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Two pipes P and Q can fill a tank in 20hrs and 25hrs respectively while a third pipe R can empty the tank in 30hrs. If all the pipes are opened together for 10hrs and then pipe R is closed then in what time the tank can be filled.
(1/20 + 1/25 – 1/30)*10 + (1/20 + 1/25)*x = 1
We get x = 130/27, so total time to fill the tank = 130/27 + 10 = 400/27 hrs
There are three taps A, B and C which can fill a tank in 12hrs, 15hrs and 30 hrs respectively. If the tap A is opened first, after one hour tap B was opened and after 2 hours from the start of A, tap C is also opened. Find the time in which the tank is full.
In first hour only A is opened, in the next hour A and B are opened and in the third hour A, B and C are opened.
So, in three hours (3/12 + 2/15 + 1/30) = 25/60 tank is already filled.
Now, 25/60 = (1/12 + 1/15 + 1/30)*t
T = 25/11. Total time = 3 + 25/11 = 58/11 hours
Three pipes P, Q and R can fill the tank in 5, 10 and 15 minutes respectively. If all the pipes are opened together and pipe Q is turned off 5 minutes before the tank is fill. Then find the time in which the tank will full.
Let total time taken by the pipes is T hrs, then
(1/5 + 1/10 + 1/15)*(T – 5) + (1/5 + 1/15)*5 = 1
A pipe can fill a tank in 20 minutes but due to a leak develop at the bottom of the tank, 1/5 of the water filled by the pipe leaks out. Find the time in which the tank is filled.
Amount of tank filled by the pipe in one minute = 1/20 and due to leakage 1/5 of 1/20 leaks out so, [1/20 – (1/5)*(1/20)]*T = 1
We get T = 25
A bathing tub can be filled by a cold pipe in 15 minutes and by a hot pipe in 10 minutes. Ramesh opened both the tap and leaves the bathroom and returns at the time when the tub should be full. He observed that a waste pipe is opened at the bottom, he now closes it. Now the tub will take more 5 minutes to fill the tank, find the time in which the leak can empty the tank.
(1/15 + 1/10 – 1/x)*6 + (1/15 + 1/10)*5 = 1
x = 36/5
There are 10 taps connected to a tank. Some of them are waste pipe and some of them are water pipe. Water pipe can fill the tank in 15 hours and waste pipe can empty the tank in 30 hours. Find the number of waste pipes if the tank is filled in 6 hours.
Let water pipes are x and waste pipe are Y.
x + y = 10
(x/15 – y/30)*6 = 1
Solve both equation to get x and y
Pipe A is 4 times as fast as B in filling a tank. If A takes 20 minutes to fill a tank, then what is the time taken by both the pipe A and B to fill the tank?
A takes 20 minutes and it is 4 times faster than B, it means B will take 80 minutes to fill the tank.
(1/20 + 1/80)*t = 1. We get t = 16
Pipe A is 4 times faster than pipe B and takes 45 minutes less to fill a tank. When both the pipes are opened together than the time in which the tank will be full.
Let A take X minute to fill a tank then B will take 4x time. 4x – x = 45 (given), X = 15.
Time taken to fill the tank together = (1/15 + 1/60)*t =1
T = 12 minute
Two pipes P and Q can fill a tank in 20 minutes and 30 minutes respectively. There is a waste pipe which withdraws water at the rate of 8 litres per minute. Now the tank is full and If all the pipes are opened simultaneously the tank is emptied in 60 minutes. Find the capacity of the tank.
(1/20 + 1/30 – 1/t)*60 = 1
‘1’ is taken because the work is negative. T is the time taken by the waste pipe to empty the tank alone. We will t = 10
So capacity = 10*8 = 80ltr
There are 4 filling pipes and 3 emptying pipes capable of filling and emptying in 12 minutes and 15 minutes respectively. If all the pipes are opened together and as a result they fill 10 litres of water per minute. Find the capacity of the tank.
(4/12 – 3/15)*t = 1
t = 15/2 minute – in this time the tank will be filled. So the capacity = (15/2)*10 = 75 litre
Two taps can separately fill the tank in 10m and 15min respectively. They fill the tank in 12 minutes when a third pipe which empties the tank is also opened. What is the time taken by the third pipe to empty the whole tank?
1/10 + 1/15 – 1/x = 1/12
Solve, x = 12
Two pipes A and B can fill a tank in 12 hours and 15 hours respectively. If they are opened on alternate hours with pipe A opened first, then in how many hours the tank will be full?
A = 12 hours, B = 15 hours
Total work = LCM(12,15) = 60
So efficiency of A = 60/12 = 5, efficiency of B = 60/15 = 4
2 hrs work of (A+B) = 5+4 = 9
2*6(12) hours work of (A+B) = 9*6 = 54
So remaining work = 6054 = 6
Now A’s turn at 13th hour, he will do remaining work(6) in 6/12 hr
So total 12 1/2 hrs
Pipes P and Q can fill the tank in 24 minutes and 32 minutes respectively. Both piped are opened together. To have the tank full in 18 minutes, after how many minutes the pipe P must be closed?
P is to be closed before 18 minutes, let it is closed after x minutes, then Q worked for all 18 minutes. So,
(1/24)*x + (1/32)*18 = 1
Solve, x = 10.5
Three pipes, A, B and C are opened to fill a tank such that A and B cam fill the tank alone in 36 min. and 45 min. respectively and C can empty it in 30 min. After 6 minutes the emptying pipe is closed. In how many minutes the tank will be full in this way?
Let the tank full in x minutes, then A and B opened for x minutes and C for 6 minutes.
(1/36 + 1/45)*x – (1/30)*6 = 1
(1/20)*x = 6/5
Solve, x = 24
A and B are pipes such that A can empty the tank in 60 minutes and B can fill in 30 minutes. The tank is full of water and pipe A is opened. If after 18 minutes, pipe B is also opened, then in how much total time the tank will be full again?
Emptying pipe A is opened first for 18 minutes, so in 18 minutes the part of tank it has emptied is (1/60)*18 = 9/30
Now filling pipe is also opened, now since only 9/30 of the tank is empty so 9/30 is only to be filled by both pipes, let it take now x minutes, so
(1/30 – 1/60)*x = 9/30
Solve, x= 18
So total = 18+18 = 36 minutes [total time is asked – 18 minutes when emptyimh pipe was only opened, 18 minutes when both were operating.]
Two pipes A and B can alone fill a tank in 20 minutes and 30 minutes respectively. But due to a leak at the bottom of tank, it took 3 more minutes to fill the tank. In how many hours, the leak can alone empty the full tank?
A and B can fill tank in (1/20 + 1/30) = 1/12 so 12 minutes
But it took 3 more minutes, this means the tank got full in 12+3 = 15 minutes
So (1/20 + 1/30 – 1/x) = 1/15
Solve, x = 60
Pipes A and B can fill a cistern in 15 hours together. But if these pipes operate separately A takes 40 hours less than B to fill the tank. In how many hours the pipe A will fill the cistern working alone?
Let A takes x hours, then B = (x+40) hours
1/x + 1/(x+40) = 1/15
Solve, x = 20
Three pipes A, B and C can fill the cistern in 10, 12, and 15 hours respectively. In how much time the cistern will be full if A is operated for the whole time and B and C are operated alternately which B being first?
In first hour, part of cistern filled is (1/10 + 1/12) = 11/60
In second hour, part of cistern filled is (1/10 + 1/15) = 1/6
So in 2 hours, part of cistern filled is 11/60 + 10/60 = 21/60 = 7/20
now in 2*2 (4) hours, part of cistern filled is (7/20)*2 = 14/20 = 7/10
now in the 5th hour, A+B’s turn which fill 11/60 in that hour, but the cistern remaining to be filled is (1 – 7/10) = 3/10, since 3/10 is more than 11/60, so after 5th hour remaining part to be filled is 3/10 – 11/60 = 7/60
now in 6th hour, (A+C)’s turn, it will fill remaining 7/60 in (7/60)*(6/1) = 7/10 so total 5 7/10 hours
A cistern is 1/4th full. Two pipes which fill the cistern in 15 minutes and 20 minutes respectively are opened simultaneously. After 5 minutes, a third pipe which empties the full cistern in 30 minutes is also opened. In how many minutes the cistern will be full?
Since 1/4th is already filled, 3/4th is to filled now.
So
(1/15 + 1/20)*(5+x) – (1/30)*x = 3/4
(7/60)*5 + (7/60 – 1/30)*x = 3/4
(5/60)*x = 2/12
Solve, x = 2 mins
So total 7 minutes
Pipes A, B and C which fill the tank together in 6 hours are opened for 2 hours after which pipe C was closed. Find the number of hours taken by pipe C to fill the tank if the remaining tank is filled in 7 hours.
1/A + 1/B + 1/C = 1/6
Now given that first all open for 2 hours, then C closed and A+B completes in 7 hours, so
(1/A + 1/B + 1/C) *2 + (1/A + 1/B)*7 = 1
Put 1/A + 1/B = 1/6 – 1/C
(1/6 – 1/C + 1/C) *2 + (1/6 – 1/C)*7 = 1
2/6 + 7/6 – 7/C = 1
Solve, C = 14
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