Let the original volume be $v$ liters. Since the solution was originally $12\%$ acid, the amount of acid was $0.12v$ liters. Adding $8$ liters of pure water does not change the amount of acid, so the new total volume is $v+8$ and the acid amount is still $0.12v$. The new solution is $9\%$ acid, so

Now solve:

So the original volume was $24$ liters.

First convert the rate from gallons per $5$ hours to gallons per hour: $\frac{1.6}{5}=0.32$ gallons per hour. Then convert gallons to liters: $0.32\times 3.8=1.216$ liters per hour. Finally, convert hours to days: $1.216\times 24=29.184$ liters per day. Therefore, the equivalent rate is $29.184$ liters per day.

From $(2,900)$ to $(8,1,350)$, the elevation increases by $1,350-900=450$ feet over $8-2=6$ miles. So the increase is $\frac{450}{6}=75$ feet per mile. The question asks for the percent increase per mile relative to the elevation at 2 miles, which is 900 feet. That percent is $\frac{75}{900}\times 100\%=8.33\%$. Therefore, the correct answer is $8.33\%$.

Let $x$ be the amount of 10% solution and $y$ be the amount of 25% solution. The total volume is 500 mL, so

The total amount of pure acid must be 18% of 500, which is

The acid from the two solutions gives the second equation:

Now substitute $y=500-x$ into the acid equation:

So the number of milliliters of the 10% solution is $233\frac{1}{3}$.

Let the original price be $p$. After a $20\%$ discount, the price is $0.80p$. Then taking $15\%$ off that sale price gives $0.85(0.80p)=0.68p$. So the final price is $68\%$ of the original price. That means choice D must be true. Equivalently, the total decrease is $100\%-68\%=32\%$.