With uniform flow, taps A and B can fill the tank in 4 and 3 hours respectively. Tap C can empty the

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08 Jun 2022 Pipes & Cisterns

With uniform flow, taps A and B can fill the tank in 4 and 3 hours respectively. Tap C can empty the completely filled tank in one and half hour. Tap A is opened at 9:30 am, tap B is opened at 10:15 am and tap C is opened at 11 am. At what exact time will the tank be empty?

OPtion

1) 7:00 pm
2) 6:15 pm
3) 7:30 pm
4) 5:45 pm
5) 6:30 pm
6) 5:00 pm
7) 4:30 pm
8) 6:45 pm
9) 5:30 pm
10) None of these

Solution

Let total capacity of tank be LCM(4, 3, 1.5) = 12 units
Tap A fills 12/4 =3 units per hour.
Tap B fills 12/3=4 units per hour
Tap R empties 12/1.5=8 units per hour.
Upto 11 am, tap A fills for 3/2 hours and B for 3/4 hour .'. Filling upto 11 am = (3/2)x3 + (3/4)x4 = 7.5 units
Now, when each tap is opened together after 11 am, filling per hour 3 + 4 - 8 = -1 units
i.e. 1 unit will be drained per hour, so to empty tank completely it require 7.5 hours after 11 am.
Hence, tank will be completely empty at 6:30 pm

Correct Option: 5 Best Solution (1)

G.S.Kantharao   2 years AGO

A,B fill in 4h,3h
A filled 3/2h=(1/4)(3/2)=3/8th
B filled 3/4h =(1/3)(3/4)=1/4th
Total filled =3/8+1/4=5/8th
A+B-C=(1/4)+(1/3)-(2/3)=-1/12
Let in x hours it emptied.
Then (5/8)-(x/12)=0
x/12=5/8=>x=5*12/8=15/2=7&1/2hr
So from
11.00 AM+7&1/2hr=6.30 PM
My option is 5👈

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