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Let xx be the volume (mL) of 20% solution, mixed with (500−x)(500 - x) mL of water (0% solute). The mass (or equivalent) of solute contributed is 0.20x0.20x grams per 100 mL, but percentage math works directly with volumes here since both sides are in mL and percent (w/v). The target solution contains 0.12×500=600.12 \times 500 = 60 units of solute (e.g., grams per 100 mL scaling). Set 0.20x=60⇒x=300 mL0.20x = 60 \Rightarrow x = 300\ \text{mL}. Options A and B underdeliver solute, producing ≤12% when diluted to 500 mL, while D overshoots the required solute mass, exceeding 12%. A quick check: 300 mL of 20% contains 60 units; adding 200 mL water totals 500 mL with 60/500 = 12%. On the DAT, set up mixture equations by equating total solute before and after mixing, and remember that adding pure solvent scales volume but not solute amount—keeping the calculation straightforward.