Pipes and Cisterns Part 7

Problem 1: 12 taps are fitted in a tank, some are inlet water taps and rest are outlet taps each water tap can fill the tank in 6 hours and outlet can empty the tank in 12 hours. If all taps are open together then the tank is full in 4 hours. find the number of inlet water taps?

Solution:

Total number of taps are=12

Let the number of inlets taps=X

The number of outlet taps =12-X

A———->6                                      +2

L.C.M12

B———->12                                     -1

X*2*4 + (12-X)*(-1)*4=12

8X-48+4X = 12

12X = 60

X = 5

The inlet water taps are 5 and outlet taps are 7.

Problem 2: 8 taps are fitted in a tank, some are inlet water taps and rest are outlet taps each water tap can fill the tank in 12 hours and outlet can empty the tank in 36 hours. lf all taps are open together then the tank is full in 3 hours. Find the number of inlet water taps?

Solution:

Total number of taps are=8

Let the number of inlet taps=X

The number of outlet taps =8-X

A———->12                                      +3

L.C.M 36

B———->36                                     -1

X*3*43 + (8-X)*(-1)*3=36

9X-24+3X=36

12X = 60

X=5

The inlet water tapes are 5 and outlet taps are 3

 

Pipes and Cisterns Part 6

Problem 1: The inlet to the tank can fill it in 4 hours while the outlet can empty it in 5 hours both the pipes are opened at 9 AM but after some time the outlet closed and it is observed that the tank was closed at 5 PM. At what time was the outlet closed?

Solution:

A———>4                                  +5

L.C.M 20

B———->5                                  -4

A+B—>5+(-4)=1

Difference between 9 AM to 5 PM is 8 hours

1*X+5(8-X)=20

X+40-5X=20

-4X=-20

X=5

Count 5 hours from 9 AM

At 2 PM the outlet was closed

Problem 2: Two pipes can fill a tank with water in 15 hours and 12 hours respectively and a third pipe can empty it in 4 hours. If the pipe is opened in order at 8,9 and 11 AM respectively, the tank will be emptied at?

Solution:

A———–>15                                  +4

B————>12     L.C.M=60         +5

C————->4                                  -15

Difference between 8 AM and 11 AM IS 3hrs so 4*3=12

Difference between 9 AM and 11 AM IS 3hrs so 5*2=10

now third also opened then A+B+C=4+5-15=-6

12+10=22

\frac{22}{-6}=3\tfrac{4}{6}\times 60 =3 hours 40 min

Count 3 hours 40 min from 11 AM.

So the tank will be emptied in2 hours 40 minutes

Pipes and Cisterns Part 5

Problem 1: Capacity of tap B is 80% more then that of tap A. If both the taps are opened simultaneously it takes 45 hours to fill the tank. How long Will B take to fill the tank alone?

Solution:

A               B

100         180

5        :          9

\frac{14\times 45}{9}=70 hours

Problem 2: Three taps A, B, and c may fill a tank in 26h, 52h,78h respectively. Tap A and B are opened initially but after 10 hours C is also opened, but A and B started working at 80% of their initial efficiency. Find the total time in which tank will be filled completely?

Solution:

A————>26                                             6

B————->52           L.C.M 156              3

C————->78                                              2

For 10 hours tap A and B are opened

9*10=90

156-90=66

A and B started working at 80% of their initial efficiency

9\times \frac{80}{100}=7.2

A+B+C=7.2+2=9.2

\frac{6.6}{9.2}+10=17 hours

Pipes and Cisterns Part 4

Problem 1: A leak in the bottom of a tank can empty it in 12 hours. A tap which can be added 20 lit per minute is turn on. Both the taps are opened now, then the tank is emptied in 20 hours. Find the capacity of the tank?

Solution:

A———–>12                                            -5

L.C.M 60

A+B———>20                                         -3

A+B——->-3

B=-3+5

B=2

\frac{60}{2}=30 hours

B takes 30 hours to fill the tank

1min——>20lit

1hour——->20*60*30

Capacity=30*60*20=36000

In the above sum, 20 is liters 60 is hours are converted into minutes 30 is hours.

Problem 2: A tank has a leak which would empty in 8 hours. A tap is turned on which admits 3 lit per min into the tank and it is now emptied in 12 hours. How many liters can the tank hold?

Solution:

A  = -8                                               -3

LCM  24

A+B=-12                                          -2

A+B=-2

A-3=-2

A=1

1hr——–> 1lit

\frac{24}{1}=24

24*60*3=72*60=4320

Problem 3: A tank has a leak which would empty the completely filled the tank in 10 hours. if the tank is full of water and the tap is opened which admits 4 lit of water per min in the tank, the leak 15 hours to empty the tank. How many liters of water does the tank hold?

Solution:

A———–>10                                          -3

L.C.M 30

A+B——–>15                                         -2

A+B=-2

B=-2+3

B=1

\frac{30}{1}=30

Capacity=30*60*4=7200

Problem 4: Two pipes M and N can fill a tank in 8\frac{1}{3}  minutes and 12\frac{1}{2} min respectively and an exhaust pipe can drain 162-liter water in 1 min. If the tank is full all the three pipes are opened then the tank will empty in 4 minutes then the capacity of the tank will be?

Solution:

M—————–>\frac{25}{3}                                                   +12(\frac{100\times 3}{25}=12)

N—————–> \frac{25}{2}                   L.C.M  100            +8 (\frac{100\times 2}{25}=8)

M+N+O——–>4                                                    -25

M+N+O=-25

O=-25-20

O=-45

45————>162 lit

100————>?

\frac{162\times 100}{45}=360 liters

Pipes and Cisterns Part 3

Problem 1: Two pipes A and 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 minutes?

Solution:

A———->24                                                  +4

L.C.M  96

B———–>32                                                 +3

First, both taps are open for some time so we write 7x after B tap is closed so we mentioned 4 in the equation.

7*X + 4(18-X) = 96

7X+72-4X = 96

3X = 24

X = 8minutes

Shortcut: We should think in a logical way

A———->24                                                  +4

L.C.M  96

B———–>32                                                 +3

whole time tap A is worked so 18*4=72lit are filled by tank A

96-72=24 lit are remaining should be filled by tank B

\frac{24}{3}=8 minutes

After 8 minutes tap B is closed

Problem 2: Two taps A and B can fill a tank in 48 min and 36 min. if both taps are opened together after how much time tap A is closed so that the whole tank fill in 25 min 30 sec?

Solution:

A———->48                                                  +3

L.C.M  144

B———–>36                                                 +4

B will work for the whole time so 4\times \frac{51}{2}=102 lit are filled by tap B

144-102=42 lit are filled by tap A

\frac{42}{3}=14 min

After 14 min tap, A is closed.

Problem 3: A cistern can be filled by two supply pipes in 20 min and 30 min respectively. But a waste pipe can empty it in an hour. Both the supply pipes were opened together to fill the empty cistern, but by mistake, the waste pipe was also open. How much later was the waste pipe closed, if the cistern is filled in 14 minutes?

Solution:

A————->20                                          +3

B————->30             L.C.M   60          +2

C————->60                                            -1

Waste pipe is closed

3+2=5

Both A and B pipes are worked whole time so 5*14=70

60-70=-10

\frac{10}{1}=10 min

After 10 minutes the waste pipe was closed.