If each tank requires less volume, the same volumetric flow can fill more tanks. Divide by the replacement tank's capacity rather than multiplying by it.

Express $q$ minutes as a fraction of an hour before multiplying by the hourly filling rate.

Enter $p=3$ and $q=150$ on separate lines. These values can be used to compare the expressions numerically.

Enter $N=(1.5)(0.8)(18/p)(q/60)/0.75$. This combines the change in the number of pumps, the change in each pump's rate, the smaller tank capacity, and the time conversion.

Enter each answer-option expression using the assigned values of $p$ and $q$. The value of $N$ is $24$, and the expression that also evaluates to $24$ is $\frac{12q}{25p}$. Therefore, the answer is $\frac{12q}{25p}$.

Let $C$ be the capacity of one original tank. The original pumps move a total volume of $18C$ in $p$ hours, so their original volumetric flow rate is $\frac{18C}{p}$ per hour.

Using 50% more pumps multiplies the total flow rate by $1.5$, while reducing each pump's rate by 20% multiplies it by $0.8$. Therefore, the adjusted volumetric flow rate is

Each replacement tank has capacity $0.75C$, so divide the adjusted volumetric flow rate by $0.75C$ to find the number of replacement tanks filled per hour. Since $q$ minutes equals $\frac{q}{60}$ hours, the required number is

Cancel $C$ and simplify:

Therefore, the answer is $\frac{12q}{25p}$.