 30% acid and the rest water. The solution undergoes the following three processes sequentially:
1. 20% of the water content is evaporated.
2. From the remaining mixture, 10% of the acid content is chemically extracted and removed.
3. Finally, 15% of the resulting solution is removed and replaced with water.
What is the volume of acid in the final solution?
A 200-litre container holds a solution that is 30% acid and the rest water. The solution undergoes the following three processes sequentially:
1. 20% of the water content is evaporated.
2. From the remaining mixture, 10% of the acid content is chemically extracted and removed.
3. Finally, 15% of the resulting solution is removed and replaced with water.
What is the volume of acid in the final solution?
1. 20% of the water content is evaporated.
2. From the remaining mixture, 10% of the acid content is chemically extracted and removed.
3. Finally, 15% of the resulting solution is removed and replaced with water.
What is the volume of acid in the final solution?
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In multi-step mixture problems, always keep a clear record of the volume of each component (e.g., acid, water) and the total volume after each step. Pay close attention to what each percentage refers to—the total solution, a specific component, etc.
Updated On: Jul 4, 2026
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Correct Answer: 45.9
Approach Solution - 1
Approach: Track only the acid through each step. Acid changes only when acid itself is removed; evaporating water or adding water doesn't touch it, and removing a uniform 15% of the mixture removes 15% of the acid too.
Step 0: Start with \(200\) L: acid \(= 200 \times 0.30 = 60\) L, water \(= 140\) L.
Step 1 (evaporate 20% of water): Water becomes \(140 \times 0.8 = 112\) L. Acid unchanged at \(60\) L. Total \(= 172\) L.
Step 2 (remove 10% of acid): Acid \(= 60 \times 0.9 = 54\) L. Total \(= 172 - 6 = 166\) L.
Step 3 (remove 15% of the mixture, replace with water): A uniform 15% removal keeps \(85\%\) of whatever is in solution, including the acid. So acid \(= 54 \times 0.85 = 45.9\) L. The replacement is water, which adds no acid.
Answer: Volume of acid in the final solution \(= \boxed{45.9 \text{ litres}}\).
Step 0: Start with \(200\) L: acid \(= 200 \times 0.30 = 60\) L, water \(= 140\) L.
Step 1 (evaporate 20% of water): Water becomes \(140 \times 0.8 = 112\) L. Acid unchanged at \(60\) L. Total \(= 172\) L.
Step 2 (remove 10% of acid): Acid \(= 60 \times 0.9 = 54\) L. Total \(= 172 - 6 = 166\) L.
Step 3 (remove 15% of the mixture, replace with water): A uniform 15% removal keeps \(85\%\) of whatever is in solution, including the acid. So acid \(= 54 \times 0.85 = 45.9\) L. The replacement is water, which adds no acid.
Answer: Volume of acid in the final solution \(= \boxed{45.9 \text{ litres}}\).
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Approach Solution -2
Approach: Track only the fraction of the ORIGINAL acid that survives each step, since evaporation only touches water/total volume, not the acid amount, while a proportional removal step takes away the same percentage of whatever acid remains.
Initial acid \(=30\% \times 200=60\) L.
Step 1 (evaporating water only): acid is untouched, retention factor \(=1\).
Step 2 (10% of the acid extracted): acid retention factor \(=0.9\).
Step 3 (15% of the whole solution removed and replaced by water): the removal is uniform across the mixture, so exactly \(15\%\) of the remaining acid also leaves, retention factor \(=0.85\).
Multiplying the three factors gives the final acid volume: \[ 60 \times 1 \times 0.9 \times 0.85 = 60\times0.765=\boxed{45.9 \text{ L}} \]
Initial acid \(=30\% \times 200=60\) L.
Step 1 (evaporating water only): acid is untouched, retention factor \(=1\).
Step 2 (10% of the acid extracted): acid retention factor \(=0.9\).
Step 3 (15% of the whole solution removed and replaced by water): the removal is uniform across the mixture, so exactly \(15\%\) of the remaining acid also leaves, retention factor \(=0.85\).
Multiplying the three factors gives the final acid volume: \[ 60 \times 1 \times 0.9 \times 0.85 = 60\times0.765=\boxed{45.9 \text{ L}} \]
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