Question 8·50 Super-Hard SAT Problem Solving and Data Analysis Questions·Problem Solving and Data Analysis

The four pumps together fill $18$ original tanks in $p$ hours. A rate is output divided by time, so their combined rate is $\frac{18}{p}$ original-tank volumes per hour. All four pumps are already counted in the $18$.

The later job uses 50% more pumps. Add that increase to the original $100\%$ of pumps:

So the combined volume rate becomes $\frac{18}{p}(1.50)$ original-tank volumes per hour.

Each pump loses $20\%$ of its flow, so each keeps 80% of its former flow:

The combined rate is now $\frac{18}{p}(1.50)(0.80)$ original-tank volumes per hour.

A replacement tank holds $25\%$ less, so its capacity is:

of an original tank. Divide by the smaller tank’s capacity to count how many tanks the same volume fills. That gives $\frac{18(1.50)(0.80)}{0.75p}$ replacement tanks per hour. Multiplying by $0.75$ would incorrectly make the tank count smaller.

The rate is per hour, but the new time is in minutes. Since one hour has $60$ minutes, convert the time:

Multiply replacement tanks per hour by $\frac{q}{60}$ hours, so the hours cancel. Type $\frac{18(1.50)(0.80)}{0.75(60)}$ into Desmos. It prints $0.48$; tap the fraction button to read $\frac{12}{25}$. So the number of replacement tanks filled is $\frac{12q}{25p}$. Choice A.