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?

Hint:

Each step of this problem tells you what a percentage refers to — sometimes the water, sometimes the acid, and sometimes the whole solution. Underline what each percentage applies to before doing any arithmetic, and remember that removing a uniformly mixed sample takes the same percentage from every substance inside it.

Answer 1

Correct Answer: 45.9

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 \(= \boxed{45.9 \text{ litres}}\).

Answer 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: \[ 60 \times 1 \times 0.9 \times 0.85 = 60\times0.765=\boxed{45.9 \text{ L}} \]

Answer 3

Concept:

• In mixture problems, tracking the acid concentration (percentage of the total solution that is acid) at each stage shows directly what a uniform removal step takes away.
• Evaporating pure water changes the total volume but leaves the acid volume unchanged, so it raises the acid concentration.
• Removing a fixed percentage of a well-mixed solution removes that same percentage of every substance present, including the acid, so the concentration itself stays the same during such a removal; only the total amount changes.

Step 1: Find the concentration right after evaporation.
Start: acid $= 60$ L, water $= 140$ L, total $= 200$ L. Evaporating $20\%$ of the water removes $0.2 \times 140 = 28$ L of pure water. New total $= 172$ L, acid is still $60$ L.

Step 2: Find the concentration after acid extraction.
Removing $10\%$ of the acid takes away $0.1 \times 60 = 6$ L: new acid $= 54$ L, new total $= 172 - 6 = 166$ L.

Step 3: Apply the final $15\%$ removal.
Removing $15\%$ of the $166$ L solution takes a representative sample with the same concentration as the bulk, so the remaining volume is $166 \times 0.85 = 141.1$ L, and the remaining acid is $141.1 \times \dfrac{54}{166} = 54 \times 0.85 = 45.9$ L. Water is then added back to restore the total to $166$ L, without adding any acid.

Final Answer: Volume of acid in the final solution $= 45.9$ litres.