: Become a pro in Pipes and Cisterns topic in arithmetics #IndiaNEWS #Education Today Hyderabad: This article is in continuation to the last article on preparation for the Sub-Inspector of Police recruitment

Become a pro in Pipes and Cisterns topic in arithmetics #IndiaNEWS #Education Today
Hyderabad: This article is in continuation to the last article on preparation for the Sub-Inspector of Police recruitment exam. Here are some practice questions, answers along with explanations on the Pipes and Cisterns topic.
1. Pipe A can fill a tank in 36 minutes and pipe B can fill it in 45 minutes. If both the pipes are opened to fill an empty tank, in how many minutes will it be full?
A. 15 B. 18 C. 20 D. 25
Ans: C
Explanation:
Part filled by A in 1 hour = 1/( 36)
Part filled by B in 1 hour =1/( 45)
Part filled by (A B) in 1 hour = 1/( 36) 1/( 45) = 9/180 = 1/20
Hence, both the pipes together will fill the tank in 20 minutes.
2. Tap A can fill the empty tank in 12 hours, but due to a leak in the bottom it is filled in 15 hours. If the tank is full and then tap A is closed then in how many hours the leak can empty it?
A. 45 hours B. 48 hours C. 52 hours D. 60 hours
Ans: D
Explanation:
The part of tank is filled by the tap in 1 hour = 1/12. But due to a leak in the bottom it is filled in 15 hours.
Thus, the part of tank is filled in one hour when both tap and leakage are open simultaneously 1/15.
Thus, the part is emptied in one hour by the tap in one hour when the tap is closed = 1/12 1/15 = (5-4)/60 = 1/60
The whole filled tank is emptied by the leakage alone = 60 hours.
3. A tank can be filled by a tap in 20 minutes and by another tap in 60 minutes. Both the taps are kept open for 10 minutes and then the first tap is shut off. After this, the tank will be completely filled in what time?
A. 10 B. 15 C. 20 D. 25
Ans: C
Explanation:
Part filled by Tap A in 1 min = 1/20
Part filled by Tap B in 1 min = 1/60
(A B)s 10 minutes work = 10× (1/20 1/60) = 10×4/60 = 2/3
Remaining work = 1- 2/3 = 1/3
1/60 Part filled by B in = 1 min
1/3P art will be filled in = (1???3)/(1???60) = 20 minutes.
4. Three pipes A, B and C can fill a tank in 15 minutes, 20 minutes and 30 minutes respectively. The pipe C is closed 6 minutes before the tank is filled. In what time the tank will be full?
A. 6 minutes B. 8 minutes C. 12 minutes D. 14 minutes
Ans: B
Explanation:
Pipe A and B can fill
T*(1/15 1/20)*6 = 0. 7 T in the last 6 minutes
So 0. 3T must be filled in the remaining time.
T*(1/15 1/20 1/30)*t = 0. 3T
t = 2
Total time taken is 2 6 = 8 minutes.
5. Two pipes A and B can fill a tank in 6 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours, the tank shall be full?
A. 3 hours B. 5 hours C. 7 hours D. 10 hours
Ans: B
Explanation:
(A B)s 2 hours work when opened = 1/6 1/4 = 5/12
(A B)?s 4 hours work = 5/12×2 = 5/6
Remaining work = 1- 5/6 = 1/6
Now, it’s A turn in 5th hour
1/6 Work will be done by A in 1 hour
Total time = 4 1 = 5 hours.