1)Two pipes can fill a cistern in 14 hours and 16 hours respectively. The pipes
are opened simultaneously and it is found that due to leakage in the bottom it
took 32 min more to fill the cistern. When the cistern is full, in what time will
the leak empty it?
Sol: Work done by the two pipes in 1 hour= 1/14+1/16=15/112
Time taken by these two pipes to fill the tank=112/15 hrs.
Due to leakage, time taken = 7 hrs 28 min+ 32 min= 8 hours
Therefore, work done by (two pipes + leak) in 1 hr= 1/8
work done by leak n 1 hour=15/112 -1/8=1/112
Leak will empty full cistern n 112 hours.
2)Two pipes A&B can fill a tank in 30 min. First, A&B are opened. After 7 min,
C also opened. In how much time, the tank s full.
Sol: Part filled n 7 min = 7*(1/36+1/45)=7/20
Remaining part= 1-7/20=13/20
Net part filled in 1 min when A,B and C are opened=1/36 +1/45- 1/30=1/60
Now, 1/60 part is filled in 1 min.
13/20 part is filled n (60*13/20)=39 min
Total time taken to fill the tank=39+7=46 min
3)Two pipes A&B can fill a tank in 24 min and 32 min respectively. If both the
pipes are opened simultaneously, after how much time B should be closed so that the tank is full in 18 min.
Sol: Let B be closed after x min, then part filled by (A+B) in x min+
part filled by A in (18-x) min=1
x(1/24+1/32) +(18-x)1/24 => x=8
Hence B must be closed after 8 min.
4)Two pipes A& B together can fill a cistern in 4 hours. Had they been opened
separately, then B would have taken 6 hours more than A to fill the cistern.
How much time will be taken by A to fill the cistern separately?
Sol: Let the cistern be filled by pipe A alone in x hours.
Pipe B will fill it in x+6 hours
1/x + 1/x+6=1/4
Solving this we get x=6.
Hence, A takes 6 hours to fill the cistern separately.
5)A tank is filled by 3 pipes with uniform flow. The first two pipes operating
simultaneously fill the tan in the same time during which the tank is filled by
the third pipe alone. The 2nd pipe fills the tank 5 hours faster than first pipe
and 4 hours slower than third pipe. The time required by first pipe is
Sol: Suppose, first pipe take x hours to fill the tank then
B & C will take (x-5) and (x-9) hours respectively.
Therefore, 1/x +1/(x-5) =1/(x-9)
On solving, x=15
Hence, time required by first pipe is 15 hours.
6)A large tanker can be filled by two pipes A& B in 60 min and 40 min
respectively. How many minutes will it take to fill the tanker from empty
state if B is used for half the time & A and B fill it together for the
other half?
Sol: Part filled by (A+B) n 1 min=(1/60 +1/40)=1/24
Suppose the tank is filled in x minutes
Then, x/2(1/24+1/40)=1
=> (x/2)*(1/15)=1
=> x=30 min.
7)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 s opened first, in how
many hours, the tank shall be full.
Sol: (A+B)'s 2 hours work when opened alternatively =1/6+1/4 =5/12
(A+B)'s 4 hours work when opened alternatively=10/12=5/6
Remaining part=1 -5/6=1/6.
Now, it is A's turn and 1/6 part is filled by A in 1 hour .
So, total time taken to fill the tank=(4+1)= 5 hours.
8)Three taps A,B and C can fill a tank in 12, 15 and 20 hours respectively.
If A is open all the time and B and C are open for one hour each alternatively,
the tank will be full in.
Sol: (A+B)'s 1 hour's work=1/12+1/15=9/60=3/20
(A+C)'s 1 hour's work=1/20+1/12=8/60=2/15
Part filled in 2 hours=3/20+2/15=17/60
Part filled in 2 hours=3/20+2/15= 17/60
Part filled in 6 hours=3*17/60 =17/20
Remaining part=1 -17/20=3/20
No
