From plots 1 through 3, the fertilizer-to-area ratio is the same: $\frac{12}{8}=1.5$, $\frac{18}{12}=1.5$, and $\frac{27}{18}=1.5$ pounds per acre. So the common application rate is $1.5$ pounds per acre. For plot 4, which has $25$ acres, the amount of fertilizer that would be used at this rate is $25(1.5)=37.5$ pounds. The table shows that plot 4 actually used $45$ pounds, so the number of fewer pounds would be $45-37.5=7.5$. Therefore, the correct answer is $7.5$.

Let $x$ be the number of liters of liquid A and $y$ be the number of liters of liquid B. Then

and the amount of pure acid must satisfy

Substitute $y=40-x$ into the second equation:

So $y=24$. The original $40$-liter mixture has acid concentration $42\%$, so it contains

liters of acid.

When $4$ liters of the mixture are spilled, $42\%$ of those $4$ liters is acid, so acid lost is

liters. Acid remaining is

liters.

Then $4$ liters of liquid A, which is $30\%$ acid, are added. That adds

liters of acid.
So the final amount of acid is

liters.

The total volume is again $40$ liters, so the final concentration is

Therefore, the correct answer is $40.8\%$.

Because equal volumes of the two solutions are combined, the amounts of A and B must be found as fractions of each original solution, then added. In the first solution, the ratio $5:3$ means $8$ total parts, so the fraction that is A is $\frac{5}{8}$ and the fraction that is B is $\frac{3}{8}$. In the second solution, the ratio $7:5$ means $12$ total parts, so the fraction that is A is $\frac{7}{12}$ and the fraction that is B is $\frac{5}{12}$. If the same volume of each solution is used, the total amount of A in the mixture is proportional to $\frac{5}{8}+\frac{7}{12}$, and the total amount of B is proportional to $\frac{3}{8}+\frac{5}{12}$. Therefore, the ratio of A to B is

which is choice C.

Use the two points on the graph to find the machine's constant production rate. From $(12,45)$ to $(30,105)$, the number of parts increases by $105-45=60$ while the time increases by $30-12=18$ minutes. So the rate is $\frac{60}{18}=\frac{10}{3}$ parts per minute.

Now find the additional number of parts needed to go from 85 parts to 165 parts: $165-85=80$ parts.

At a rate of $\frac{10}{3}$ parts per minute, the time needed for 80 parts is

minutes.

So the machine needs $24$ more minutes.

Let $x$ be the number of liters of 25% solution and $y$ be the number of liters of 55% solution.

The total amount of mixture is 30 liters, so

The final mixture must be 40% concentrate, so the total amount of concentrate must be

Set up the concentrate equation:

Use $x=30-y$ from the first equation:

Now check the constraints: using 15 liters of the 55% solution means the 25% solution is also 15 liters. These satisfy the requirements of at least 8 liters of the 25% solution and at least 6 liters of the 55% solution.

Therefore, the lab assistant should use $15$ liters of the 55% solution.