Pipes and Cistern- Aptitude Questions and Answers

44 pipes => 27 hours
66 pipes=> ?

Amount of work done is same for filling a pipe, So 44*27 = 66*X
X = 44*27 / 66
X = 18

66 pipes can fill one tanks in 18 hours.
Hence, they will fill 4 tanks in 18*4 = 72 hours

Three Pipes A,B and C together filled a tank in 6 hours
So 1 hour work of A,B and C =1/6
C closed after 2 hour's, Part filled in 2 hours working together(A,B and C) = 2/6 = 1/3
Remaining part = (1-1/3) = 2/3

A and B can fill remaining in 8 hours, (A + B)'s 8 hour's work = 2/3

So, (A+B)'s 1 hour work = 2/24=1/12

Therefore C's 1 hour's work
= [(A+B+C)'s 1 hour]-[(A+B)'s 1 hr work]
=(1/6 - 1/12) = 1/12

C alone can fill the tank in 12 hours

As the pipes are operating alternatively for one minute, thus (A+B)'s 2 minutes job is =(1/4 +1/6) =5/12
in next, 2 minutes both pipe can fill another 5/12 part.
So in 4 minutes A and B are operating alternatively will fill 5/12 + 5/12 = 5/6

Remaining part of cistern= 1 - 5/6 =1/6

Pipe A can fill 1/4 part of the cistern in 1 minute,and can fill 1/6 part of the cistern in = 4*(1/6) =2/3 minutes

Total time taken to fill the Cistern = time taken to fill (5/6 part + 1/6 part)
=> 4 minutes + 2/3 minutes = 4(2/3) minutes Or, 4 minutes 40 seconds

If a pipe can fill a tank in x hours, then : part filled in 1 hour = 1/x

Part filled A in 1 hr= (1/45)
Part filled B in 1 hr= (1/36)
Part filled by (A+B) together in 1 hr=(1/45)+(1/36)=1/20

So, The tank will be full in 20 hr.

Let the pipe can fill a tank in x hours, then part filled in 1 hour = 1/x

A pipe fills a tank in 6 h, then the part of tank filled in 1 h = 1/6
Hence, required part of the tank to be filled in 1h = 1/6 part

Let a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where x > y), when opening both the pipes, the net part emptied in 1 hour = (1/y)-(1/x)
=> (1/8 - 1/16) = 1/16

Time taken to empty the full tank = 16 min.
Hence, time taken to empty the half tank = 8 min

Let a pipe can fill a tank in x hours, and another pipe can empty the full tank in y hours (where y> x), then on opening both the pipes, the net part filled in 1 hour = (1/x)-(1/y)
=> (1/2 - 1/3) = 1/6

Hence, time taken to fill the cistern = 6 hours

Work done by first tap in 1 min is 1/10
Work done by second tap in 1 min is 1/30

In 5 minutes work done by both pipes is = 5*(1/10 + 1/30)
=5*(4/30)
=2/3

After 5 min first tap is closed,
Now, remaining work is = 1 - 2/3 = 1/3

So, remaining work done by second tap is = 30*(1/3) = 10
After closing first tap, the second tap will be filled the tank completely in 10 minutes.

Part filled by A in 1 hour=1/36
Part filled by B in 1 hour= 1/45
Part filled by (A+B)'s in 1 hour=1/36 +1/45
= 9/180
= 1/20
Hence, both the pipes together will fill the tank in 20 hours

Three pipes 1 hour work =1/10 +1/12 -1/20
=8/60
=2/15
The tank be filled in 15/2 hours= 7 hrs 30 min

Net part filled in 1 hour= 1/4 -1/9= 5/36
Therefore the cistern will be filled in 36/5 hours or 7.2 hours

Let the reservoir be filled by the 1st pipe in x hours
The second pipe will fill it in (x+10) hours
1/x + (1/(x+10))= 1/12
=> (2x+10)/((x)*(x+10))= 1/12
=> x2-14x-120=0
=> (x-20)(x+6)=0
=> x=20 (value can not be -ve )
So, the second pipe will take 30 hours to fill the reservoir

Work done by a waste pipe in 1 min
=(1/12+1/15) - 1/20 = 1/10

Waste pipe will empty the cistern in 10 minutes.

Time taken by one tap to fill the half of the tank =3 hours
Part filled by the four taps in 1 hour=4/6=2/3
Remaining part=1 -1/2=1/2

2/3 of the tank is filled by four taps in 1 hour
let the time taken to fill Remaining part is x

Therefore, 2/3:1/2 :: 1:x
or x=(1/2)*1*(3/2)=3/4 hours
ie 45 min

So, the total time taken = 3 hrs + 45 min = 3 hrs 45 min

Clearly, pipe B is faster than A and So, the tank will be emptied.

Part to be emptied=2/5
Part emptied by (A+B) in 1 min= 1/6 - 1/10=1/15
Therefore, 1/15:2/5::1:x
or x=((2/5)*1*15)=6 min

So, the tank be emptied in 6 min

Let the capacity of P be x liters
Then capacity of Q=x/3 liter
Capacity of the drum=60x liter
Required number of turns= 60x/(x+(x/3))= 60x*3/4x=45

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