Two pipes A and B can fill a cistern in 30 and 45 minutes respectively. Both are opened simultaneously, but after 10 minutes pipe A is closed. 10 minutes after that pipe B is closed and pipe A is reopened. How much more time does it take to fill the tank?
Cistern part filled by pipe A in 1 min = 1/30
Similary Cistern part filled by pipe B in 1 min = 1/45
Cistern part filled by both (A+B) in 1 min = (1/30+1/45) = 5/90 = 1/18
Cistern part filled by both(A+B) in 10 min = 10 * 1/18 = 10/18 ----- (1)
Now Cistern part filled by pipe B in 10 min = 10/45 = 2/9 ------- (2)
Total cistern part filled so far = (10/18+2/9) = 14/18 = 7/9
remaining part = (1-7/9) = 2/9
Time required to fill 1/30 part of tank by A = 1 min
so for 2/9 part of tank, time needed = 30 * 2/9 = 20/3 = 6 2/3 min
+I7When both the pipes are open (1/30 + 1/45) of the tank is filled in 1 minute.
In the first 10 minutes (1/30 + 1/45) x 10 = 10/18 of the tank gets filled ....(1).
In the next 10 minutes, 10/45 of the tank is filled by pipe B ....(2)
Thus, adding (1) and (2) we get 7/9
Thus, 2/9 of the tank is to be filled by A
A takes 2/9 x 30 = 6 2/3 minutes to fill the tank
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Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:
Tank part filled by pipes (A+B+C) in 1 hrs = 1/6 ------- (1)
so tank part filled by (A+B+C) in 2 hrs = 2*1/6 = 1/3
Now find the remaining part = (1-1/3) = 2/3
=> (A+B) 7 hs work = 2/3
so (A+B) 1 hrs work = 2/21 ------ (2)
To find the C 1 hrs work use eq. 1 & 2
=> 1/6-2/21 = 1/14
so C alone can fill the tank in 14 hrs.
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A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?
Time taken by one tap to fill half tank = 3hrs.
Part filled by the four taps in 1 hrs = (4*1/6) = 2/3.
Remaining part = (1-1/2) = 1/2.
so 2/3 : 1/2 :: 1:x => x = 1/2*1*3/2) = 3/4 hrs. => 45 min.
=> Total time taken = 3 hrs 45 min.
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13 buckets of water fill a tank when the capacity of each bucket is 51 litres. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 17 litres?
Number of buckets required when capacity of each bucket is 17 liters =13*51/17=39
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A leak in bottom of tank can empty a full tank in 12 minutes. If the leak is not there then two pipes A and B can fill the tank in 15 minutes and 20 minutes respectively. If the leak is active and both the pipes are open then How much time will be required to fill the tank:
Tank part filled by both (A+B) in 1 min = (1/15+1/20) = 7/60
Tank part can be empty by leak in 1 min = 1/12
So Tank part will be filled when Both taps and leak are active in 1 min = (7/60 - 1/12) = 2/60 = 1/30
=> 1/30 part of tank filled by both tap when leak is active = 1 min
=> Whole tank can be filled by both tap when leak is active = 30 min
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Two taps can separately fill a cistern 10 minutes and 15 minutes respectively and when the waste pipe is open, they can together fill it in 18 minutes. The waste pipe can empty the full cistern in?
Work done by third tap in 1 min = (1/18-1/10+1/15) = -10/90 = -1/9 (The negative value denote that emptying)
so the third tap alone can empty the cistern in 9 min.
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There are two pipes A and B. If A filled 10 liters in hour B can fills 20 liters in same time. Likewise B can fill 10, 20, 40, 80,160. If B filled in (1/16) th of a tank in 3 hours, how much time will it take to fill completely?
B fills double amount than in previous hour (from Given series 10,20,40,80,160)
So in 4 hours tank filled=2/16
In 5 hours tank filled=2*2/16=4/16
In 6 hours tank filled=2*4/16=8/16
In 7 hours tank filled=16/16=1
So, it took 7 hours to fill the tank
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One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in:
Let the slower pipe alone fill the tank in x minutes.
Then, faster pipe will fill it in x/3 minutes.
So, 1/x + 3/x = 1/36
=> 4/x = 1/36
=> x = 144 min.
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A tank is fitted with 8 pipes, some of them that fill the tank and others that are waste pipe meant to empty the tank. Each of the pipes that fill the tank can fill it in 8 hours, while each of those that empty the tank can empty it in 6 hours. If all the pipes are kept open when the tank is full, it will take exactly 6 hours for the tank to empty. How many of these are fill pipes?
Let the number of fill pipes be x and empty pipes be 8-x
Let one pipe in 1 hr fills 1/8th part of the tank then x pipes will fill x/8 part
Similarly 8-x pipes will empty (8-x)/6th part of the tank in 1 hour
Hence x/8 - (8-x)/6 =1/6 (All the pipes operating simultaneously so finding the result of 1 hr)
Solving for x we get as x= 4
So no. of fill pipes =no. of empty pipes=4
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Pipe A usually fills a tank in 2 hours. On account of a leak at the bottom of the tank, it takes pipe A 30 more minutes to fill the tank. How long will the leak take to empty a full tank if pipe A is shut?
Pipe A takes 2 hours to fill the tank. Therefore, it fills half the tank in an hour or 50% of the tank in an hour.
When there is a leak it takes 2 hours 30 minutes for the tank to fill. i.e 5/2 hrs hours to fill the tank or 2/5th or 40% of the tank gets filled in 1 hr.
On account of the leak, (50−40)%=10% of the water gets wasted every hour. Therefore, the leak will take 10 hours to drain a full tank.
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