A chemist has $40$ mL of a $25\%$ acid solution. How many mL of a $60\%$ acid solution must be added to obtain a final mixture that is $40\%$ acid?

Let $x$ = mL of $60\%$ solution added. The mixture has $40 + x$ total mL. Acid balance: $0.25(40) + 0.60(x) = 0.40(40 + x)$. $10 + 0.60 x = 16 + 0.40 x$ $0.20 x = 6$ $x = 30$ mL. Check: total volume $70$ mL; acid $= 10 + 18 = 28$ mL; concentration $= 28/70 = 0.40$ ✓. The answer must lie *between* $25\%$ and $60\%$ — $40\%$ is closer to $25\%$ than to $60\%$, so we need *less* than $40$ mL of the strong solution. $30$ mL fits.