working at a constant rate, pump x pumped out half of the water in a flooded basement in 4 hours. then pump y was started and the two pumps, working independently at their respective constant rates, pumped out the rest of the water in 3 hours. how many hours would it have taken pump y, operating alone at its own constant rate, to pump out all of the water that was pumped out of the basement?
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Pump X took 4 hours to pump out half of the volume.. Let the total volume be V
in 1 hour .. Pump X will pump V/8 water
If pump X and Y both take 3 hours together to pump out the remaining V/2 water.. this means.. For 3 hours Pump X has also been pumping water along with Y.
Water pumped by Y in 3 Hours = V/2-3V/8=V/8 (remaining volume- volume removed by pump X in 3 hours)
this means.. Pump Y takes 3 hours to pump out V/8 water
in 1 hour.. Pump Y can pump V/24 Water...(This is the Rate of pumping of Pump Y)
therefore in order to pump out V water..
pump Y will take V/V/24 = 24 hours :)
in 1 hour .. Pump X will pump V/8 water
If pump X and Y both take 3 hours together to pump out the remaining V/2 water.. this means.. For 3 hours Pump X has also been pumping water along with Y.
Water pumped by Y in 3 Hours = V/2-3V/8=V/8 (remaining volume- volume removed by pump X in 3 hours)
this means.. Pump Y takes 3 hours to pump out V/8 water
in 1 hour.. Pump Y can pump V/24 Water...(This is the Rate of pumping of Pump Y)
therefore in order to pump out V water..
pump Y will take V/V/24 = 24 hours :)
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