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Working alone at their respective constant rates, an outlet pump can e
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22 Jan 2021, 06:43
22 Jan 2021, 06:43
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Working alone at their respective constant rates, an outlet pump can empty a tank in 7 hours when the tank is full, and an inlet pump can fill the same tank in 10 hours when the tank is empty. Beginning with the full tank. Both the pumps work simultaneously at their respective rates for one 1hour before the inlet pump is shut off. How many more hours would it take the outlet to empty the tank?
A) 6.1
B) 6.3
C) 6.5
D) 6.7
E) 6.9
A) 6.1
B) 6.3
C) 6.5
D) 6.7
E) 6.9
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Re: Working alone at their respective constant rates, an outlet pump can e
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22 Jan 2021, 06:57
22 Jan 2021, 06:57
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Ashok2345 wrote:
Working alone at their respective constant rates, an outlet pump can empty a tank in 7 hours when the tank is full, and an inlet pump can fill the same tank in 10 hours when the tank is empty. Beginning with the full tank. Both the pumps work simultaneously at their respective rates for one 1hour before the inlet pump is shut off. How many more hours would it take the outlet to empty the tank?
A) 6.1
B) 6.3
C) 6.5
D) 6.7
E) 6.9
A) 6.1
B) 6.3
C) 6.5
D) 6.7
E) 6.9
For work questions such as this, I like to assign a "nice" value (a value that works well with the given information of 7 hours and 10 hours).
So for this question, let's say the tank holds 70 gallons of water
The OUTLET pump can empty a tank in 7 hours when the tank is full
Rate = output/time = 70/7 = 10 gallons per hour
The INLET pump can fill the same tank in 10 hours when the tank is empty
Rate = output/time = 70/10 = 7 gallons per hour
Both the pumps work simultaneously at their respective rates for one 1hour before the inlet pump is shut off. How many more hours would it take the outlet to empty the tank?
In 1 hour, the INLET pump adds 7 gallons to the tank, and the OUTLET pump removes 10 gallons from the tank.
10 - 7 = 3
So, the NET effect is that 3 gallons are REMOVED from the tank during that first hour.
70 - 3 = 67
This means that, after the first hour, 67 gallons of water remain in the tank.
Time to empty the remaining 67 gallons = output/rate = 67/10 = 6.7 hours.
Answer: D
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Re: Working alone at their respective constant rates, an outlet pump can e
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03 Feb 2021, 08:31
03 Feb 2021, 08:31
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Time required by Outlet pump to empty the tank = 7 hours
Time required by Inlet pump to fill the tank = 10 hours
Now, if they work together for an hour;
Work done = 1/7 - 1/10 = 3/70
Work left for the Outlet pump= 1 - 3/70 = 67/70
Rate of Outlet pump = 1/7
Therefore, T = W/R = (67/70) / (1/7) = 67/10 = 6.7 hours
Hence, option D
[Note: The tank can be emptied faster than filing, that is why 1/7 - 1/10]
Time required by Inlet pump to fill the tank = 10 hours
Now, if they work together for an hour;
Work done = 1/7 - 1/10 = 3/70
Work left for the Outlet pump= 1 - 3/70 = 67/70
Rate of Outlet pump = 1/7
Therefore, T = W/R = (67/70) / (1/7) = 67/10 = 6.7 hours
Hence, option D
[Note: The tank can be emptied faster than filing, that is why 1/7 - 1/10]
