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Let xx be liters of the 30% solution and 20−x20-x be liters of the 10% solution. Match total acid:

0.30x+0.10(20−x)=0.18⋅20=3.6.0.30x + 0.10(20 - x) = 0.18 \cdot 20 = 3.6.

Simplify: 0.30x+2−0.10x=3.6⇒0.20x+2=3.6⇒0.20x=1.6⇒x=8.0.30x + 2 - 0.10x = 3.6 \Rightarrow 0.20x + 2 = 3.6 \Rightarrow 0.20x = 1.6 \Rightarrow x = 8.

So, 8 L of 30% and 12 L of 10% yield the 18% target. A quick alligation check gives the same result: place 30 and 10 at the ends with target 18; distances from the target are 12 (30→18) and 8 (18→10). The ratio of amounts is inverse to these distances: 8 (30%) : 12 (10%) → simplifies to 2:3; out of 20 L, that’s 20⋅25=820 \cdot \frac{2}{5} = 8 L of 30%. Common pitfalls include averaging the percentages (which would be valid only for equal volumes) or forgetting the total-volume constraint. Always base on total pure acid, not on the percentages alone.