## Pipes and Cisterns Part 1

in this chapter one this we should remember i.e. admits means we should use ‘+’ sign and empty means we should use ‘-‘ sign?

Problem 1: Two pipes A and B can fill a tank in 20 min and 30 min respectively. If both pipes are opened together, the time taken to fill the tank is?

Solution:

A———> 20                                  +3

L.C.M 60

B———–>30                                 +2

3+2=5

$\frac{60}{5}=12$

Problem 2: A pipe can fill a tank with water in 3 hours. Due to leakage in the bottom, it takes  $3\frac{1}{2}$  hours to fill it. in what time the leak will empty the fully filled tank?

Solution:

A———–> 3hrs                                           $\frac{7}{2}$

L.CM  $\frac{21}{2}$

A+B———->$3\frac{1}{2}=\frac{7}{2}$                                    3

A+B=3

B=3-$\frac{7}{2}=-\frac{1}{2}$

$\frac{\frac{21}{2}}{\frac{1}{2}}=21$ hrs

Shortcut: $\frac{a\times&space;b}{a-b}$ = $\frac{3\times&space;\frac{1}{2}}{\frac{1}{2}}=21$

Problem 3: A cistern is normally filled in 9 hours but due to the leak in the bottom, it takes one hour more to be filled. if the cistern is full, in what time will the leak empty it?

Solution:

A———>9                                                10

L.C.M 90

A+B—–>10                                                9

A+B=9

B=9-10

B=-1

$\frac{90}{1}=90$

Shortcut: $\frac{a\times&space;b}{a-b}$=$\frac{9\times&space;10}{1}=90$

## Pipes and Cisterns Part 2

Problem 1: 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?

Solution:

A——->12                                              +5

B——->15           L.C.M 60                  +4

C——–>20                                             +3

A    B|    A        C|      A    B|      A        C|

5     4|    5         3|       5     4|    5         3|

5+4=9         5+3=8

9+8=17

2 hrs ———>   17

3*2————->17*3

6h—————>51

In 6 hours 51 parts completed remaining 60-51=9 parts will be complete by A and B in 1 hour

so 6+1=7 hrs

In 7 hrs the tank will be full.

Problem 2: Two taps can fill a tub in 5 minutes and 7 minutes respectively. A pipe can empty it in 3 min. If all the three taps are kept open simultaneously, when will the tub be full?

Solution:

A———->5                                         +21

B———–>7          L.C.M  105         +15

C———–>3                                         -35

21+15-36=1

1———->A+B+C

$\frac{105}{1}=105$ minutes

Problem 3: A pipe can fill a tank in 12 minutes and another can fill it in 15 minutes, but a third pipe can empty it in 6 minutes. The first two pipes are kept open for 5 minutes in the beginning and then the third pipe is also opened. Number minutes taken to empty the cistern is?

Solution:

A———–>12                             +5

B———–>15       L.C.M 60     +4

C————>6                               -10

For 5 minutes first two pipes are opened

5+4=9

9*5=45 lit fill the tank

If we open the third pipe it will empty the tank

5+4+(-10)=-1

$\frac{45}{1}=45min$

## 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 $4\times&space;\frac{51}{2}=102$ 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.

## 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($\frac{100\times&space;3}{25}=12$)

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

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

M+N+O=-25

O=-25-20

O=-45

45————>162 lit

100————>?

$\frac{162\times&space;100}{45}=360$ liters

## 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

$\frac{14\times&space;45}{9}=70&space;hours$

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

$9\times&space;\frac{80}{100}=7.2$

A+B+C=7.2+2=9.2

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