INLET AND OUTLET WORD PROBLEMS WITH SOLUTIONS

Problem 1 :

A pipe can fill up an empty tank in 12 minutes. Another pipe flows out 14 liter of water per minute. If the two pipes are opened together and the empty tank is filled up in 96 minutes, how much water does the tank contain ?

Solution :

Let x liter of water filled in the tank per minute. So, y be the quantity of water in the tank

According to the given information, the empty tank is filled in 12 minutes by pumping x liter of water per minute.

y  =  12x   ---(1)

Now inlet and outlet both are opened and it is filled in 96 minutes. 

y  =  96x - 96(14)  ------(2)

(1)  =  (2)

12x  =  96x - 1344

84x  =  1344

x  =  16

By applying the value of x in (1), we get

y  =  12 (16)

y  =  192

So, the tank will contain 192 liter of water.

Problem 2 :

Two pipes are connected with a tank. The empty tank is filled up in t₁ minutes by the first pipe and it becomes empty in t₂ minutes by the second pipe. If the two pipes are opened together, how much time will the tank be filled up ? (Here t₂ > t₁)

Solution :

Part of the tank filled by the first pipe in 1 hour  =  1/t₁

Part of the tank emptying by the second pipe in 1 hour  =  1/t₂

Part of the tank filled in 1 hour when both pipes are opened :

=  1/t₁ - 1/t₂

=  (t₂-t₁)/t₁t₂

Time taken by both pipes to fill the tank 

=  t₁t₂ / (t₂-t₁)

Problem 3 :

A tank is filled up in 12 minutes by a pipe. Another pipe flows out 15 liter of water in 1 minute. When the tank is empty, the two pipes are opened together and the tank is filled up in 48 minutes. How much water does the tank contain ?

Solution :

Let x liter of water filled in the tank in minute. So, y be the quantity of water in the tank.

According to the question

y  =  12x  ----(1)

When both pipes are opened and the tank is empty, if it filled in 48 minutes.

y = 48x - 15(48) ----(2)

(1)  =  (2)

12x  =  48x - 720

12x-48x  =  -720

36x  =  720

x  =  720/36

x  =  20 

20 liter of water is pumped in the tank per minute.

By applying x  =  20 in (1), we get

y  =  12(20)

y  =  240

So, 240 liter of water in the tank.

Problem 4 :

Two pipes, Pipe X and Pipe Y can empty a cistern in 24 hours and 40 hours respectively while working alone. If both pipes are opened together, then the time taken to empty the cistern is given by

Solution :

Quantity of water discharging in one hour by

Pipe X = 1/24

Quantity of water discharging in one hour b

Pipe Y = 1/40

Now the pipes are opened together,

quantity of water discharging from two pipes

= 1/24 + 1/40

LCM (24 and 40) = 2 x 2 x 2 x 3 x 5

= 120

= (5 + 3)/120

= 8/120

Quantity of water discharged by two pipes X and Y in one hour

= 8/120

Quantity of water discharged by two pipes X and Y totally

= 120/8

= 15 hours

So, to empty the tank the pipes will take 15 hours.

Problem 5 :

Two pipes, pipe X can fill a tank in 20 hours and pipe Y can empty a tank in 30 hours while working alone. If both pipes are opened together, then the time taken to fill/empty the tank is?

Solution :

Time taken by pipe X to fill the tank = 20 hours

Work done by pipe X in 1 hour = 1/20 hours

Time taken by pipe Y to empty the tank = 30 hours

Work done by pipe Y in 1 hour = -1/30

Since it is emptying the tank, we use the negative sign.

Work done by these two pipes in 1 hour, when they are opened together

= 1/20 - 1/30

= (3 - 2) / 60

= 1/60

Total time taken = 60 hours.

Since we receive the answer with positive, we understand that the tank can be filled in 60 hours.

Problem 6 :

If pipe X is 2 times fast as pipe Y and can fill a tank 15 min less than pipe Y. Then find the time in which both the pipes can fill the tank together? 

Solution :

Let t be the time taken by pipe Y, then time taken by pipe X = 2t

Work done by pipe X in 1 hour = 1/2t

Work done by pipe Y = 1/t

1/t - 1/2t = 1/15

(2 - 1)/2t = 1/15

1/2t = 1/15

2t = 15

t = 15/2

When pipes are opened together = 1/t + 1/2t

= (2 + 1)/2t

= 3/2t

= 3/2(15/2)

= 3/15

= 1/5

So, 5 hours is the answer.

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