All questions of Pipes & Cistern for Civil Engineering (CE) Exam

Lets assume Total capacity 1000 Litres. 1 Pipe will fill 33.333 litres in 1 minute while 2nd pipe will fill 5 times faster i.e. 166.6666 litres . so total tank filled in 1minute is equal to 200 litres. so to fill 1000 litre tank it will take 1000/200= 5 Minutes. so to overflow the tank it requires more than 5 minutes.

To solve this problem, we can calculate the rates at which each pipe fills the tank and then determine the combined rate at which both pipes fill the tank.

Let's start by finding the rate at which each pipe fills the tank.

Pipe 1: Fills 1/4 of the tank in 4 minutes
Pipe 2: Fills 1/5 of the tank in 4 minutes

To find the rates, we can divide the fraction filled by the time taken for each pipe.

Pipe 1 rate = (1/4) / 4 = 1/16 of the tank per minute
Pipe 2 rate = (1/5) / 4 = 1/20 of the tank per minute

Next, we need to determine the combined rate at which both pipes fill the tank. We can add the rates of the two pipes together.

Combined rate = Pipe 1 rate + Pipe 2 rate
Combined rate = 1/16 + 1/20
Combined rate = (5/80) + (4/80)
Combined rate = 9/80 of the tank per minute

Now we can find the time taken by both pipes to fill half the tank. Since the combined rate is given in terms of minutes per tank, we can invert the rate to find the time taken.

Time taken = 1 / (combined rate)
Time taken = 1 / (9/80)
Time taken = 80/9 minutes

Therefore, the time taken by both pipes together to fill half the tank is 80/9 minutes, which is equivalent to option A.

To solve this problem, we can consider the rates at which the two pipes fill the cistern. Let the rate at which pipe P fills the cistern be x liters per hour, and the rate at which pipe Q fills the cistern be y liters per hour.

Rate of pipe P = 1 cistern / 10 hours = 1/10 cistern per hour = x liters per hour
Rate of pipe Q = 1 cistern / 20 hours = 1/20 cistern per hour = y liters per hour

Since the rates are given in terms of cisterns per hour, we can equate the rates to find the values of x and y:

x = 1/10 cistern per hour
y = 1/20 cistern per hour

Simultaneously filling the cistern:
When both pipes P and Q are opened simultaneously, their rates of filling are additive. Therefore, the combined rate of filling the cistern is:

x + y = 1/10 + 1/20 = 3/20 cistern per hour

After some time, pipe Q is closed. Let's assume that pipe Q was closed after t hours. So, for the first t hours, both pipes P and Q were open, and for the remaining 8 hours, only pipe P was open.

Total time taken to fill the cistern = t + 8 hours
Rate of pipe P = x liters per hour (as pipe Q is closed)
Rate of pipe Q = 0 liters per hour (as pipe Q is closed)

Using the rates, we can set up the following equation based on the principle of work:

(t + 8)(x) = 1 cistern

Simplifying the equation, we get:

(t + 8)(1/10) = 1
(t + 8)/10 = 1
t + 8 = 10
t = 10 - 8
t = 2

Therefore, pipe Q was closed after 2 hours (option C).

Problem:
Two pipes P and Q can fill a tank in 10 min and 12 min respectively and a waste pipe can carry off 12 litres of water per minute. If all the pipes are opened when the tank is full and it takes one hour to empty the tank. Find the capacity of the tank.

Solution:
Let the capacity of the tank be 'x' litres.
Given, Pipe P can fill the tank in 10 minutes. So, the amount of water it can fill in 1 minute is x/10 litres.
Similarly, Pipe Q can fill the tank in 12 minutes. So, the amount of water it can fill in 1 minute is x/12 litres.
The waste pipe can carry off 12 litres of water per minute. So, the net amount of water filled in 1 minute when all pipes are opened is (x/10 + x/12 - 12) litres.
It takes 1 hour to empty the tank. So, the amount of water emptied in 1 minute is x/60 litres.
Therefore, the net amount of water filled in 1 minute is equal to the amount of water emptied in 1 minute. Hence, we can write the equation as follows:
x/10 + x/12 - 12 = x/60
Solving this equation, we get x = 60 litres.

Answer:
The capacity of the tank is 60 litres. Therefore, the correct answer is option (c).

20

Given:
Pipe A fills the tank alone in 48 hours.
Pipe B fills the tank alone in 24 hours.
Pipe C empties the tank alone in 36 hours.

In the first hour, only Pipe A is opened, so it fills 1/48th of the tank.
In the second hour, only Pipe B is opened, so it fills 1/24th of the tank.
In the third hour, only Pipe C is opened, so it empties 1/36th of the tank.

We can observe that in the first three hours, the net amount of water filled in the tank is:
1/48 - 1/24 - 1/36 = (1/48) - (2/48) - (3/48) = -4/48 = -1/12

Since the tank is initially empty, the net amount of water in the tank after the first three hours is negative, which means the tank is not filled yet.

Let's assume that after x hours, the tank is filled. We can write the equation as:

(x/48) - (x/24) - (x/36) = 1

Simplifying the equation, we get:

(3x - 6x - 4x) / (48 * 24 * 36) = 1

-7x / (48 * 24 * 36) = 1

Solving for x, we get:

x = -48 * 24 * 36 / 7

Since x represents the number of hours, it cannot be negative. Therefore, we can ignore the negative sign and calculate the value of x as:

x = 48 * 24 * 36 / 7 = 82971.4286 hours

Since x represents the number of hours, it cannot be in decimal form. Therefore, we round it up to the nearest whole number, which is 82972 hours.

To find the number of hours Pipe B is opened, we subtract the first three hours from the total time:

82972 - 3 = 82969 hours

Therefore, Pipe B is opened for 82969 hours, which is equivalent to 28 hours and 10 minutes.

Hence, the correct answer is option B) 28 hours 10 minutes.

Problem:

Two pipes P and Q can fill a cistern in 12 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours will the tank be full?

Solution:

Let us assume the capacity of the tank as 'C' and the rate of filling of pipe P and Q as 'x' and 'y' respectively.

Given,

Pipe P can fill the tank in 12 hours.

Therefore, the rate of filling of pipe P = C/12

Pipe Q can fill the tank in 4 hours.

Therefore, the rate of filling of pipe Q = C/4

Let the time taken by both the pipes working alternatively to fill the tank be 't' hours.

In the first hour, pipe P will fill the tank with a rate of C/12.

In the second hour, pipe Q will fill the tank with a rate of C/4.

In the third hour, pipe P will fill the tank with a rate of C/12.

In the fourth hour, pipe Q will fill the tank with a rate of C/4.

This process will continue until the tank gets filled.

As per the given condition, pipe P is opened first.

Therefore, the tank will be filled by pipe P in the first hour.

So, the total time taken to fill the tank will be:

C/12 + C/4 + C/12 + C/4 + C/12 + C/4 +... (t terms)

= [t/2][(2C/12) + (2C/4)]

= [t/2][(C/6) + (C/2)]

= t(C/3)

As per the question, the tank should be filled in 't' hours.

Therefore,

t(C/3) = C

t = 3 hours

Hence, the tank will be full in 6 hours.

Thus, option (D) is the correct answer.