jose.mario.amaya wrote:S and T are two water pumps that run at constant rates. If pump T pumped alone, how many more hours would it take for it to finish pumping a large container than it would take for pump S to accomplish the same task by itself?
(1) When both the water pumps work together, they finish pumping a large container in  2/3rd the time it takes for pump S to finish the task alone.
(2) Pump T is capable of pumping a large container in twice the time that it takes pump S to accomplish the same task alone.
Statement 1: When both water pumps work together, they finish pumping a large container in 2/3rd the time it takes S to finish the task alone.
Time and rate are RECIPROCALS.
Since S and T's time together = 2/3 of S's time alone, S and T's combined rate = 3/2 of S's rate alone.
Let S's rate alone = 2 units per hour.
Then S and T's combined rate = (3/2) * 2 = 3 units per hour.
Thus, T's rate alone = (S and T's combined rate) - (S's rate alone) = 3-2 = 1 unit per hour.
Case 1: Job = 2 units
Time for T = w/r = 2/1 = 2 hours.
Time for S = w/r = 2/2 = 1 hour.
Difference = 2-1 = 1 hour.
Case 2: Job = 20 units
Time for T = w/r = 20/1 = 20 hours.
Time for S = w/r = 20/2 = 10 hours.
Difference = 20-10 = 10 hours.
Since the time difference can be different values, INSUFFICIENT.
Case 1 and Case 2 also satisfy statement 2.
Thus, the two statements combined are INSUFFICIENT.
The correct answer is
E.
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