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08 Aug 2023, 01:30
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D
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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
08 Aug 2023, 09:50
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Rate at which the tank would fill up = 1/10. [Full tank = Job = 1]
Similarly, tank would be emptied at rate = 1/7.
Simultaneously on, outlet rate is higher, hence work done in the first hour is (1/7-1/10)*1 = 3/70
The remaining tank would be emptied at full outlet rate, 1/7*x (x= required time).
Thus Job= 1= 3/70 + 1/7x
x=6.7h i.e Option D
Pray and bless this user with a 720 please, thanks.
Similarly, tank would be emptied at rate = 1/7.
Simultaneously on, outlet rate is higher, hence work done in the first hour is (1/7-1/10)*1 = 3/70
The remaining tank would be emptied at full outlet rate, 1/7*x (x= required time).
Thus Job= 1= 3/70 + 1/7x
x=6.7h i.e Option D
Pray and bless this user with a 720 please, thanks.
General Discussion
10 Aug 2023, 08:36
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In 1 hour, the inlet pump working alone would fill up 1/10 of the total volume of the tank.
In 1 hour, the outlet pump working alone would empty 1/7 of the total volume of the tank.
When both the pumps are working simultaneously, the total work done will be (1/7-1/10) = 3/70
Since the outlet pump rate was higher, 3/70 of the tank is empty and (1- 3/70) = 67/70, is the remaining volume to be emptied.
So, 1/7 work by the outlet pump was done in 1 hour.
67/70 work will take = 1/(1/7) * 67/70 = 7 * 67/70 = 6.7 hours.
Option D
In 1 hour, the outlet pump working alone would empty 1/7 of the total volume of the tank.
When both the pumps are working simultaneously, the total work done will be (1/7-1/10) = 3/70
Since the outlet pump rate was higher, 3/70 of the tank is empty and (1- 3/70) = 67/70, is the remaining volume to be emptied.
So, 1/7 work by the outlet pump was done in 1 hour.
67/70 work will take = 1/(1/7) * 67/70 = 7 * 67/70 = 6.7 hours.
Option D
