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03 Aug 2026, 09:53
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At a water-treatment facility, a pump was used to fill an empty storage tank. If the pump operated at a constant rate throughout the process, how many minutes did it take the pump to fill the tank?
(1) If the pump’s flow rate had been 25% greater, the tank would have been filled 16 minutes sooner.
(2) If the pump’s flow rate had been 20% lower, the tank would have taken 20 minutes longer to fill.
(1) If the pump’s flow rate had been 25% greater, the tank would have been filled 16 minutes sooner.
(2) If the pump’s flow rate had been 20% lower, the tank would have taken 20 minutes longer to fill.
03 Aug 2026, 09:53
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Official Solution:
At a water-treatment facility, a pump was used to fill an empty storage tank. If the pump operated at a constant rate throughout the process, how many minutes did it take the pump to fill the tank?
(1) If the pump’s flow rate had been 25% greater, the tank would have been filled 16 minutes sooner.
Let the actual time be \(t\) minutes.
If the pump’s flow rate had been 25% greater, the new rate would have been \(1.25\) times the actual rate. So the filling time would have been \(\frac{t}{1.25} = 0.8t\).
The statement says this would have been 16 minutes sooner, so:
\(t - 0.8t = 16\)
\(0.2t = 16\)
\(t = 80\)
Sufficient.
(2) If the pump’s flow rate had been 20% lower, the tank would have taken 20 minutes longer to fill.
Again let the actual time be \(t\) minutes.
If the pump’s flow rate had been 20% lower, the new rate would have been \(0.8\) times the actual rate. So the filling time would have been \(\frac{t}{0.8} = 1.25t\).
The statement says this would have increased the time by 20 minutes, so:
\(1.25t - t = 20\)
\(0.25t = 20\)
\(t = 80\)
Sufficient.
Answer: D
At a water-treatment facility, a pump was used to fill an empty storage tank. If the pump operated at a constant rate throughout the process, how many minutes did it take the pump to fill the tank?
(1) If the pump’s flow rate had been 25% greater, the tank would have been filled 16 minutes sooner.
Let the actual time be \(t\) minutes.
If the pump’s flow rate had been 25% greater, the new rate would have been \(1.25\) times the actual rate. So the filling time would have been \(\frac{t}{1.25} = 0.8t\).
The statement says this would have been 16 minutes sooner, so:
\(t - 0.8t = 16\)
\(0.2t = 16\)
\(t = 80\)
Sufficient.
(2) If the pump’s flow rate had been 20% lower, the tank would have taken 20 minutes longer to fill.
Again let the actual time be \(t\) minutes.
If the pump’s flow rate had been 20% lower, the new rate would have been \(0.8\) times the actual rate. So the filling time would have been \(\frac{t}{0.8} = 1.25t\).
The statement says this would have increased the time by 20 minutes, so:
\(1.25t - t = 20\)
\(0.25t = 20\)
\(t = 80\)
Sufficient.
Answer: D
