hemant_rajput wrote:Q17 . A tank in X apartment is full of water is fitted with pumps of equal capacity. Besides it is also fitted with a certain no. of inlet pipes which are always kept open. When tank is full, 10 pumps of equal capacity empty the tank in 12 hrs. , while 15 pumps of the same capacity empty the tank in 6 hrs. In how much time would 25 pump of the same capacity empty the tank, if the tank is initially full?(All the pipes pouring water into the tank are always open)
a. 3
b. 2.5
c. 4
d. 3.5
e. 5.5
[spoiler]oa:d[/spoiler]
Every hour:
The pumps pump units OUT of the tank.
The inlet pushes units INTO the tank.
When the pumps and the inlet work simultaneously for a certain amount of time, the ENTIRE VOLUME OF THE TANK is removed.
Thus:
Tank volume = total amount PUMPED OUT - total amount PUSHED IN.
Let x = the amount PUSHED IN by the inlet each hour.
Case 1: 10 pumps empty the tank in 12 hrs
In 12 hours:
The amount PUMPED OUT by 10 pumps = r*t = 10*12 = 120.
The amount PUSHED IN by the inlet = 12x.
Thus:
Tank volume = total pumped out - total pushed in = 120 - 12x.
Case 2: 15 pumps empty the tank in 6 hrs
In 6 hours:
The amount PUMPED OUT by 15 pumps = r*t = 15*6 = 90.
The amount PUSHED IN by the inlet = 6x.
Thus:
Tank volume = total pumped out - total pumped in = 90 - 6x.
Since the volume is the same in each case, we get:
120 - 12x = 90 - 6x
30 = 6x
x = 5.
Thus:
Each hour, the amount PUSHED IN by the inlet = 5 units.
Tank volume = 90 - 6x = 90 - 6*5 = 60 units.
When 25 pumps work:
Amount PUMPED OUT by 25 pumps each hour = 25 units.
Amount PUSHED IN by the inlet each hour = 5 units.
Thus:
Net amount removed each hour = 25-5 = 20 units per hour.
Time to empty the tank = w/r = 60/20 = 3 hours.
The correct answer is
A.
Same answer as Tommy's.
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