EgmatQuantExpert wrote:
Q.)
Operating at a constant rate, 2 pumps of Type 1 can completely fill a reservoir with water in 24 hours. A pump of Type 2 can also fill the same reservoir with water but it operates at a different constant rate. At 1 PM on 21 March, 3 pumps of Type 2 start filling the empty reservoir. At 3 PM, 6 pumps of Type 1 also start filling the reservoir with water. The reservoir is completely filled with water at 7 PM. If the 6 Type 1 pumps were not started, at what time would the reservoir have been completely filled with water?
A. 1 AM on 22 March
B. 7 AM on 22 March
C. 1 AM on 23 March
D. 7 AM on 23 March
E. 7 AM on 25 March
Thanks,
Saquib
Quant Expert
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Let rate of Type 2 =B
& Type 1 =A
thus 2 type 1 can fill with rate= 1/24
thus A=1/48
similarly work done by 3 Pump Type 2 is 3/B
thus after starting type-2
1 PM it worked alone upto 3PM(2 hrs)
work done by type-2 for 2 hrs = (2*3)/B
for the next 4 hrs (7PM) both types of pump run.
contribution of 6 pump type 1 for 4 hrs = (6*4)/48 =1/2
contribution of 3 pump type 2 for 4 hrs = (3*12)/B
all three combined = work done by type-2 for initial 2 hrs + contribution of 6 pump type 1 for 4 hrs + contribution of 6 pump type 2 for 4 hrs
6/B + 12/B + 1/2 = 1
18/B = 1/2
or ( 3*6)/B = 1/2
3/B = 1/12
thus 3 type -2 pump can fill in 12 hrs
1 PM on 21st March +12 hrs
=1 AM on 22nd march
Ans A