How many liters of pure water must be added to $30$ liters of a $20\%$ saline solution to dilute it to a $12\%$ saline solution?

Adding pure water *preserves* the absolute amount of salt but **dilutes** the percentage. Salt amount: $0.20 \cdot 30 = 6$ L of salt. Let $x$ = liters of water added. New volume $= 30 + x$, new percentage $= \dfrac{6}{30 + x} = 0.12$. Solve: $6 = 0.12 (30 + x) = 3.6 + 0.12 x \Rightarrow 0.12 x = 2.4 \Rightarrow x = 20$ L. Check: final volume $50$ L; salt $6$ L; concentration $6/50 = 0.12$ ✓. - Trap C ($25$) misreads the equation. - Trap A is half the answer. - Trap D doubles it.