Pipes and Cisterns Questions and Answers

Solution:

Total capacity of cistern = 60*15 = 900 liters

Total vessels needed to fill the cistern if the total volume of each vessel is reduces to 20 liters = 60*15 / 20 = 900/20 = 45 vessels

Correct answer B

Solution:

Container filled by pipe M in 1 hour = 1 / 9

Container filled by pipe N in 1 hour = 1 / 3

Container filled by both if open alternatively = 1/9 + 1/3 = 4/9 (in 2 hours)

Container filled by both in another 2 hours = 4/9 + 4/9 = 8/9

Remaining 1/9^th part will be filled by pipe M in another one hour.

Total time = 2+2+1 = 5 hours

Correct answer B

Solution:

Let us assume the filling capacity of the pump be x liter/minute

The emptying capacity will be 20+x liter/minute

Time needed to fill the storage chamber = 3200 / x

Time needed to empty the storage chamber = 3200 / x+20

Given, the pump requires 8 minutes less in emptying the storage chamber,

= 3200/x – 3200/x+20 = 8

= 400/x – 400/x+20 = 1

= 400(x+20) – 400x = x(x+20)

= 400x + 8000 – 400x = x² + 20x

= 8000 = x² + 20x

= x² + 20x – 8000 = 0

= (x +  100) (x - 80) = 0

x = -100, x = 80

Since liters cannot be negative, therefore the filling capacity is 80 liters/ minute.

Correct answer B

Solution:

Let us assume the time taken by II pipe be x

Then the time in which Pipe I will fill the container be (x-5)

Given, 1/x + 1/x-5 = 1/6

x =2 or x = 15

x cannot be 2 as together pipe A and B takes 6 minutes, so alone pipe B cannot fill the container in 2 minutes.

Therefore, x will be 15.

Hence, Pipe B will take 15 minutes to fill the container and pipe A will take 10 minutes to fill the container.

Correct answer B

Solution:

The complete tank is filled in 8 hours, so in 1 hour = 1/8^th tank will be filled

Due to outlet pipe in 1 hour = 1/12^th part will be filled

Part of the tank emptied in 1 hour = 1/8 – 1/12 = 1/24

Therefore the tank will be emptied in 24 hours.

Correct answer B

Solution:

The cistern is filled in 20 hours

In 1 hour = 1/20^th portion of the cistern is filled

Due to crack the cistern is filled in 25 hours

In 1 hour = 1/25^th portion of the cistern is filled

Part of the tank emptied in a hour = 1/20 – 1/25 = 1/100

Therefore, the cistern will be completely unfilled in 100 hours.

Correct answer A

Solution:

25 km = 25*1000 = 25000 m

In 1 hour the level of the water will be = 0.5*0.4*25000 = 5000m³

Let us assume that, in x hours the water level will reach to 18m.

Therefore, 5000x = 250×200×18

5000x = 180 hours.

Correct answer C

Solution:

Pipe R fills the tanker in 40 mins

Pipe R will fill the tanker in 1 minute = 1/40^th of the tank

Pipe S fills the tank in 5/40 = 8 minutes

Pipe S will fill the tanker in 1 minute = 1/8^th of the tank

Together they can fill the tanker in a single minute = (1/8) – (1/40) = 1/10^th part

Therefore, the tanker will be filled in 10 minutes.

Correct answer D

Solution:

Let us assume the leakage will take x hours to empty the tank

Then, 1/x = 1/21 – 1/14

1/x = 1/42

Therefore x = 42 hours.

Correct answer D

Solution:

Let us assume the time taken by slower pipe as x

Then, the time taken by faster pipe = x/4 minutes.

Given, 1/x + 4/x = 1/42

5/x = 1/42

x = 210 minutes

Correct answer D

 

Explanation:   Part filled by (U+V+W) in 3 minutes = (1/30 + 1/20 + 1/ 10)* 3 = 3* 11/60 = 11/ 20

Part filled by W in 3 minutes =     3/ 10

Required ratio =        (3  / 10) x     (20 /11  )     =     6 / 11

 

Explanation:

Work done by the leak in 1 hour =   ½ -     3/7        =     1 /14    .

Leak will empty the tank in 14 hrs

Explanation:

Suppose Hose D alone takes x hours to fill the tank.

Then, Hoses E and F will take  x /2  and  x/4     hours respectively to fill the tank.

1 /x    +  2 /x  +     4/x   =     1/  5

7 /x    =     1/5

x = 35 hrs.

Let the time taken by Hose I to fill the tank be x minutes

Time is taken by Hose J to fill the tank = x+5 minutes

So, 1/x + 1/(x+5) = 1/6

⇒ x = 10

Thus, time taken by J alone to fill the tank is 10+5, i.e., 15 minutes

144 minutes

Solution:

Let the time taken by Hose B be x minutes

So, the time taken by Hose A = x/3 minutes

Thus, 1/3 + 3/x = 1/36

⇒ 4/x = 1/36

⇒ x = 4×36

⇒ x = 144 minutes

 

120 gallons

Solution:

Work done by the outlet Hose in 1 minute = {1/15 – (1/24)+(1/20)} = 1/15 – 11/120 = -(1/40)

Here, the negative sign indicates the negative work done, that is the loss of water from the outlet

The capacity of 1/40 part = 3 gallons

So, Capacity of whole tank = 40×3 = 120 gallons

Let capacity of tank be 1 unit

⇒ Part of tank emptied by Hose C in an hour = 1/6

As, B and D do the opposite work, they cancel each other out. Also, Hose D is twice as efficient as Hose C

⇒ Part of tank emptied by Hose D in an hour = 2 × (1/6) = 1/3

⇒ Part of tank filled by Hose B in an hour = 1/3

Now, Hose A is twice as efficient as Hose B

⇒ Part of tank filled by Hose A in an hour = 2 × (1/3) = 2/3

⇒ Part of tank filled in an hour = (2/3) – (1/6) = 1/2

⇒ Time taken by all the Hoses to fill the tank = 1/(1/2) = 2 hour